Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} for $n\ge 2$. Otherwise by going back to the original integral representation we have: \begin{equation} {\bf H}^{(1)}_1(-1) = -\frac{\pi^2}{12} + \frac{1}{2} \log(2)^2 \end{equation} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} for $n\ge 2$. Otherwise by going back to the original integral representation we have: \begin{equation} {\bf H}^{(1)}_1(-1) = -\frac{\pi^2}{12} + \frac{1}{2} \log(2)^2 \end{equation} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Update: We start from the middle equation in the rhs above. We have: \begin{eqnarray} {\bf H}_n^{(q)}(t) &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \sum\limits_{l=1}^n \sum\limits_{k=1}^\infty \sum\limits_{m=1}^\infty \frac{t^{m+k}}{m^{q-1} k^l (k+m)^{n-l+1}}\\ &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \\ &&\sum\limits_{l=1}^n\sum\limits_{l_1=1}^l \binom{n-l_1}{n-l} (-1)^{l-l_1} Li_{n+q-l_1}(t) Li_{l_1}(t) - \sum\limits_{l=1}^n\sum\limits_{l_1=1}^{n-l+1}\binom{n-l_1}{l-1} (-1)^l{\bf H}^{(n+q-l_1)}_{l_1}(t) \end{eqnarray} In the top line we expanded both poly-logarithms in the integrand in series and we integrated term by term. Then in the bottom line we decomposed the denominator into partial fractions in the variable $k$ by using Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. and subsequently we re-summed the result firstly over $k$ and then over $n$ by using the series expansions of poly-logarithms.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Update: We start from the middle equation in the rhs above. We have: \begin{eqnarray} {\bf H}_n^{(q)}(t) &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \sum\limits_{l=1}^n \sum\limits_{k=1}^\infty \sum\limits_{m=1}^\infty \frac{t^{m+k}}{m^{q-1} k^l (k+m)^{n-l+1}}\\ &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \\ &&\sum\limits_{l=1}^n\sum\limits_{l_1=1}^l \binom{n-l_1}{n-l} (-1)^{l-l_1} Li_{n+q-l_1}(t) Li_{l_1}(t) - \sum\limits_{l=1}^n\sum\limits_{l_1=1}^{n-l+1}\binom{n-l_1}{l-1} (-1)^l{\bf H}^{(n+q-l_1)}_{l_1}(t) \end{eqnarray} In the top line we expanded both poly-logarithms in the integrand in series and we integrated term by term. Then in the bottom line we decomposed the denominator into partial fractions in the variable $k$ by using Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. and subsequently we re-summed the result firstly over $k$ and then over $n$ by using the series expansions of poly-logarithms.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations. Work on this topic is underway
Update: We start from the middle equation in the rhs above. We have: \begin{eqnarray} {\bf H}_n^{(q)}(t) &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \sum\limits_{l=1}^n \sum\limits_{k=1}^\infty \sum\limits_{m=1}^\infty \frac{t^{m+k}}{m^{q-1} k^l (k+m)^{n-l+1}}\\ &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \\ &&\sum\limits_{l=1}^n\sum\limits_{l_1=1}^l \binom{n-l_1}{n-l} (-1)^{l-l_1} Li_{n+q-l_1}(t) Li_{l_1}(t) - \sum\limits_{l=1}^n\sum\limits_{l_1=1}^{n-l+1}\binom{n-l_1}{l-1} (-1)^l{\bf H}^{(n+q-l_1)}_{l_1}(t) \end{eqnarray} In the top line we expanded both poly-logarithms in the integrand in series and we integrated term by term. Then in the bottom line we decomposed the denominator into partial fractions in the variable $k$ by using Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. and subsequently we re-summed the result firstly over $k$ and then over $n$ by using the series expansions of poly-logarithms.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations. Work on this topic is underway.
Please allow me the privilege of using a different notation so that I can make use of the results that I have derived myself without falling into a danger of making mistakes.We have: \begin{eqnarray} {\bf H}_n^{(1)}(t):= \sum\limits_{m=1}^\infty H_m^{(1)} \frac{t^m}{m^n} &=& PolyLog[n-1,2,t] + Li_{n+1}(t)\\ &=&\frac{(-1)^n}{2!(n-2)!} \int\limits_0^1 \frac{[\log(\eta)]^{n-2}}{\eta} [\log(1-t \eta)]^2d \eta + Li_{n+1}(t) \end{eqnarray} Here PolyLog[,,] is the Nielsen generalised poly-logarithm. In the first line above we used the results from the answer to Closed form expressions for harmonic sums and in the second line we wrote down explicitely the definition of the Nielsen poly-logarithm. Now are are going to set $t=-1$ and then we use the identity $\log(1+\eta) = \log(1-\eta^2) - \log(1-\eta)$. Therefore we have: \begin{eqnarray} &&\left({\bf H}_n^{(1)}(t)- Li_{n+1}(-1)\right) \cdot \frac{2!(n-2)!}{(-1)^n} = \\ && \int\limits_0^1 \frac{[\log(\eta)]^{n-2}\left(\log(1-\eta^2)^2 - 2 \log(1+\eta)\log(1-\eta) - \log(1-\eta)^2\right)}{\eta} d \eta=\\ && \left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(\eta)^{n-2} \frac{\log(1-\eta)^2}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}}) \int\limits_0^1 \log(1-\eta)^{n-1} \frac{\log(\eta)^{1}}{\eta} d\eta - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)=\\ &&\left((\frac{1-2^{n-1}}{2^{n-1}})(-1)^n (n-2)!\left(n \zeta(n+1) - \sum\limits_{j=1}^{n-2} \zeta(1+j) \zeta(n-j)\right) - 2 \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \right)= \end{eqnarray} In the top line we just expanded the square of the difference of the two logs. In the next line changed variables accordingly and regrouped the whole expression into two different types of integrals. In the subsequent line we integrated by parts in the first integrals and left the second integral unchanged and finally in the bottom line we used Compute an integral containing a product of powers of logarithms. . Now, the only thing that remains is to compute the remaining integral on the rhs. We have: \begin{eqnarray} {\mathcal I}^{(n-2)} &:=& \int\limits_0^1 \log(\eta)^{n-2} \log(1+\eta) \frac{\log(1-\eta)}{\eta}d\eta \\ &=& \sum\limits_{r=1}^\infty \frac{(-1)^{r-1}}{r} \cdot \frac{\partial^{n-2}}{\partial \theta^{n-2}} \left.\left( \frac{\Psi^{(0)}(1) - \Psi^{(0)}(1+r+\theta)}{\theta+r}\right)\right|_{\theta=0}\\ &=& \sum\limits_{r=1}^\infty\sum\limits_{l=0}^{n-2} \frac{(-1)^{r-1}}{r} \cdot \binom{n-2}{l} \frac{(n-2-l)! (-1)^{n-2-l}}{r^{n-2-l+1}} (\Psi^{(0)}(1) \delta_{l,0} - \Psi^{(l)}(1+r)) \\ &=&(-1)^n (n-2)! \left(- \sum\limits_{l=1}^{n-2} Li_{n-l}(-1) \zeta(l+1)+ \sum\limits_{l=0}^{n-2} \sum\limits_{r=1}^\infty \frac{(-1)^r}{r^{n-l}} H_r^{(l+1)}\right)\\ &=&(-1)^n (n-2)! \left(+ \sum\limits_{l=1}^{n-2} \frac{2^{n-l-1}-1}{2^{n-l-1}}\zeta(n-l)\zeta(l+1)+ \sum\limits_{l=0}^{n-2} {\bf H}_{n-l}^{(l+1)}(-1) \right) \end{eqnarray} In the top line we expanded the term $\log(1+\eta)$ in a series and then integrated term by term by using Compute an integral containing a product of powers of logarithms. .In the following line we computed the partial derivative using the chain rule and in the last line we simplified the result. Bringing everything together we get the following: \begin{eqnarray} {\bf H}_n^{(1)}(-1) = \left(\frac{n+1}{2^n} - \frac{n}{2} - 1\right) \zeta(n+1) - \sum\limits_{j=1}^{n-2} \left(\frac{1}{2^n} + \frac{1}{2} - \frac{1}{2^{n-j-1}}\right) \zeta(1+j) \zeta(n-j) - \sum\limits_{l=0}^{n-2} {\bf H}^{(l+1)}_{n-l}(-1) \end{eqnarray} It is clearly seen that calculating our sum in question requires knowledge of sums that involve generalized harmonic numbers. It is quite likely that there exists a whole hierarchy of recurrence relations that interwine the generating functions ${\bf H}_n^{(q)}(t)$. In order to get some insight into this topic we at least write down a following identity for another sum in the hierarchy. We have: \begin{eqnarray} &&{\bf H}_n^{(q)}(t) := \sum\limits_{m=1}^\infty H_m^{(q)} \cdot \frac{t^m}{m^n} =\\ &&Li_{n+q}(t) + Li_n(t) Li_q(t) - \sum\limits_{l=1}^n \frac{1}{(n-l)!}\int\limits_0^t \frac{[\log(t/\eta)]^{n-l}}{\eta} \cdot Li_{q-1}(\eta) Li_l(\eta) d \eta =\\ &&Li_{n+q}(t) + Li_{n}(t) Li_{q}(t) - \sum\limits_{l=1}^n Li_{n-l+q}(t) Li_{l}(t) + \\ &&\sum\limits_{l=0}^{n-1} \sum\limits_{l_1=q}^{n-l-1+q} \frac{1}{(n-l-l_1+q-1)!} \int\limits_0^t \frac{[\log(t/\eta)]^{n-l-l_1+q-1}}{\eta} Li_{l_1}(\eta) Li_l (\eta) d \eta \end{eqnarray} Here $n\ge 2$ and $q \ge 1$. The above expression follows from the general integral representation given in the answer to Closed form expressions for harmonic sums and from integration by parts once (middle) and twice (bottom). We believe that the last integral on the rhs above will be easy to calculate since because it satisfies certain recurrence relations.
Update: We start from the middle equation in the rhs above. We have: \begin{eqnarray} {\bf H}_n^{(q)}(t) &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \sum\limits_{l=1}^n \sum\limits_{k=1}^\infty \sum\limits_{m=1}^\infty \frac{t^{m+k}}{m^{q-1} k^l (k+m)^{n-l+1}}\\ &=& Li_{n+q}(t) + Li_n(t) Li_q(t)- \\ &&\sum\limits_{l=1}^n\sum\limits_{l_1=1}^l \binom{n-l_1}{n-l} (-1)^{l-l_1} Li_{n+q-l_1}(t) Li_{l_1}(t) - \sum\limits_{l=1}^n\sum\limits_{l_1=1}^{n-l+1}\binom{n-l_1}{l-1} (-1)^l{\bf H}^{(n+q-l_1)}_{l_1}(t) \end{eqnarray} In the top line we expanded both poly-logarithms in the integrand in series and we integrated term by term. Then in the bottom line we decomposed the denominator into partial fractions in the variable $k$ by using Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. and subsequently we re-summed the result firstly over $k$ and then over $n$ by using the series expansions of poly-logarithms.