Linked Questions

I found the following formula $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$ and it is cited that Euler proved the ...

I saw the two identities $$ -\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2) $$ and $$ -\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2) $$ here: twist on classic log of ...

I proved the following result $$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$ After ...

Question Is there a closed form of this harmonic approximation sum $$s=\sum _{k=1}^{\infty } \left(H_k^{(2)}-\zeta (2)\right){}^2\tag{1}$$ The notation is standard. Motivation This question ...

Prove that $$S=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$ where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the n$th$ generalized harmonic ...

An interesting investigation started here and it showed that $$ \sum_{k\geq 1}\left(\zeta(m)-H_{k}^{(m)}\right)^2 $$ has a closed form in terms of values of the Riemann $\zeta$ function for any ...

Are there any general formula for the following series $$\tag{1}\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$$ Where we define $$H^{(p)}_n= \sum_{k=1}^n \frac{1}{k^p}\,\,\,\,\,H^{(1)}_n\equiv H_n =\...

Here is an Euler sum I ran into. $$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(4)} \mathcal{H}_n^2}{n^6}$$ where $\mathcal{H}_n^{(s)}$ is the generalised harmonic number of order $s$. I ...

Could someone help find identities for these two? I started with $ζ^\star(2,1,2)$ = $\sum_{k=1}^{\infty}(\frac{1}{k^2})\sum_{l=1}^{k}(\frac{1}{l})\sum_{m=1}^{l}(\frac{1}{m^2})$ and $ζ^\star(\bar{...

I've noticed that there are lots of similar-looking (at least to my untrained eye) questions on Math Stackexchange involving polylogarithms, all equally delicious, in which the OP asks about how to ...