how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$$$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*}\begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac n{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac1{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also found out thatnumerically evaluated $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$$$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$$$\frac32\sum_{n=1}^\infty \frac{H_n^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$$$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac32\sum_{n=1}^\infty\frac{H_n^2}{n^3}?$$ how to prove this thanks?
Find $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^$\sum\limits_{\inftyi,j,n\ge1} \frac\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$
Find $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$
how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also found out that $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$ how to prove this thanks
Find $\sum\limits_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$
how to find the following series: $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac n{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac1{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also numerically evaluated $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac32\sum_{n=1}^\infty \frac{H_n^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac32\sum_{n=1}^\infty\frac{H_n^2}{n^3}?$$ how to prove this?
how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*} alsobut i don't know how to proceed. also found out that $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$ how to prove this thanks
how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*} also found out that $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$ how to prove this thanks
how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also found out that $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$ how to prove this thanks
Find $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$
how to find the following series: $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac{n}{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac{1}{i j (n + i)(n + j)(i + j)}. \end{align*} also found out that $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac{3}{2} \sum _{n=1}^{\infty } \frac{\left(H_n\right){}^2}{n^3}?$$ how to prove this thanks