Some Triple Sum

I replace $i$ and $j$ with $m$ and $n$ as I am more comfortable working with them.

Note that

$\begin{aligned} \frac{k!}{(n+k+1)!}-\frac{(k+1)!}{(n+k+2)!} &=(n+k+2)\frac{k!}{(n+k+2)!}-(k+1)\frac{k!}{(n+k+2)!}\\\ &=(n+1)\frac{k!}{(n+k+2)!}\tag{1} \end{aligned}$

Sum $(1)$ for $k≥0$ and divide by $n+1$ to get \displaystyle\sum_{k=0}^\infty\frac{k!}{(n+k+2)!}=\frac1{(n+1)(n+1)!}\tag{2}

One application of $(2)$ gives \begin{aligned} \sum_{n=0}^\infty\sum_{k=0}^\infty\frac{n!k!}{(n+k+2)!} &=\sum_{n=0}^\infty\frac{n!}{(n+1)(n+1)!}\\ &=\sum_{n=0}^\infty\frac1{(n+1)^2}\\ &=\frac{\pi^2}{6}\tag{3} \end{aligned}

Another application of (2) yields $\begin{aligned} \sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{k=0}^\infty\frac{n!m!k!}{(n+m+k+2)!} &=\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{n!m!}{(n+m+1)(n+m+1)!}\tag{4} \end{aligned}$

Now, consider \begin{aligned} a_n &=\sum_{m=0}^nm!(n-m)!\\ &=n!+\sum_{m=0}^{n-1}m!(n-m-1)!\,((n+1)-(m+1))\\ &=n!+(n+1)\sum_{m=0}^{n-1}m!(n-m-1)!-\sum_{m=0}^{n-1}(m+1)!(n-m-1)!\\ &=n!+(n+1)a_{n-1}-(a_n-n!)\tag{5}\\ a_n&=n!+\frac{n+1}{2}a_{n-1}\tag{6}\\ b_n&=\frac{2^n}{n+1}+b_{n-1}\quad\text{where }b_n=\frac{2^n}{(n+1)!}a_n\tag{7} \end{aligned}

Thus $a_n=\frac{(n+1)!}{2^n}\sum_{k=0}^n\frac{2^k}{k+1}\tag{8}$ Using (8), the sum in (4) is $\begin{aligned} \sum_{n=0}^\infty\frac{a_n}{(n+1)(n+1)!} &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{2^{k-n}}{(n+1)(k+1)}\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{2^{k-n}}{(n+1)(k+1)}\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{2^{-n}}{(n+k+1)(k+1)}\\ &=\frac{\pi^2}{6}+\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{2^{-n}}{n}\left(\frac1{k+1}-\frac1{n+k+1}\right)\\ &=\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{H_n}{2^nn}\\ &=\frac{\pi^2}{4}\tag{9} \end{aligned}$

This can be achieved by noting that $\begin{aligned} \frac{\pi^2}{12} &=\sum_{n=0}^\infty2^{-n-1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^2}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k+1}\frac{(-1)^k}{k+1}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\sum_{j=0}^n\binom{j}{k}\frac{(-1)^k}{k+1}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\sum_{k=0}^n\binom{j+1}{k+1}\frac{(-1)^k}{j+1}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\frac1{j+1}\\ &=\sum_{n=1}^\infty2^{-n}\frac1{n}\sum_{j=1}^n\frac1j\\ &=\sum_{n=1}^\infty\frac{H_n}{n2^n} \end{aligned}$