FInding the exact value of a series I would like to see if anyone can find the EXACT value of
$$5(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+...)+7(\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+...)+9(\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+...)+...$$
I have tried to regroup the terms and obtain
$$5(\zeta(4)-\frac{1}{1^4})+7(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4})+9(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4}-\frac{1}{3^4})+11(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4}-\frac{1}{3^4}-\frac{1}{4^4})+...$$
Seems like this is a diverging series. Any idea how to resolve this?

Since all summands are positive, you can rearrange and get
$$
\sum_{n=2}^\infty (2n+1)\sum_{k=n}^\infty \frac1{k^4}=\sum_{n=2}^\infty\frac1{n^4}\sum_{k=2}^n (2k+1)
$$
The coefficient of $\frac1{n^4}$ is $5+7+\dots+(2n+1)=n^2+2n-3$, so by absolute convergence,
\begin{align*}
\sum_{n=2}^\infty\frac{n^2+2n-3}{n^4}
&=\sum_{n=1}^\infty\frac{n^2+2n-3}{n^4}\\
&=\sum_{n=1}^\infty\frac1{n^2}+2\sum_{n=1}^\infty\frac1{n^3}+(-3)\sum_{n=1}^\infty\frac1{n^4}\\
&=\zeta(2)+2\zeta(3)-3\zeta(4).
\end{align*}
