Asking about $\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]$ $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{n^2+n+4372}{(2n+1)^7(n+1)}\right]=\frac{61}{184320}\pi^7\tag1$$
Step 1:
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{n^2+n}{(2n+1)^7(n+1)}\right]\tag2$$
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]+\sum_{n=2}^{\infty}(-1)^n\left[\frac{1}{(2n+1)^7}\right]\tag3$$
Recall $$\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n+1)^7}=\frac{61}{184320}\pi^7$$
It looks like that this sum $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\left[\frac{4372}{(2n-1)^7(n-1)}+\frac{4372}{(2n+1)^7(n+1)}\right]$$
is a rational number
I am not able to show that sum $(1)=\frac{61}{184320}\pi^7$
Any help.Thank you!

It comes from telescoping series:
$$
\frac{(-1)^n}n\left(\frac{1}{(2 n+1)^7 (n+1)}+\frac{1}{(2 n-1)^7 (n-1)}\right) = (-1)^n\left(\frac{1}{n-1} + \frac{1}{n}\right) +
(-1)^n\left(\frac{1}{n}+\frac{1}{n+1}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^7}+\frac{1}{(2 n+1)^7}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^5}+\frac{1}{(2 n+1)^5}\right)
- 4(-1)^n\left(\frac{1}{(2 n-1)^3}+\frac{1}{(2 n+1)^3}\right)
- 4(-1)^n\left(\frac{1}{2 n-1}+\frac{1}{2 n+1}\right)
$$
Each parenthesis has two terms. The second term of $n=n_0$ is equal to the first term of $n=n_0+1$ with the opposite sign (coming from $(-1)^n$). So only the first term of $n=2$ survives.
Edit. P.S. By the way, it's not necessary to calculate partial fractions, if you just want to show the rationality. We can see that the function is odd ($n\to -n$), so there are no fractions with even exponent and all $(2n\pm1)^{-k}$ parentheses have the same coefficient. We only need to calculate the coefficients of $n^{-1}$, $(n+1)^{-1}$ and $(n-1)^{-1}$. We can easily do it with multiplying by corresponing monomial and letting $n=-1,0,1$.
