Find $ \sum \limits^{\infty }_{n=1}\left( \sum \limits^{n}_{k=0}k^{5}\right) ^{-1}$ $$ \sum \limits^{\infty }_{n=1}\left( \sum \limits^{n}_{k=0}k^{5}\right) ^{-1}$$
How can we put the series above in a nice closed form? 
This problem was proposed on Brilliant.
 A: So, we want to compute:
$$ \sum_{n\geq 1}\frac{12}{n^2(n+1)^2(2n^2+2n-1)}.\tag{1}$$
By using the digamma reflection formula we have:
$$ \sum_{n\geq 1}\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)\frac{1}{2n^2+2n-1}=7-\frac{\pi^2}{3}+\frac{4\pi}{\sqrt{3}}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right) \tag{2}$$
hence the problem boils down to computing:
$$ \sum_{n\geq 1}\frac{1}{n(n+1)(2n^2+2n-1)}=1+\frac{\pi}{\sqrt{3}}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right)\tag{3} $$
with the same technique, i.e. through the identities:
$$ \sum_{n\geq 1}\left(\frac{1}{n+a}-\frac{1}{n+b}\right)= \psi(a+1)-\psi(b+1),\qquad \sum_{n\geq 1}\frac{1}{(n+a)^2}=\psi'(a+1)$$
and the well-known $\psi(z)-\psi(1-z)=-\pi\cot(\pi z)$. By putting all together, we reach:

$$ \sum_{n\geq 1}\left(\sum_{k=1}^{n}k^5\right)^{-1}=60-4\pi^2+8\pi\sqrt{3}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right).\tag{4}$$

We may compute the non-trivial series:
$$\sum_{n\geq 1}\frac{1}{(2n+1)^2-3}$$
also by considering the logarithmic derivative of the Weierstrass product for the cosine function. That gives:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-a^2}=\frac{\pi}{4a}\,\tan\left(\frac{\pi a}{2}\right).\tag{5}$$
A: *

*Write the inside summation in a closed form.

*Invert this expression.

*Use partial fraction decomposition.

*Evaluate the infinite summation. There should be some telescoping.

A: Hint:
let S be the closed form of our summation .
$$\sum_{k=1}^nk^5=\frac{1}{6}\sum_{j=0}^{5}\binom{6}{j}B_jn^{6-j}=\frac{1}{36}(6n^6-18n^5+15n^4-3n^2)=\frac{1}{12}(n^2(1+n)^2(2n(1+n)-1))$$
where is $B_j$ is Bernoulli number.
so
$$S=12\sum_{n=1}^{\infty}\frac{1}{n^2(1+n)^2(2n(1+n)-1)}$$
$$S=12\sum_{n=1}^{\infty}\left (\frac{-1}{n^2}-\frac{1}{(n+1)^2}+\frac{4}{(2n(1+n)-1)} \right )$$
$$S=4(15-\pi^2+2\sqrt{3}\tan(\frac{\sqrt{3}\pi}{2}))$$
