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.

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
$$S=12\sum_{n=1}^{\infty}\frac{1}{n^2(1+n)^2(2n(1+n)-1)}$$
$$S=12\sum_{n=1}^{\infty}\left (\frac{-1}{n^2}-\frac{1}{(n+1)^2}+\frac{4}{(2n(1+n)-1)} \right )$$
$$S=4(15-\pi^2+2\sqrt{3}\tan(\frac{\sqrt{3}\pi}{2}))$$
