# Closed form of $\sum^{\infty}_{n=1} \dfrac{1}{n^a{(n+1)}^a}$ where $a$ is a positive integer

Recently, I bought a book about arithmetic. I saw a question is like that:

Given that $$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots=\dfrac{\pi^2}{6}$$, find the value of $$\dfrac{1}{1^32^3}+\dfrac{1}{2^33^3}+\dfrac{1}{3^34^3}+\cdots$$

Then, I did this: $$\sum_{n=1}^\infty \dfrac{1}{n(n+1)}=\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1 \\ \sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2}=\sum_{n=1}^\infty \dfrac{n^2+(n+1)^2-2n(n+1)}{n^2(n+1)^2}\\=\sum_{n=1}^\infty \left[\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}-\dfrac{2}{n(n+1)}\right]=\zeta(2)+\zeta(2)-1-2=\dfrac{\pi^2}{3}-3 \\ \sum_{n=1}^\infty \dfrac{1}{n^3(n+1)^3}=\sum_{n=1}^\infty \dfrac{(n+1)^3-n^3-3n(n+1)}{n^3(n+1)^3} \\= \sum_{n=1}^\infty\left[\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}-\dfrac{3}{n^2(n+1)^2}\right]=1-3\left(\dfrac{\pi^2}{3}-3\right)=10-\pi^2$$ Though I finished the problem, but I found something interesting, so I wanted to discover more. I let $$f(a)=\sum_{n=1}^\infty \dfrac{1}{n^a(n+1)^a}$$. Then, I found something special: $$f(4)=2\zeta(4)-2f(3)-2f(2)-1 \\ f(5)=1-5f(4)-5f(3) \\ f(6)=2\zeta(6)-6f(5)-9f(4)-2f(3) \\ f(7)= 1-7f(6)-14f(5)-7f(4)$$

It seems to have a sequence, but I can't tell what it is. If anyone knows what is the sequence or even the closed form of $$f(a)$$, please tell me. Thank you very much!

Let $$h(z)= \frac1{(z+1)^k},g(z)=\frac{1}{z^k}$$ then $$\frac{1}{z^k(z+1)^k} -\frac{\sum_{m=0}^{k-1} \frac{h^{(m)}(0)}{m!} z^m}{z^k}- \frac{\sum_{m=0}^{k-1} \frac{g^{(m)}(-1)}{m!} (z+1)^m}{(z+1)^k}$$
is a rational function with no pole thus it is a polynomial and since it vanishes at $$\infty$$ it is $$0$$.
$$\frac{h^{(m)}(0)}{m!} =(-1)^m {m+k-1 \choose m}$$, $$\frac{g^{(m)}(-1)}{m!} = (-1)^{k}{m+k-1 \choose m}$$
Thus $$\sum_{n=1}^\infty \frac{1}{n^k(n+1)^k}= \sum_{m=0}^{k-2}\zeta(k-m) {m+k-1 \choose m} ((-1)^m+ (-1)^{k})-\sum_{m=0}^{k-1} (-1)^{k}{m+k-1 \choose m}$$