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!
from Hot Weekly Questions - Mathematics Stack Exchange
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