Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$

Question

Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2} {6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$

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Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$ but nothing seems to lead to anything!

Answer 1

HINT:

Use the partial fraction decomposition $$\frac{1}{n^3(n+1)^3} = \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}+ \frac{6}{n} - \frac{6}{n+1}\right) - \left(\frac{3}{n^2} + \frac{3}{(n + 1)^2}\right) $$

The sum equals $(1+6) -( 3 \zeta(2) +3 \zeta(2)- 3)= 10 - \pi^2$.

Answer 2

I'm basically just going to use the identity you noted in the comments $\frac 1n-\frac 1{n+1}=\frac 1{n(n+1)}$ repeatedly and then the closed form for $\zeta(2)$.

$$\sum_{n=1}^\infty \frac 1{n^3(n+1)^3}=\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^3=\sum_{n=1}^\infty \frac 1{n^3}-\frac{3}{n^2(n+1)}+\frac{3}{n(n+1)^2}-\frac 1{(n+1)^3}$$

The first and last terms telescope so you get

$$=1-3\sum_{n=1}^\infty \frac 1{n(n+1)}\left(\frac 1n-\frac 1{n+1}\right)=1-3\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^2$$

$$=1-3\sum_{n=1}^\infty \frac 1{n^2}-\frac{2}{n(n+1)}+\frac 1{(n+1)^2}$$

The first and last terms are $\zeta(2)$ and $\zeta(2)-1$ respectively

$$=4-6\zeta(2)+6\sum_{n=1}^\infty \frac 1{n(n+1)}=4-6\zeta(2)+6\sum_{n=1}^\infty\frac 1n-\frac 1{n+1}=10-6\zeta(2)=10-\pi^2$$

Answer 3

Hint: $$ \frac{1}{n^3} - \frac{1}{{(n+1)}^3} \;=\; \frac{3n(n+1) + 1}{n^3{(n+1)}^3} $$