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

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

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!

• What have you tried? Sep 18, 2021 at 21:27
• Please, follow this link math.meta.stackexchange.com/questions/9959/… The current form of your question lacks a whole swath of things. Do consider improvements, and at the very least, show attempts, this is not a "do-it-for-me" site.
– user956717
Sep 18, 2021 at 21:36
• i have tried using the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$ Sep 18, 2021 at 21:37
• @Math3147 you are right.Sorry for that. Sep 18, 2021 at 21:41
• Retracted my downvote, and upvoted. :)
– user956717
Sep 18, 2021 at 21:44

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

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

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

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