Sum of series with addition: $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$

Question

I am looking at some homework where I have:

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

How can I sum this?

I know that

$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$

and also that

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

But I just can't see or find the connection

See also: Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ — Martin Sleziak, Sep 10 '20 at 04:09
Does this answer your question? Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ — amWhy, Mar 02 '21 at 20:25

score 9 · Accepted Answer · answered Sep 27 '19 at 13:55

9

Hint:

$$\dfrac1{n^2(n+1)}=\dfrac{n+1-n}{n^2(n+1)}=\dfrac1{n^2}-\dfrac1{n(n+1)}$$

$$\dfrac1{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=?$$

answered Sep 27 '19 at 13:55

lab bhattacharjee

score 7 · Answer 2 · answered Sep 27 '19 at 13:55

7

HINT

$$\frac{1}{n^2(n+1)}=\frac1{n^2}+\frac1{n+1}-\frac1{n}$$

answered Sep 27 '19 at 13:55

user

score 5 · Answer 3 · answered Sep 27 '19 at 13:56

5

Very simple $\dfrac{1}{n^2(n+1)} =\dfrac{(n+1)-n}{n^2(n+1)} =\dfrac{1}{n^2} -\dfrac{1}{n(n+1)} $

Can you take it from here?

answered Sep 27 '19 at 13:56

Akshaj Bansal

Thanks, but can it really be true that this sums up to some strange real number? I would think homework would be some friendly number – Daniel Sep 27 '19 at 14:38
I dont whether it is special or not but the answer to question is $0.644934067$ – Akshaj Bansal Sep 27 '19 at 14:43
Ok thanks, seems to be $\pi^2/6-1$ – Daniel Sep 27 '19 at 14:49
Yeah sure its that only – Akshaj Bansal Sep 27 '19 at 14:51

3 Answers3