Calculating $\sum\frac{1}{n(n+1)(n+2)}$

Question

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

I know it converges via comparison test, but how to actually find the series value?

score 5 · Accepted Answer · answered Mar 07 '16 at 03:35

5

Hint: $\dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2}\left(\dfrac{1}{n(n+1)} - \dfrac{1}{(n+1)(n+2)}\right)$

answered Mar 07 '16 at 03:35

DeepSea

1

Out of curiosity, is there a way you knew to manipulate the summand like this? – Chio Mar 07 '16 at 03:35
3

@Chio it's a really common method called partial fraction decomposition. – YoTengoUnLCD Mar 07 '16 at 03:37
Ah, ok. I think I understand. Then I just compute each term's sum, yes? – Chio Mar 07 '16 at 03:39
Since I can break each term into another partial fraction decomposition! – Chio Mar 07 '16 at 03:42
And then those telescope. Thanks! – Chio Mar 07 '16 at 03:43
@Chio You needn't even split the terms given inasmuch as the right-hand side already is telescoping. ;-)) -Mark – Mark Viola Mar 07 '16 at 04:28

1 Answers1