How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify?

Question

$\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$

$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$

My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$?

$$\frac1{n(n+1)}=\frac1n-\frac 1{n+1}$$ – Nov 15 '14 at 19:33 — , Nov 15 '14 at 19:33
Do you know partial fraction decomposition ? – Claude Leibovici Nov 15 '14 at 19:33 — Claude Leibovici, Nov 15 '14 at 19:33

score 1 · Answer 1 · answered Nov 15 '14 at 19:33

1

It is because $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$

answered Nov 15 '14 at 19:33

BS.

abnry · Accepted Answer · 2014-11-17T23:57:01.593

1

Write $$\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1},$$ split the sum into two sums, shift the index of one, recombine, simplify, and you are done.

N.B. You need to do this for partial sums and only at the end take the limit to be rigorous.

edited Nov 17 '14 at 23:57

answered Nov 15 '14 at 19:33

abnry

2 Answers2