Question

Sum the following series to n terms and to infinity $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....$.

Answer 1

General term ($n^{th}$ term) $=\frac{1}{n(n+1)}$

We can rewrite it as $=\frac{n+1-n}{n(n+1)}$

$=\frac{1}{n}-\frac{1}{n+1}$

Sum of $n$ terms is $\sum _{n=1}^n[\frac{1}{n}-\frac{1}{n+1}]$

$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{n}.........-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{n}-\frac{1}{n+1}$

$=1-\frac{1}{n+1}$

Sum to infinity will get by putting $n\to \infty$

Then, $1-\frac{1}{n+1}$ $=1-0=1$ as $n\to \infty$