Question

Find the sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-.....\infty$

Answer 1

Say $S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{n(n+1)}$

Now the n-th term of the series, $t=1n(n+1)$

or, $t=\frac{1}{n}-\frac{1}{(n+1)}$

So, $S=$ $\sum t$, as $n$ runs from $1$ to $n$

or, $S=\sum \frac{1}{n}-\frac{1}{(n+1)}$ , as $n$ runs from $1$ to$n$

or, $S=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+.....+(\frac{1}{n}-\frac{1}{(n+1)})$

or, $S=1-\frac{1}{(n+1)}$

As n tends to infinity, $\frac{1}{(n+1)}$ tends to $0$

Thus the required value of the sum is $1$