+1 vote
121 views

The sum of the series $\dfrac{1}{1.2} + \dfrac{1}{2.3}+ \cdots + \dfrac{1}{n(n+1)} + \cdots$ is

1. $1$
2. $1/2$
3. $0$
4. non-existent

recategorized | 121 views

## 2 Answers

+6 votes
$\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}+\ldots$

$=\sum _{k=1}^{\infty}\frac{1}{k(k+1)}$

$=\sum _{k=1}^{\infty}\frac{(k+1)-k}{k(k+1)}$

$=\sum _{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)$

$\require{cancel}=\left(\frac{1}{1}-\cancel{\frac{1}{2}}\right)+\left(\cancel{\frac{1}{2}}-\cancel{\frac{1}{3}}\right)+\left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right)+\ldots$

$=1-\frac{1}{\infty}$

$=1$

$\therefore$ Option (A).
by Boss (13.4k points)
0 votes
the series can be expanded as 1-1/2+1/2-1/3...........1/n-1/n+1, all the terms will cancel out each other except first and last term so at the end we will get 1-1/n+1, and for larger N this term will be getting smaller and smaller so when n is very large 1/n+1 will be equal to 0, therefore, the final answer is 1.
by (299 points)

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