Question

Find the sum of the first 100 terms in the series

Answer 1

Hello,

$\dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{1}{n(n+1)} \\ \dfrac{1}{1*2} = \dfrac{1}{1} - \dfrac{1}{2} \\ \dfrac{1}{2*3} = \dfrac{1}{2} - \dfrac{1}{3} \\ \dfrac{1}{3*4} = \dfrac{1}{3} - \frac{1}{4} \\ ...\\ \dfrac{1}{n*(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$

Adding member by member, we have

$\dfrac{1}{1*2} + \dfrac{1}{2*3} +\dfrac{1}{3*4} +...\dfrac{1}{n*(n+1)}=\\ \dfrac{1}{1} - \dfrac{1}{n+1} \\ = \dfrac{n}{n+1}$

if n=100 sum $\boxed{= \dfrac{100}{101} }$