Recall that given a geometric series, we were able to establish convergence by deriving an expression for the partial sum, $s_{n}$, and by determining the value of

$\lim _{n\rightarrow \infty }s_{n}.$

However, given a general infinite sum, this approach is not always convenient and sometimes impossible because we cannot always find an expression for $s_{n}$. But there is another class of infinite series where this approach is feasible.

Consider the following example. Suppose we would like to determine whether the series

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

converges, and determine its sum. One approach is to use the definition of convergence, which requires an expression for the partial sum, $s_{n}$. We see that

${\begin{aligned}s_{n}&=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}\\&=\sum _{k=1}^{n}{\Big (}{\frac {1}{k}}-{\frac {1}{k+1}}{\Big )},\end{aligned}}$

by using partial fractions. Expanding the sum yields

${\begin{aligned}s_{n}&={\Big (}{\frac {1}{1}}-{\frac {1}{2}}{\Big )}+{\Big (}{\frac {1}{2}}-{\frac {1}{3}}{\Big )}+{\Big (}{\frac {1}{3}}-{\frac {1}{4}}{\Big )}+\ldots +{\Big (}{\frac {1}{n-1}}-{\frac {1}{n}}{\Big )}+{\Big (}{\frac {1}{n}}-{\frac {1}{n+1}}{\Big )}\end{aligned}}$

Rearranging the brackets, we see that the terms in the infinite sum cancel in pairs, leaving only the first and lasts terms.

${\begin{aligned}s_{n}&={\Big (}{\frac {1}{1}}\mathbf {\color {Purple}{-{\frac {1}{2}}}} {\Big )}+{\Big (}\mathbf {\color {Purple}{\frac {1}{2}}} \mathbf {\color {Blue}{-{\frac {1}{3}}}} {\Big )}+{\Big (}\mathbf {\color {Blue}{\frac {1}{3}}} -{\frac {1}{4}}{\Big )}+\ldots +{\Big (}{\frac {1}{n-1}}\mathbf {\color {Brown}{-{\frac {1}{n}}}} {\Big )}+{\Big (}\mathbf {\color {Brown}{\frac {1}{n}}} -{\frac {1}{n+1}}{\Big )}\\&={\frac {1}{1}}+{\Big (}\mathbf {\color {Purple}{-{\frac {1}{2}}+{\frac {1}{2}}}} {\Big )}+{\Big (}\mathbf {\color {Blue}{-{\frac {1}{3}}+{\frac {1}{3}}}} {\Big )}+\ldots +{\Big (}\mathbf {\color {Brown}{-{\frac {1}{n}}+{\frac {1}{n}}}} {\Big )}-{\frac {1}{n+1}}{\Big )}\\&={\frac {1}{1}}+{\Big (}0{\Big )}+{\Big (}0{\Big )}+\ldots +{\Big (}0{\Big )}-{\frac {1}{n+1}}\\&=1-{\frac {1}{n+1}}\end{aligned}}$

Hence,

${\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}&=\lim _{n\rightarrow \infty }s_{n}\\&=\lim _{n\rightarrow \infty }{\Big (}1-{\frac {1}{n+1}}{\Big )}\\&=1\end{aligned}}$

Therefore, by the definition of convergence for infinite series, **the above telescopic series converges and is equal to 1**.