# How do you find the n-th partial sum of an infinite series?

Sep 19, 2014

The nth partial sum ${S}_{n}$ of a series ${\sum}_{k = 1}^{\infty} {a}_{k}$ is
${S}_{n} = {\sum}_{k = 1}^{n} = {a}_{1} + {a}_{2} + {a}_{3} + \cdots + {a}_{n}$

Let us find ${S}_{n}$ for the telescoping series

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

The partial sum is

${S}_{n} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n + 1}\right)$

by regrouping,

$= 1 + \left(- \frac{1}{2} + \frac{1}{2}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n}\right) - \frac{1}{n + 1}$

$= 1 + 0 + \cdots + 0 - \frac{1}{n + 1}$

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

Hence, ${S}_{n} = \frac{n}{n + 1}$.