# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1+1/3+1/9+...+(1/3)^n+...?

Jul 30, 2017

${S}_{n} = {\sum}_{k = 0}^{n} {\left(\frac{1}{3}\right)}^{k} = \frac{1}{2} \frac{{3}^{n} - 1}{3} ^ \left(n - 1\right)$

$S = {\sum}_{k = 0}^{\infty} {\left(\frac{1}{3}\right)}^{k} = \frac{3}{2}$

#### Explanation:

This is a geometric series of ratio $q = \frac{1}{3}$:

$S = {\sum}_{k = 0}^{\infty} {\left(\frac{1}{3}\right)}^{k} = {\sum}_{k = 0}^{\infty} {q}^{k}$

Consider the $n$-th partial sum of the series:

${S}_{n} = {\sum}_{k = 0}^{n - 1} {q}^{k} = 1 + q + {q}^{2} + \ldots + {q}^{n - 1}$

and note that:

$\left(1 + q + {q}^{2} + \ldots + {q}^{n - 1}\right) \left(1 - q\right) = 1 - \cancel{q} + \cancel{q} - \cancel{{q}^{2}} + \cancel{{q}^{2}} + \ldots - {q}^{n} = 1 - {q}^{n}$

so that:

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

As ${\lim}_{n \to \infty} {q}^{n} = {\lim}_{n \to \infty} {\left(\frac{1}{3}\right)}^{n} = 0$ we have

${\lim}_{n \to \infty} {S}_{n} = \frac{1 - {q}^{n}}{1 - q} = \frac{1}{1 - q} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}$

And in general we can see that a geometric series converges when:

${\lim}_{n \to \infty} {q}^{n} = 0$

that is for $\left\mid q \right\mid < 1$