# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1+3/4+9/16+...+(3/4)^n+...?

Jun 8, 2018

${\sum}_{k = 0}^{n} {\left(\frac{3}{4}\right)}^{k} = 4 - 3 {\left(\frac{3}{4}\right)}^{n - 1}$

${\sum}_{k = 0}^{\infty} {\left(\frac{3}{4}\right)}^{k} = 4$

#### Explanation:

This is a geometric series of ratio $q = \frac{3}{4}$.

Consider the series:

${\sum}_{k = 0}^{\infty} {q}^{k}$

and its partial sum:

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

Then, if $\left\mid q \right\mid < 1$:

${\lim}_{n \to \infty} {s}_{n} = {\lim}_{n \to \infty} \frac{{q}^{n + 1} - 1}{q - 1} = \frac{1}{1 - q}$

For $q = \frac{3}{4}$:

${s}_{n} = \frac{{\left(\frac{3}{4}\right)}^{n} - 1}{\frac{3}{4} - 1} = \frac{{3}^{n} - {4}^{n}}{{4}^{n} \left(- \frac{1}{4}\right)} = \frac{{4}^{n} - {3}^{n}}{4} ^ \left(n - 1\right) = 4 - 3 {\left(\frac{3}{4}\right)}^{n - 1}$

and:

${\sum}_{k = 0}^{\infty} {\left(\frac{3}{4}\right)}^{k} = 4$