# Compute the exact value of the following sum: \sum_{k=1}^{\infty} \frac{2^k}{5^{k+1}}

## Question:

Compute the exact value of the following sum:

{eq}\sum_{k=1}^{\infty} \frac{2^k}{5^{k+1}} {/eq}

## Geometric Series:

A series of the form {eq}\sum_{n=1}^{\infty} ar^{n-1} {/eq} is said to be a geometric series. The series converges if {eq}|r|<1 {/eq} and diverges if {eq}|r|\geq 1 {/eq}. When the series converges, the sum is

{eq}\dfrac{a}{1-r} {/eq}. The geometric series has a bunch of important properties. In particular, we can write the function {eq}f(x)=\frac{1}{1-x} {/eq} in the form {eq}\sum_{n=0}^{\infty} x^n {/eq}.

We first rewrite the series

{eq}\begin{align} \sum_{k=1}^{\infty} \frac{2^k}{5^{k+1}}&=\sum_{k=1}^{\infty} \frac{2^k}{55^k}\\ &=\sum_{k=1}^{\infty} \frac{22^{k-1}}{5^25^{k-1}}\\ &=\sum_{k=1}^{\infty} \frac{2}{25}(\frac{2}{5})^{k-1} \end{align} {/eq}

This is a geometric series with {eq}a=\frac{2}{25} {/eq} and {eq}r=\frac{2}{5} {/eq}. Since {eq}|\frac{2}{5}|<1 {/eq}, the series converges to

{eq}\begin{align} \dfrac{a}{1-r}&=\dfrac{2/25}{1-2/5}\\ &=\dfrac{2/25}{3/5}\\ &=\dfrac{2}{15} \end{align} {/eq}