# Geometric Series Sum (Hoping somebody can check my work)

• Feb 28th 2013, 10:06 PM
AZach
Geometric Series Sum (Hoping somebody can check my work)
$\displaystyle \sum_{n=0}^\infty \frac{2^{n}}{5^{2n+1}}$

I broke it up like $\displaystyle {\frac{1}{5}}\sum_{n=0}^\infty ({\frac{2}{25}})^{n}$

I calculated the sum $\displaystyle S = \frac{1}{5}[\frac{\frac{1}{5}}{\frac{25}{25}-\frac{2}{25}}]$

Which ends up being $\displaystyle S = \frac{1}{23}$ ?
• Mar 1st 2013, 03:54 AM
Plato
Re: Geometric Series Sum (Hoping somebody can check my work)
Quote:

Originally Posted by AZach
$\displaystyle \sum_{n=0}^\infty \frac{2^{n}}{5^{2n+1}}$

I broke it up like $\displaystyle {\frac{1}{5}}\sum_{n=0}^\infty ({\frac{2}{25}})^{n}$

I calculated the sum $\displaystyle S = \frac{1}{5}[\frac{\frac{1}{5}}{\frac{25}{25}-\frac{2}{25}}]$

Which ends up being $\displaystyle S = \frac{1}{23}$ ?

It should be
$\displaystyle S = \frac{1}{5}\left[\frac{1}{\frac{25}{25}-\frac{2}{25}}\right]$

Recall that $\displaystyle S=\frac{a}{1-r}$ where $\displaystyle a$ is the first term.
• Mar 1st 2013, 04:43 AM
ibdutt
Re: Geometric Series Sum (Hoping somebody can check my work)