Geometric Series Sum (Hoping somebody can check my work)

Printable View

February 28th 2013, 10:06 PM
AZach

Geometric Series Sum (Hoping somebody can check my work)

$\sum_{n=0}^\infty \frac{2^{n}}{5^{2n+1}}$

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

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

Which ends up being $S = \frac{1}{23}$ ?
March 1st 2013, 03:54 AM
Plato

Re: Geometric Series Sum (Hoping somebody can check my work)

Quote:

Originally Posted by AZach

$\sum_{n=0}^\infty \frac{2^{n}}{5^{2n+1}}$

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

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

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

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

Recall that $S=\frac{a}{1-r}$ where $a$ is the first term.
March 1st 2013, 04:43 AM
ibdutt

1 Attachment(s)

Re: Geometric Series Sum (Hoping somebody can check my work)

Attachment 27313

All times are GMT -8. The time now is 10:08 PM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.