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 $\displaystyle S = \frac{1}{23}$ ?

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

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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**.

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