Evalulating $\sum\limits_{n=1}^\infty \left\{ (\frac{5}{13})^n \right\}$ How do I evaluate: 
$$\sum_{n=1}^\infty \left\{ \left(\frac{5}{13}\right)^n \right\}$$
using mathematical techniques?  I wrote a line of Powershell to do it:
@(1..100) | % { [Math]::Pow(5/13, $_) } | Measure-Object -Sum

and the answer is $0.625$, but how do I do it with maths?
 A: Let $$S = \sum_{n=1}^{\infty}\left(\frac{5}{13}\right)^n$$
Then we have
$$S - \frac{5}{13}S = \sum_{n=1}^{\infty}\left(\frac{5}{13}\right)^n - \sum_{n=2}^{\infty}\left(\frac{5}{13}\right)^n$$
Note that this works because $\left(\frac{5}{13}\right)^n \to 0$ as $n\to \infty$. The base of the exponential expression has to have an absolute value $ < 1$, for the series to converge. It won't work for things like $2^n$, or $(-1.1)^n$.
Continuing, we have:
$$\frac{8}{13}S = \frac{5}{13}$$
$$S = \frac{5}{8}$$
We can, using the same approach, generalize (for $|x| < 1$) this to derive the summation of a convergent infinite geometric series:
$$\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$$
A: This is a geometric series:
$\displaystyle \sum_{n=0}^\infty x^n = \frac{1}{1-x}$, or
$\displaystyle \sum_{n=1}^\infty x^n = \frac{x}{1-x}$.
These expressions are valid for $|x| < 1$, in particular for $x = \frac{5}{13}$, which is your case.
A: Using the formula of summation of Infinite Geometric Series, the sum will be $$\frac{\dfrac5{13}}{1-\dfrac5{13}}=\frac5{13-5}$$ 
Here the common ratio is $\displaystyle\frac5{13}$
