Sum of series

I'm struggling with finding the sum of the series [math]\sum_{n=0}^{\infty}(n+1)(\frac{1}{4})^n [\math]
So far I've got:
\sum_{n=0}^{\infty}(n+1)(\frac{1}{4})^n = \sum_{n=0}^{\infty}n(\frac{1}{4})^n + \sum_{n=0}^{\infty}(\frac{1}{4})^n = \sum_{n=0}^{\infty}n(\frac{1}{4})^n + \frac{4}{3}
By a numerical approach I've found \sum_{n=0}^{\infty}n(\frac{1}{4})^n \approx \frac{4}{9} , but I have little clue on how to derive it. Any ideas?

Let's try that again:

I'm struggling with finding the sum of the series \sum_{n=0}^{\infty}(n+1)(\frac{1}{4})^n
So far I've got:
\sum_{n=0}^{\infty}(n+1)(\frac{1}{4})^n = \sum_{n=0}^{\infty}n(\frac{1}{4})^n + \sum_{n=0}^{\infty}(\frac{1}{4})^n = \sum_{n=0}^{\infty}n(\frac{1}{4})^n + \frac{4}{3}
By a numerical approach I've found \sum_{n=0}^{\infty}n(\frac{1}{4})^n \approx \frac{4}{9} , but I have little clue on how to derive it. Any ideas?

Two ways:
1) Differentiate \sum_{n=0}^\infty r^n=\frac{1}{1-r} with respect to r.
2) Continue in the way you simplifed your original sum, by spliting the sum into two repeatedly.  This gives \sum_{n=1}^\infty nr^n=\sum_{m=1}^\infty\sum_{n=m}^\infty r^n, which can be evaluated by summing a geometric series twice.

No matter what the sums are, the process is always the same : http://www.damianmarksmyth.com/no-matter-what-the-sums-are-the-process-is-always-the-same

Bitch, put it in MathCad, Sagemath or MATLAB or whatever is here on WikiFaggotia: http://en.m.wikipedia.org/wiki/List_of_numerical_analysis_software

Motherfucker