# The sum of an infinite series

I'm working on a probability question and I want to find the precise summation of:

$$\sum_{p=1}^\infty \frac{n^{p-1} - \sum_{x=0}^{p-2} n^x}{n^p}$$

Where $n$ is an integer greater than $1$. More precisely, I am wondering how variations of $n$ may affect the infinite sum (given this series is not diverging, for I think it converges).

As you may have guessed, I have failed so far.

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What exactly would you like to know? Where does this come from? There might be an alternative approach to find your answer... – vonbrand Feb 11 '14 at 16:11
The case $n=2$ is OK, larger integers are not. – André Nicolas Feb 11 '14 at 16:14
The series diverges except when $n=2$ (and then it sums to $1$). – Did Feb 11 '14 at 16:15
oh, for n=2, that was obvious :P Thanks! Btw, does it diverge for a greater n due to the increasing discrepancy between n^(p-1) and n^(p-2) as p approaches infinity? @ vonbrand, there isn't any concrete problem I got this from, there were some loosely related ones and I got inspired from them to create this, in retrospect rather boring series. – Just_a_fool Feb 11 '14 at 16:30
@AndréNicolas, why does it diverge for $n > 2$? Probably something silly I'm overlooking... and I'd ask you to make your comment into a complete answer, so this doesn't stay open. – vonbrand Feb 11 '14 at 17:45

$$\sum_{p = 1}^{\infty} \frac {(n - 2) n^{p - 1} + 1} {n^p (n - 1)} \sim \sum_{p = 1}^{\infty} \frac {1} {n}$$