How do derive this result in stat-mech style I'm going through (well, at least I'm planning to) Rief's book about statistical mechanic (I want to improve my knowledge). I want to be serious about this so I'm trying to solve as much problem as I can (and understand them). I came upon this one (1.5):

In a game of a Russian roulette (not recommended by the author), one
  inserts a single cartridge into the drum and of a revolver, leaving
  the other five chambers of the drum empty. One then spins the drum,
  aims at one's head, and pulls the trigger.
  (a) What it the probability
  of being still alive after playing the game $N$ times?
  (b) What it the
  probability of surviving ($N-1$) turns in this game and then being
  shot the $N$th time one pulls the trigger?
  (c) What is the mean number
  of times a player gets the opportunity of pulling the trigger in this
  macabre game?

I believe that the starting point for (c) is:
$\left \langle n\right \rangle = \sum_{n=1}^\infty n \left(\frac{5}{6}\right)^{(n-1)} \frac{1}{6}$
Wolfram alpha tells me this is an exact sum which equals to 6 (logical). I want however, to try and imply the standard derivative trick (and this is why I'm asking this here and not in the math thread):
$\left \langle n\right \rangle = \frac{1}{6}\sum_{n=1}^\infty \frac{\partial}{\partial n} \left(\frac{5}{6}\right)^{n} = \frac{1}{6} \frac{\partial}{\partial n} \sum_{n=1}^\infty  \left(\frac{5}{6}\right)^{n}$
But this is an exact sum (that equals to 1) and I can't take it's derivative. I know I could use some table to solve this, the question is if there some way to solve this in a way that a student can do this in the exam?
I hope any of what I wrote is clear and approved by the site policy. 
 A: The trick works, you just need to apply it differently. The problem is that you are differentiating with respect to $n$, which won't work because you are summing over $n$.
Instead, the sum you wish to evaluate is,
$$\langle n \rangle = \sum_{n=1}^\infty n x^{n-1}(1-x) = (1-x) \sum_{n=1}^\infty n x^{n-1}$$
with $x=5/6$. Now this can be written as,
$$\langle n \rangle = (1-x)\frac{\partial}{\partial x} \sum_{n=1}^\infty x^{n}  = (1-x)\frac{\partial}{\partial x}\left(\frac{1}{1-x} -1\right) = 1/(1-x)$$
which gives 6 for x = 5/6.
A: $$\left \langle n\right \rangle ~=~ \frac{1}{6}\sum_{n=1}^\infty n \left(\frac{5}{6}\right)^{n-1}~=~ \left.\frac{1}{6}\sum_{n=1}^\infty n x^{n-1} \right|_{x=\frac{5}{6}}
~=~ \left.\frac{1}{6}\sum_{n=1}^\infty \frac{d}{dx} x^n \right|_{x=\frac{5}{6}} $$
$$~=~ \left.\frac{1}{6}\frac{d}{dx}\sum_{n=1}^\infty  x^n \right|_{x=\frac{5}{6}}
~=~ \left.\frac{1}{6}\frac{d}{dx}\frac{x}{1-x} \right|_{x=\frac{5}{6}}~=~ \left.\frac{1}{6}\frac{1}{(1-x)^2} \right|_{x=\frac{5}{6}}~=~6.$$
