What is the expected value of the number of die rolls necessary to get a specific number? Given a discrete random number generator, such as a six-sided die, what is the expected value of the number of rolls necessary to roll a specific number (e.g. a six)? 
I think the result should be given by E$\langle$rolls$\rangle$ = $\frac{1}{6}\sum_{n=0}^\infty{(\frac{5}{6})^n(n+1)}$, but I don't know how to calculate the convergence of that sum.
Also, how do I calculate the variance?

Elliott's answer is surely the nicest. To sum the series, we can use a method similar to the geometric series.
Let $$S = 1 + 2 \cdot \left(\frac{5}{6}\right) + 3 \cdot \left(\frac{5}{6}\right)^2 + \cdots $$
Then
$$\frac{5}{6}S = \frac{5}{6} + 2 \cdot \left(\frac{5}{6}\right)^2 + 3 \cdot \left(\frac{5}{6}\right)^3 + \cdots $$
$$S - \frac{5}{6}S = 1+ \left(\frac{5}{6}\right) +\left(\frac{5}{6}\right)^2 + \cdots $$
This is just a geometric series and hence we have that 
$$S = 36$$
Now we have that $$\frac{1}{6} \sum_{n=0}^\infty \left(\frac{5}{6}\right)^n (n+1) = \frac{1}{6} \cdot S = 6,$$ as expected.
(Convergence of the series can be seen by the ratio test)
