Suppose that there are $n$ different coupons, equally likely, from which coupons are being randomly drawn with replacement. Find the expected number of draws for some coupon to be drawn twice.
I've attempted to solve this from the basic definition of expectation. If $N$ is a randomly variable for the number of draws for some coupon to be drawn twice, then $1\lt N\le n+1$ and
$$ E[N]=\sum_{i=2}^{n+1} n*Pr(N=n)\\ =2*\left(\frac{1}{n}\right)+3*\left(\frac{n-1}{n}\right)\left(\frac{2}{n}\right)+4*\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\left(\frac{3}{n}\right)+... $$
(Sorry for not finishing the equation, hopefully you get the idea, it goes up to $n+1$, my latex maths is pretty crap.)
Is this correct? How do I evaluate this sum or is there a closed-form solution to this sum? Is there a better way to solve this?