Here's yet another proof. This one uses probability theory. Let's warm up with a probabilistic proof of 1 + p + p^2 + ... = 1/(1-p) in the form (1-p

Here's yet another proof. This one uses probability theory. Let's warm up with a probabilistic proof of 1 + p + p^2 + ... = 1/(1-p) in the form (1-p)(1 + p + p^2 + ...) = 1 This says that if you have a Bernoulli process with failure probability p and success probability 1-p, you will eventually succeed with probability 1. Indeed, (1-p) p^n is the probability that you will succeed after n failures. To see this is true (it is certainly intuitively plausible), note that the probability of taking longer than n tries is p^n, which goes to zero as n goes to infinity. We can apply the same kind of reasoning to (1-p)^2 (1 + 2p + 3p^2 + ...) = 1 As before, we have a process with failure probability p and success probability p-1. However, rather than stopping after the first success, we will continue until we have two successes. Now take a look at the terms of the equation's left-hand side: (1-p)^2 (n+1) p^n This is the probability that we will have gotten 2 successes (and hence n failures) after n+2 steps. The second success must always be the final trial, so there are n+1 possible positions for the first success. That explains the factor of n+1. With that understood, the equation simply says that we will eventually get 2 successes with probability 1. To see why, note that the probability of never getting any successes in n tries is p^n as before, which tends to zero. So we will eventually get at least one success with probability 1. But once we have that first success, getting the second success is just our first problem repeated, so we know that happens eventually with probability 1. This proof technique generalizes easily. If we keep trying until we have 3 successes, we get a factor of C(n+2, 2) = (n+2)(n+1)/2 for the number of ways the first 2 successes can be placed before the third and final success when there are n failures and 3 successes. We show that a first success occurs eventually with probability 1 as before, which reduces the problem to getting two subsequent successes, a problem we already solved. You can see this yields a proof by induction for this whole class of series identities, but it's not of those unenlightening proofs by induction that shows you something is true but does not tell you why.