I was hoping someone would come up with a nice combinatorial argument! I know of two ways of doing it. The simplest way is to rewrite \binom{n}{k}/(k+1) as \binom{n+1}{k+1}/(n+1) and using the binomial theorem (since the denominators no longer depend on k). You can turn this into a combinatorial argument, but it seems kind of clunky to me and I don't see how to make it elegant. The other way, which sounds like basically the way you did it, is how I originally came across this. I was idly thinking about things like "if you take n random numbers (with respect to some distribution), what is the expected value of the largest one?" This sum came out of doing an integral to compute the following: if you pick n numbers in [1,\infty) with respect to the probability density function 1/x^2, what is the probability that the last one is the largest of them? In this formulation, of course, the answer is obvious.