A slightly more intuitive argument. If you have a half pill and k - 1 whole pills remaining, then because we stop once all the whole pills are broken, then thi

A slightly more intuitive argument. If you have a half pill and k - 1 whole pills remaining, then because we stop once all the whole pills are broken, then thinking about the order in which the pills will get picked next/again, by symmetry the probability that the half pill will remain at the end (i.e. be the last one to get picked next/again) is 1 / k. If p_k is the k'th last pill broken (so p_n is the first pill), then because at the point p_k is broken there are k - 1 whole pills remaining we get E[N] = sum_k (probability p_k has half a pill remaining at the end) = sum_k (1 / k)