A summing problem

Here’s a nice little problem that somebody brought along to the Skills Centre recently. (Thanks to Dr Peter Davidson for the tip-off!)

We define $S_n = \displaystyle\sum_{k=1}^n\dfrac{1}{k}$ for natural numbers $n$. The task is to prove that

$S_{n+1} = 1 + \dfrac{1}{n+1}\displaystyle\sum_{k=1}^nS_k \quad \forall \ n \in \mathbb{N}$.

As so often, there’s an easy way to do this and there’s a hard way. Which is which may not be the same for everybody…

(DP)

1. It is easy to check $(n+1)(S_{n+1}-1) = n(S_{n}-1)+S_n$, then iteration will finish the proof.