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)

Advertisements
This entry was posted in Puzzles. Bookmark the permalink.

2 Responses to A summing problem

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

  2. Pingback: Solution to the summing problem | Degree of Freedom

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s