Cliff Pickover twitted a fun puzzle: Which series is bigger?

\[\sum_{n=0}^\infty \frac{1}{2^{n+1}} \quad \text{or} \quad \sum_{n=0}^\infty \frac{n}{2^{n+1}}\]

The first one is the famous geometric series which sum is equal to 1. The second one seems to be bigger because 1 < n, except for the 0th term, but that 0th term makes a big difference.

First of all, the second series converges. You can prove it by D’Alembert’s theorem

\[\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+2}}\cdot\frac{2^{n+1}}{n} = \frac{1}{2}\left(1 + \frac{1}{n}\right) \le \frac{3}{4} < 1, \text{for} \; n \ge 2\]

Because the terms are positive, the series converges absolutely, therefore we can rearrange the terms.

\[\begin{align*} S & = \sum_{n=0}^\infty \frac{n}{2^{n+1}} = \sum_{n=0}^\infty \left(\frac{1}{2^{n+1}} + \frac{n-1}{2^{n+1}}\right) = \sum_{n=0}^\infty \frac{1}{2^{n+1}} + \sum_{n=0}^\infty \frac{n-1}{2^{n+1}} \\ & = 1 + \frac{1}{2} \sum_{n=0}^\infty \frac{n-1}{2^n} = 1 + \frac{1}{2} \sum_{n=-1}^\infty \frac{n}{2^{n+1}} = 1 - \frac{1}{2} + \frac{1}{2} \sum_{n=0}^\infty \frac{n}{2^{n+1}} \\ & = \frac{1}{2} + \frac{1}{2}S \implies S = 1 \end{align*}\]

Both series are equal.