A weighted geometric series · Problem 267

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. Regroup the series as a sum of geometric tails: each tail is easy, and the tails themselves form a geometric series.

Step 1: regroup. Writing $\frac{n}{2 ^{n}}$ as $n$ copies of $\frac{1}{2 ^{n}}$ , $n \geq 1 \sum \frac{n}{2 ^{n}} = k \geq 1 \sum n \geq k \sum \frac{1}{2 ^{n}} .$

Step 2: the tails. $\sum_{n \geq k} \frac{1}{2 ^{n}} = \frac{1}{2 ^{k - 1}}$ .

Step 3. $k \geq 1 \sum \frac{1}{2 ^{k - 1}} = 1 + \frac{1}{2} + \frac{1}{4} + \dots = 2.$

Two other routes.

Differentiate $\sum x^{n} = \frac{1}{1 - x}$ : $\sum n x^{n} = \frac{x}{( 1 - x ) ^{2}}$ , which is 2 at $x = \frac{1}{2}$ .
A fair coin is tossed until the first head. The first head comes on toss $n$ with probability $\frac{1}{2 ^{n}}$ , so the series is the expected number of tosses, which is 2.