Question

Evaluate the sum $\sum_{r=1}^{\infty} \frac{(2r-1)}{2^r}$

Answer 1

3

Answer 2

The sum is an arithmetico-geometric series. Let $S = \sum_{r=1}^{\infty} \frac{(2r-1)}{2^r}$.

Write out the series: $S = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \dots$.

Multiply by the common ratio $\frac{1}{2}$: $\frac{1}{2}S = \frac{1}{4} + \frac{3}{8} + \frac{5}{16} + \dots$.

Subtract the second equation from the first: $\frac{1}{2}S = \frac{1}{2} + \left(\frac{3}{4} - \frac{1}{4}\right) + \left(\frac{5}{8} - \frac{3}{8}\right) + \dots$

This simplifies to $\frac{1}{2}S = \frac{1}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \dots = \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$.

The terms $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ form an infinite geometric series with first term $a=\frac{1}{2}$ and common ratio $r=\frac{1}{2}$. Its sum is $\frac{a}{1-r} = \frac{1/2}{1-1/2} = 1$.

So, $\frac{1}{2}S = \frac{1}{2} + 1 = \frac{3}{2}$.

Multiplying by 2, we get $S = 3$.