How do you prove $\sum \frac {n}{2^n} = 2$? [duplicate]

Solution 1:

One approach I find intuitive and avoids derivatives:

$$\sum_{n=1}^{\infty} \frac {n}{2^n} = \sum_{n=1}^{\infty} \frac {1}{2^n} + \sum_{n=2}^{\infty} \frac {1}{2^n} + \sum_{n=3}^{\infty} \frac {1}{2^n} + \cdots $$

Now analyze this.

Solution 2:

Here is a simple way to see this. It ignores technical aspects of rearranging infinite series.

Write down the series as follows: $$ \quad \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots \\ = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad \qquad \quad + \frac{1}{2^3} + \frac{1}{2^4} + \dots \\ \qquad \qquad + \frac{1}{2^4} + \dots \\ \vdots $$ Each row is a geometric series. The values of the rows are $1, \frac{1}{2}, \, \frac{1}{2^2}, \frac{1}{2^3}$, which is again a geometric series. Add this series and you'll get the answer $\boxed{2}$.