How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$
My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes.
Please give any hints for the first step towards a solution.
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.
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 $\boxed{2}$.
A general way to handle similar problems is to note that
$$\sum_{n=1}^{\infty}nx^n=x\frac{d}{dx}\sum_{n=1}^{\infty}x^n \tag1$$
The sum on the right-hand side of $(1)$ is the series for
$$\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}\tag 2$$
Taking a derivative of the right-hand side of $(2)$ reveals that
$$\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}$$
whereupon evaluation at $x=1/2$ provide the expected result
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=2$$
Notice that $\dfrac{1}{1-x} = \sum_{i=0}^\infty x^i$.
Now differentiate: $\dfrac{1}{(1-x)^2} = \sum_{i=1}^\infty ix^{i-1}$.
You asked for a hint, but there isn't much left to do.
Here's my hint.
This is very similar to the geometric sum $$S=\sum_{n=0}^{\infty} a^n$$ simply try expanding the terms out and analyze the term by term expansion of $$S-S \cdot a$$ and look at the derivation for the geometric sum.
