Find the sum of this series: $\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$

Question

May I know how I should go about finding the sum of this series?

$\displaystyle\sum_{n=1}^\infty$ $\dfrac{n}{2^{n-1}}$

I am really stuck. Thanks!

Answer 1

Hint

Consider the expression $$A=\sum_{n=1}^{\infty}n x^{n-1}$$ You recognize that $A$ is just the derivative with respect to $x$ of $$B=\sum_{n=1}^{\infty} x^{n}$$ which is a geometric progression. So, $$B=\sum_{n=1}^{\infty} x^{n}=\frac{x}{1-x}$$ and its derivative with respect to $x$ is then $$A=\frac{1}{(1-x)^2}$$ Now, replace $x$ by $\frac{1}{2}$.

I am sure that you can take from here.

Answer 2

Here is a way without differentiation.

Let $$S=\sum_{n=1}^\infty \frac{n}{2^{n-1}}=\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb$$ Now, let's see if we can get $\displaystyle\sum_{n=1}^\infty \frac{n-1}{2^{n-1}}$ in terms of $S$ in two different ways.

First way: \begin{align} \sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\sum_{n=1}^\infty \frac{n}{2^{n-1}}-\sum_{n=1}^\infty \frac{1}{2^{n-1}} \\ &=S-\left(1+\frac{1}{2}+\frac{1}{4}+\dotsb\right) \\ &=S-\frac{1}{1-\frac{1}{2}}=S-2 \end{align}

Second way: \begin{align} \sum_{n=1}^\infty \frac{n-1}{2^{n-1}}&=\frac{1}{2}\sum_{n=1}^\infty \frac{n-1}{2^{n-2}} \\ &=\frac{1}{2}\left(0+\frac{1}{1}+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\dotsb\right) \\ &=\frac{1}{2}S \end{align}

Since these must be equal, we have: $$S-2=\frac{1}{2}S$$ Solving, we get $S=4$, which is the answer.

Answer 3

$$s=\sum_{n=1}^\infty \dfrac{n}{2^{n-1}}=1+2/2+3/2^2+4/2^3+5/2^4+....(1)$$ By dividing whole equation by $2,$ $$\dfrac{s}{2}=\dfrac{1}{2}(\sum_{n=1}^\infty \dfrac{n}{2^{n}})=1/2+2/2^2+3/2^3+4/2^4+5/2^5+....(2)$$ Now$(1)-(2)$ gives us, $$\dfrac{s}{2}=1+1/2+1/2^2+1/2^3+1/2^4+....=1+1$$ Hence we can obtain $s=4.$