How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $

Question

I have put his on wolfram and obtained answer as follows:

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

And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$

However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?

Answer 1

Let $S$ be our sum. Then $S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$. Thus $$\begin{align}2S&=1+&\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\cdots\\ S&=&\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots\end{align} $$ Subtract. We get $$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$

Answer 2

Hint:

$$ \sum_{k=1}^{\infty} x^k = \frac{x}{1-x} \implies \sum_{k=1}^{\infty} kx^k=\frac{d}{dx} \frac{x}{1-x} . $$

Answer 3

Another way to do the problem is to note that $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 = \sum_{k=1}^\infty \frac{k-1}{2^k}$$ so that the desired series is really just $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 + \sum_{k=1}^\infty \frac{1}{2^k}$$

Answer 4

If $\displaystyle S=\sum_{n=1}^\infty\frac{n}{2^n}$, then $$S-\frac12=\sum_{n=1}^\infty\frac{n+1}{2^{n+1}}=\frac12\sum_{n=1}^\infty\frac{n+1}{2^n}=\frac12\left(S+\sum_{n=1}^\infty\frac1{2^n}\right)=\frac12(S+1),$$ so $2S-1=S+1$, or $S=2$.

All that remains is to justify that the series converges. But $n<1.1^n$ for $n$ large enough, so a tail of $S$ is bounded above by a tail of the convergent geometric series $\sum_n\left(\frac{1.1}2\right)^n$.