Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2? [duplicate]

Apparently,

$$ \sum_{n = 0}^\infty \frac{n}{2^n} $$

converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term.

Solution 1:

Besides the differentiation trick mentioned by others, here's another trick:

$$S = \sum_{n=0}^{\infty} \frac{n}{2^n} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{n}{2^{n-1}} = \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{n - 1}{2^{n-1}} + \sum_{n=0}^{\infty} \frac{1}{2^{n-1}}\right) = \frac{1}{2} \left(S + \frac{-1}{2^{-1}} + 4\right) = \frac{1}{2}(S + 2).$$

Solution 2:

$$\begin{array}{} \sum_{n\ge 0}\frac{n}{2^n}&=&\frac1{2^1}&+&\frac2{2^2}&+&\frac3{2^3}&+&\frac4{2^4}&+&\ldots&=\\ \hline &&\frac1{2^1}&+&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 1}\left(\frac12\right)^n=1\\ &&&&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 2}\left(\frac12\right)^n=\frac12\\ &&&&&&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 3}\left(\frac12\right)^n=\frac14\\ &&&&&&&&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 4}\left(\frac12\right)^n=\frac18\\ &&&&&&&&&&\ddots&\vdots&\qquad\vdots\\ &&&&&&&&&&&&\color{blue}{\sum_{n\ge 0}\frac1{2^n}=2} \end{array}$$

Solution 3:

Hint: $$\begin{align} \frac12+\frac14+\frac18+\frac1{16}+\dots&=\color{red}{1}\\ \frac14+\frac18+\frac1{16}+\dots&=\color{red}{\frac12}\\ \frac18+\frac1{16}+\dots&=\color{red}{\frac14}\\ \frac1{16}+\dots&=\color{red}{\frac18}\\ \hline \frac12+\frac24+\frac38+\frac4{16}+\dots&=2 \end{align}$$

Solution 4:

This is the derivative of a geometric series: Let $$f(x)=\sum_{n=0}^\infty x^n.$$ Then (by taking derivative summandwise) $$f'(x) = \sum_{n=1}^\infty nx^{n-1}.$$ Since $f(x)=\frac1{1-x}$ if $|x|<1$, we have $f'(x)=\frac1{(1-x)^2}$. Your sum is just $\frac12f'(\frac12)$.

Solution 5:

Consider the power series $$f(x)=\sum_{n=0}^\infty\frac{x^n}{2^n}=\sum_{n=0}\left(\frac x2\right)^n=\sum_{n=0}^\infty t^n=\frac{1}{1-t}=\frac{1}{1-\frac x2}=\frac{2}{2-x}$$ Then we have that $f(1)=2$. We also have that $$f'(x)=\sum_{n=0}^\infty\frac{nx^{n-1}}{2^n}$$

But we also have that $$\left(\frac2{2-x}\right)'=\frac2{(2-x)^2}$$

Therefore $f'(1)=2$ as wanted.