# A simple series $\sum_{i=1}^\infty \frac{i}{2^i} = 2$ [duplicate]

I don't do math a long time, so I completely don't remember how to prove that: $$\sum_{i=1}^\infty \frac{i}{2^i} = 2$$

Can anybody help me?

• Didn't you mean to write $\sum_{i=0}^\infty 2^{-i}$? – tomasz Feb 12 '14 at 20:02
• Decent guess tomasz, but the summation in the question also equals 2. – mlg4080 Feb 12 '14 at 20:04

$S = x + 2x^2 + 3x^3 + \ldots$

It can be written as

$\Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots$

$\Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots$

$\Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. )$

$\Rightarrow S = \frac{1}{1-x}\times\frac{x}{1-x}$

$\Rightarrow S = \frac{x}{(1-x)^2}$

Put $x =0.5$ you will get the answer

The series $$\sum_{i=1}^\infty x^i=\frac{x}{1-x}$$ is the geometric series and we can differentiate it term by term since it's a power series so we have $$\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1}{(1-x)^2}=\sum_{i=1}^\infty ix^{i-1}$$ so multiplying by $x$ gives $$\sum_{i=1}^\infty i x^{i}=\frac{x}{(1-x)^2}$$ and set $x=\frac 1 2$.

• One little mistake. The Summation has to start at $i=0$ in the geometric series. (But later we have $\sum_{i=0}^\infty ix^{i-1} = \sum_{i=1}^\infty ix^{i-1}$) – user127.0.0.1 Feb 12 '14 at 20:09
• Summation in the second line requires knowledge of the binomial coefficient of teh form $\binom{-2}{k}$ which may be hard for the OP, who doesn't do math for a long time – Alex Feb 12 '14 at 20:30

The standard procedure to solve this kind of series is to define a power series like $$f(x)=\sum_{i=1}^\infty\frac{i}{2^i}x^i$$ so what we want to figure out is $f(1)$.

Then we manipulate the terms using differentiation. Since $$x\frac{d}{dx}(x^i)=xix^{i-1}=ix^i$$ you have $$f(x)=\sum_{i=1}^\infty\frac{1}{2^i}x\frac{d}{dx}(x^i)$$ And by linearity of $x\frac{d}{dx}$, $$f(x)=x\frac{d}{dx}\left(\sum_{i=1}^\infty\frac{1}{2^i}x^i\right)$$ Now using a property of exponents, $$f(x)=x\frac{d}{dx}\left(\sum_{i=1}^\infty\left(\frac{x}{2}\right)^i\right)$$ We managed to get a power series out of this, which is great because $$\sum_{i=1}^\infty\left(\frac{x}{2}\right)^i=\frac{1}{1-\frac{x}{2}}-1=\frac{2}{2-x}-1$$ So $$f(x)=x\frac{d}{dx}\left(\frac{2}{2-x}-1\right)$$ $$f(x)=-\left(-x\frac{2}{(2-x)^2}\right)=\frac{2x}{(2-x)^2}$$ Finally, we evaluate at $x=1$, $$f(1)=\frac{2}{(2-1)^2}=2$$