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

Question

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.

Answer 1

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

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

Answer 2

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

Answer 3

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 $\frac12f'(\frac12)$.

Answer 4

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

Answer 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.

Answer 6

The method that says $$ \sum_n nx^n = x \sum_n \frac{d}{dx} x^n,\text{ etc, etc.,} $$ has already been mentioned.

Here's another way: $$ \begin{array}{rrrrrrrrrrrrrrrr} 1/2 \\[8pt] {}+ 1/4 & {}+ 1/4 \\[8pt] {}+ 1/8 & {}+1/8 & {}+1/8 \\[8pt] {}+\cdots{} \end{array} $$

Now find the sum of the entries in the first column. Then the second column. Then the third. And so on.

After that, find the sum of those sums.

Answer 7

Think about, in general,

$$S = 1+2 r+3 r^2 + 4 r^3 + \cdots$$

$$r S = r + 2 r^2 + 3 r^3 + \cdots$$

$$S - r S = (1-r) S = 1 + r + r^2 + r^3 + \cdots = \frac{1}{1-r}$$

Therefore

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

This, however, is not really the series you listed; that series has an extra factor of $r$; so the sum is actually $r/(1-r)^2$. Plug in $r=1/2$ and the sum is $2$.

Answer 8

Toss a fair coin until you get a head. Let $X$ be the number of tosses. Your sum is $E(X)$. Write $a$ for $E(X)$.

With probability $\frac{1}{2}$ you get a head on the first toss. Given this happened, $E(X)=1$.

With probability $\frac{1}{2}$, you get a tail on the first toss. Given this happened, $E(X)=1+a$.

Thus $$a=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot a.$$ Solve for $a$.

Answer 9

You know that $\sum_{k=0}^\infty \frac{1}{2^k}=2$ from the geometric series formula. Write \begin{align*} 1&= \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \\ \frac{1}{2}&=0 +\frac{1}{2^2}+\frac{1}{2^3}+\cdots \\ \frac{1}{4}&=0+0+\frac{1}{2^3}+\cdots \\ \vdots&= \qquad\ddots \end{align*}

Summing the total of each column on the right and using the known fact that the left hand side is $2$, you get your desired answer without any other tricks

Answer 10

Lets call your series $\Sigma$.

Lets seperate the $\Sigma$ into a sub series we'll call ${1 \over 2} \Sigma$.

${1 \over 2}\Sigma$ is a sub series of $\Sigma$, right?

Lets subtract ${1 \over 2} \Sigma$ from $\Sigma$. Will call the result as $S_{1 \over 2}$.

Well, $S_{1 \over 2} = {1 \over 2}\Sigma - \Sigma:$

$$\Sigma = {1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \dots$$

as $${1 \over 2} \Sigma = {1 \over 4} + {2 \over 8} + {3 \over 16} + {4 \over 32} + \dots $$

We've devided and subtracted half series from the original series, $\Sigma$,and got $S_{1 \over 2}$,

Well, $\Sigma$ is $2 {S_{1 \over 2}}$. (Why?)

We can see that ${S_{1 \over 2}}$ = $\Sigma$ - ${1 \over 2}\Sigma$ $=>$

$${1 \over 2} + {2 \over 4} + {3 \over 8} + {4 \over 16} + \dots$$ $$ - $$ $${1 \over 4} + {2 \over 8} + {3 \over 16} + {4 \over 32} + \dots $$

$$ = $$

$$ {1 \over 2} - {1 \over 4} + {2 \over 4} - {2 \over 8} + {3 \over 8} - {3 \over 16} + {4 \over 16} - {4 \over 18} \dots$$

which is:

$$ {1 \over 2} + {1 \over 4} + {1 \over 8} + {1 \over 16} + \dots$$

Can you notice what happened? We got a Geometric series.

Hence, we can easly get the convertange limit of $S_{1 \over 2}$: ${a_1} \over {1 - q}$ , in our case:

$$S_{1\over2} = {{1 \over 2} \over {1 - {1 \over 2}}} = {{1 \over 2} \over {1 \over 2}} = 1$$

Wait! We've devided by 2. Well, it's time to turn back :) multiply $S_{1 \over 2}$ by the inverse of ${1 \over 2}$, which is 2.

Well, $\Sigma = 2S_{1\over2} = 2*1 = 2$.

Done!