Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2? 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.
 A: Hint:
$$\begin{align}
\frac12+\frac14+\frac18+\frac1{16}+\dots&=\color{red}{1}\\
\frac14+\frac18+\frac1{16}+\dots&=\color{red}{\frac12}\\
\frac18+\frac1{16}+\dots&=\color{red}{\frac14}\\
\frac1{16}+\dots&=\color{red}{\frac18}\\ \hline
\frac12+\frac24+\frac38+\frac4{16}+\dots&=2
\end{align}$$
A: 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 $\frac12f'(\frac12)$.
A: 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.
A: 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.
A: 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 $2$.
A: 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 $a$.
A: Besides the differentiation trick mentioned by others, here's another trick:
$$S = \sum_{n=0}^{\infty} \frac{n}{2^n} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{n}{2^{n-1}} = \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{n - 1}{2^{n-1}} + \sum_{n=0}^{\infty} \frac{1}{2^{n-1}}\right) = \frac{1}{2} \left(S + \frac{-1}{2^{-1}} + 4\right) = \frac{1}{2}(S + 2).$$
A: $$\begin{array}{}
\sum_{n\ge 0}\frac{n}{2^n}&=&\frac1{2^1}&+&\frac2{2^2}&+&\frac3{2^3}&+&\frac4{2^4}&+&\ldots&=\\ \hline
&&\frac1{2^1}&+&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 1}\left(\frac12\right)^n=1\\
&&&&\frac1{2^2}&+&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 2}\left(\frac12\right)^n=\frac12\\
&&&&&&\frac1{2^3}&+&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 3}\left(\frac12\right)^n=\frac14\\
&&&&&&&&\frac1{2^4}&+&\ldots&=&\sum_{n\ge 4}\left(\frac12\right)^n=\frac18\\
&&&&&&&&&&\ddots&\vdots&\qquad\vdots\\
&&&&&&&&&&&&\color{blue}{\sum_{n\ge 0}\frac1{2^n}=2}
\end{array}$$
A: 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
A: 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! 
