# Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? [duplicate]

Can you please explain why $$\sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2$$

I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$

• Do you know about power series and differentiation? There is a proof involving $\sum x^i = \frac 1 {1-x}$ Mar 22, 2013 at 14:56

$$\begin{gather*} |x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\ xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2} \end{gather*}$$

Let $$x=\frac{1}{2}$$

• Why the downvote? Mar 22, 2013 at 14:58
• I just downvoted, due to a lack of words or explanation. For example, if you're going to define something, MAKE THAT CLEAR. As in, "Define a function $f : (-1,1) \rightarrow \mathbb{R}$ as follows. $$f(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ Then:..." I shouldn't have to wade through a sea of symbols to understand the big picture of what you are doing. Jan 13, 2015 at 15:49
• $\sum_{n=1}^{\infty} x^n=\frac{x}{1-x}$ is actually wrong and should be $\sum_{n=0}^{\infty} x^n=\frac{x}{1-x}$. Nevertheless the solution is correct. Mar 22 at 20:32
• @ThomasGassmann I don't think so? The first term of the sequence being summed needs to be $x$, not $1$, for us to get $x/1-x$. Oct 2 at 15:28

Since infinite series with nonnegative terms can be rearranged arbitrarily,

$$\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \sum_{j=1}^i \frac{1}{2^i} = \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{2^i} = \sum_{j=1}^\infty \frac{1}{2^{j-1}} = 2$$

More graphically,

  1/2 + 2/4 + 3/8 + 4/16 + ...

= 1/2 + 1/4 + 1/8 + 1/16 + ...    (= 1)
+ 1/4 + 1/8 + 1/16 + ...    (= 1/2)
+ 1/8 + 1/16 + ...    (= 1/4)
+ 1/16 + ...    (= 1/8)
....       ....

• Thanks for this insight (I got so many answers so quickly :) Mar 22, 2013 at 15:16
• If we don't believe it from the picture, we can also show that $\sum_{i=j}^\infty \frac{1}{2^i} = \frac{1}{2^{j-1}}$ by induction. Mar 21, 2021 at 21:16

Let $s = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + \dots$

Then $2s = 1 + 2/2 + 3/4 + 4/8 + 5/16 + \dots$

And then subtracting terms with similar denominators gives: $2s - s = 1 +1/2+1/4+1/8+1/16+\dots = 2$

• I think this proof is the simplest one, however, you need to prove that s is finite to do such subtraction. It can be considered obvious in this task, but that matters :) Nov 9, 2017 at 15:25

Here is a visual proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$ which is the same relation after dividing by $2$ on each side: If it is not clear how, here some areas are labeled (although I prefer the aesthetics of the unlabeled version). The rectangles fill up this $1\times1$ square, and for $i\in\mathbb{N}$, there are $i$ rectangles of area $\frac{1}{2^{n+1}}$. A bit late.. but all the calculus and double sums aren't necessary.

$$\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$

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

Such a sequence is called Arithemtico-Geomteric Progression. $$S_n=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } }$$ $$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i+1 } } }=\sum _{ i=2 }^{ n+1 }{ \frac { i-1 }{ { 2 }^{ i } } }$$ Subtracting $$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { 1 }{ { 2 }^{ i } } } -\frac { n }{ { 2 }^{ n+1 } }$$ as $$n\rightarrow \infty$$ It's easily seen that $$S_{\infty}=2$$

How to evaluate that limit

$$\sum _{ i=1 }^{ n }{ x^{i-1} } =\frac{1-x^n}{1-x}$$ If $$|x|<1$$ as $$n\rightarrow \infty$$ $$\sum _{ i=1 }^{ n }{ x^{i-1} } =\frac{1-x^n}{1-x}=\frac{1}{1-x}$$

Second one is directly from Taylor series.

Although there exist simpler proof , I have a rigorous proof of the second part of the limit

$$0<\log _{ 2 }{ x } =\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t } } dt$$ When $$x>1$$ $$\frac{1}{t}<\frac{1}{\sqrt{t}}$$ is valid

$$\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t } } dt<\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ \sqrt { t } } } dt=2\log _{ 2 }{ e } \left( \sqrt { x } -1 \right) <2\log _{ 2 }{ e }\cdot \sqrt { x }$$

$$0<\log _{ 2 }{ x } <2\log _{ 2 }{ e } \sqrt { x }$$ $$\Rightarrow 0<\frac { \log _{ 2 }{ x } }{ x } <\frac { 2\log _{ 2 }{ e } }{ \sqrt { x } } \tag{1}$$ As $$x\rightarrow \infty$$ using $$(1)$$ and squeeze principle. We get $$\lim_{x\rightarrow \infty}{\frac{\log_{2}{x}}{x}}=0\tag{2}$$ By continuity of $$2^t$$ making the sutbtituion $$x=2^t$$ and as $$x\rightarrow \infty$$ then.$$t\rightarrow \infty$$ Now $$(2)$$ is changed to $$\lim_{t\rightarrow \infty}{\frac{t}{2^t}}=0$$

• we need to prove $\lim_{n\to\infty} \frac n{2^{n+1}}=0,$ right? Mar 22, 2013 at 15:45
• @labbhattacharjee Does this suffice? $$0<\frac n{2^n}<\frac n{n^2}=\frac1n\to0$$ Jul 11, 2017 at 23:49

$$s(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$

Then

$$x \frac{d}{dx} s(x) = \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$$

Your case has $x=1/2$.

Try to show that:

$$\sum_{i=1}^{n}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}$$

You can do this by induction.

Than the step $n\rightarrow \infty$ will be easy.

to see visually what the other answerers wrote, just explode the terms of the sum in this way:

1/2
1/4  1/4
1/8  1/8  1/8
1/16 1/16 1/16 1/16
1/32 1/32 1/32 1/32 1/32
...  ...  ...  ...  ...


Since all elements are positive, you may sum them in the order you want. Start summing each column, and obtain

1    1/2  1/4  1/8  1/16 ...


and now you're done.

It's fairly simple. Look at it this way:

$$\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \frac{\sum_{k=1}^i 1}{2^i} = \sum_{i=1}^\infty \sum_{k=1}^i \frac{1}{2^i}$$ From here, we just change the order of addition. Rather than adding along $k$, and then $i$, we add along $j=i-k$, and then along $k$. This turns our double sum into

$$\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{k=1}^\infty \sum_{j=0}^\infty \frac{1}{2^{j+k}} = \sum_{k=1}^\infty \frac{1}{2^k} \sum_{j=0}^\infty \frac{1}{2^j} = 1\cdot 2 = 2$$

Consider the function $$f(z) = \sum_{i=0}^{\infty}\frac{z^i}{2^i} = \frac z{2-z}.$$ Its radius of convergence is $2$. Its derivative is $$f'(z) = \sum_{i=1}^{\infty}\frac{i}{2^i}z^{i-1} = \frac 2{(2-z)^2}.$$ Since $1<2$, it is allowed to evaluate the derivative at $z=1$ with the written formula and obtain $$\sum_{i=1}^{\infty}\frac{i}{2^i} = f'(1) = \frac 2{(2-1)^2} = 2.$$

the sequence $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+.....$$ is an $arithmetico geometric series$

the sum of an$AG$ series of the form $$S_{\infty}=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$converting the above series inti this form $$\frac{1}{2}( \frac{1}{1} +\frac{2}{2}+ \frac{3}{4}+...... )$$ here$a=1$,$d=1$,$r=\frac{1}{2}$. you get the answer $2$.