What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to?

Question

This problem comes from another equation on another question (this one). I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided.

Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$

But that's not what I wanted. Could anyone help me?

$\sqrt{k}^k \neq k$ so that's where your last line of algebra went wrong — Zach Stone, Jun 14 '15 at 18:03
Possible duplicate of How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? — Hans Lundmark, Apr 14 '18 at 15:07
The mistake is not so much that $\sqrt[k]{k}\neq k$, but the series is NOT a geometric series. Also, notice that the right-hand side of $\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$ is independent of the index $k$. (In other words, $x$ should be fixed and independent of $k$). — Mathematician 42, Apr 14 '18 at 20:56

score 28 · Accepted Answer · edited Feb 15 '23 at 00:53

28

$$S=\sum_{k=0}^\infty{k\over2^k}=2\sum_{k=0}^\infty{k\over2^{k+1}}=2\sum_{k=1}^\infty{k-1\over2^k}=2S-2\sum_{k=1}^\infty{1\over2^k}=2S-2\\$$

Using the ratio test, we show that $S$ converges. Therefore, we are able to rearrange the above result to solve for $S$ which gives us the desired result of $S=2$.

edited Feb 15 '23 at 00:53

FD_bfa

3,989

answered Jun 14 '15 at 18:13

Jack's wasted life

8,849

2

This is only correct if you already know that $S$ exists (i.e. that the sum doesn't diverge). Otherwise, very nice! – Mathmo123 Jun 14 '15 at 18:15
Solid approach! +1 – Mark Viola Jun 14 '15 at 19:00

Anurag A · Answer 2 · 2018-11-08T21:33:54.073

26

Start with: $$\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k.$$ Then take derivative with respect to $x$. $$\frac{1}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k-1}.$$ Multiply by $x$. $$\frac{x}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k}.$$ Now substitute $x=\frac{1}{2}$.

edited Nov 08 '18 at 21:33

answered Jun 14 '15 at 18:05

Anurag A

41,067

am I wrong? if you place $x=1/2$ then the summation will be negative. – kelalaka Nov 08 '18 at 20:33
@kelalaka I had a typo with a negative sign. It's fixed. Thanks for pointing it out. – Anurag A Nov 08 '18 at 21:34
I check that someone edited, and the community accepted. Your original answer was correct. Revering back is also possible. – kelalaka Nov 09 '18 at 08:16
Why did you change the index when you took the derivative of the series? – Burt Feb 04 '20 at 02:00
2

@Burt The initial series starts with $1+x+x^2+\dotsb$ so after taking the derivative we will have $0+1+2x+3x^2+\dotsb$. Since the first term is now $0$ so I changed the index to start with $1$ (the first non-zero term of the derived series). – Anurag A Feb 04 '20 at 04:45
does this only work for 0<x<1? – Xin Niu Mar 25 '22 at 15:48

score 12 · Answer 3 · edited Feb 15 '23 at 00:44

You can also view it this way, which is quite intuitive.

\begin{align*} \sum_{k=0}^\infty \frac{k}{2^k} &= \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\ &= \frac12 + \Bigl(\frac14 + \frac14\Bigr) + \Bigl(\frac18 + \frac18 + \frac18\Bigr) + \Bigl(\frac1{16} + \frac1{16} + \frac1{16} + \frac1{16}\Bigr) + \cdots \\ &= \Bigl(\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac18 + \frac1{16} + \cdots\Bigr) + \cdots \\ &= 1 + \frac12 + \frac14 + \frac18 + \cdots \\ &= 2. \end{align*}

We are able to rearrange the terms of the infinite series in this way because the series is absolutely convergent. It is important that we meet this condition, as without this we would not be able to reorder the terms of the series and group them in this way.

Because the series is absolutely converge it, you can rearrange it however you want. So it’s pretty easy to make this argument rigorous. — Mark Saving, Dec 04 '22 at 23:04

What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to?

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