Simplification of geometric series.

Question

Can someone please help simplify this series?

$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$

In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$

However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$. I understand if the problem is $\sum_{k=1}^\infty\big(\frac 12\big)^k$. However, it is the $k$ term that I don't understand. The answer is suppose to be $2$.

Can someone help me? Thank you.

Right, but I still would have that k factor. And I still can't simplify to 2. — user10843, Oct 11 '14 at 01:22
I swear. As often as this exact question gets asked, we could almost make a decent drinking game taking a shot for each time. :) — David H, Oct 11 '14 at 01:37
Then I am not sure what type of series it is. But this comes from a problem from my stats. class and I have to simplify it to 2. — user10843, Oct 11 '14 at 01:37
$$\sum^\infty_{k=1}k(0.5)^k=x\frac{{\rm d}}{{\rm d}x}\frac{x}{1-x}\Bigg{|}_{x=0.5}=2$$ — SuperAbound, Oct 11 '14 at 01:58
possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ — Hans Lundmark, Oct 11 '14 at 09:34

score 4 · Answer 1 · answered Oct 11 '14 at 01:57

4

We have the geometric series $$ \sum_{k=0}^\infty x^k=\frac1{1-x} $$ Differentiate to get $$ \sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2} $$ Multiply by $x$, and plug in $x=\frac12$.

answered Oct 11 '14 at 01:57

robjohn

345,667

score 2 · Answer 2 · answered Oct 11 '14 at 02:00

Assume $|x|<1$ and write $$f(x)\equiv\sum_{k=1}^{\infty}kx^k=x\sum_{k=1}^{\infty}kx ^{k-1} = x\left(\sum_{k=1}^{\infty}x ^k\right)' = x\left(\sum_{k=0}^{\infty}x ^k -1\right)'$$ $$=x\left(\frac{1}{1-x}-1\right)' = x\left(\frac{1}{(1-x)^2}\right) $$ $$=\frac{x}{(1-x)^2} $$

In you case, $x=\frac12$, so we have the series $$f(\tfrac12)\equiv\sum_{k=1}^{\infty}k\left(\frac12\right)^k=\frac{\frac12}{(1-\frac12)^2}=\boxed{2} $$

score 2 · Answer 3 · answered Oct 11 '14 at 07:36

\begin{align*} \sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k &= \frac{1}{2} + 2 \frac{1}{2^2} + 3 \frac{1}{2^3} + 4 \frac{1}{2^4} \dots\\ &= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + 2 \frac{1}{2^3} + 3 \frac{1}{2^4} + \dots\right)\\ &= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \right) + \left(\frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^4} + \dots \right) + \dots\\ &= 1 + \frac{1}{2} + \frac{1}{2^2} + \dots\\ &= 2 \end{align*}

Jared · Accepted Answer · 2014-10-11T01:46:12.357

We can try to solve like a geometric series:

$$ S_n = \frac{1}{2} + 2\frac{1}{2^2} + 3\frac{1}{2^3} + \dots + (n - 1)\frac{1}{2^{n - 1}} + n\frac{1}{2^n} \\ \frac{1}{2}S_n = \frac{1}{2^2} + 2\frac{1}{2^3} + 3\frac{1}{2^4} + \dots + (n - 1)\frac{1}{2^{n}} + n\frac{1}{2^{n+1}} \\ S_n - \frac{1}{2}S_n = \frac{1}{2} + (2 - 1)\frac{1}{2^2} + (3 - 2)\frac{1}{2^3} + \dots + (n - (n - 1))\frac{1}{2^n} - n\frac{1}{2^{n+1}} $$

This gives:

$$ S_n - \frac{1}{2}S_n = \left(\sum_{i = 1}^{i = n} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}} = \frac{1}{2}\left(\sum_{i = 0}^{i = n - 1} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}} $$

We know the geometric sum: $\lim_{n\rightarrow\infty}\sum_0^n \frac{1}{2^i} = \frac{1}{1 - \frac{1}{2}} = 2$, which gives:

$$ \lim_{n\rightarrow\infty} S_n = \frac{\frac{1}{2}\cdot 2}{1 - \frac{1}{2}} = 2 $$

Simplification of geometric series.

4 Answers4