How do you prove the following limit?
$$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{k}{2^k}\right)=2$$
Do you need any theorems to prove it?
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1The limit is unnecessary as what is inside of the parenthesis does not depend on $n$ at all. It is like saying $\lim\limits_{x\to\infty}(52)$... What would make more sense is to write $\lim\limits_{b\to\infty}\left(\sum\limits_{n=1}^b\frac{n}{2^n}\right)$ – JMoravitz Nov 13 '16 at 20:13
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3See this. – David Mitra Nov 13 '16 at 20:18
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$?, Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2?, What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to?. If you search in approach0 you can find more questions about the same sum. – Martin Sleziak Nov 18 '16 at 06:11
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We may start with the standard finite evaluation: $$ 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1 $$ Then by differentiating $(1)$ we have $$ 1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2 $$ by multiplying by $x$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, we get
$$ \sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}. \tag3 $$ Then put $x:=\dfrac12$.
Edit. One may observe we have avoided differentiating an infinite sum.
Olivier Oloa
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1Differentiating an infinite sum seems like the recipe of Coca-Cola. – the_candyman Nov 13 '16 at 20:45
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@the_candyman quite a lot is known about swapping sum and differentiation... – Gabriel Romon Nov 13 '16 at 20:55
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Hint: $${\displaystyle {\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}\quad {\text{ for }}|x|<1\!}$$
E.H.E
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I thought it might be useful to present two ways forward that rely on elementary pre-calculus knowledge only. To that end, we proceed.
METHODOLOGY 1:
Note that we can write $n=\sum_{m=1}^n (1)$. Therefore,
$$\begin{align} \sum_{n=1}^\infty nx^n&=\sum_{n=1}^\infty \sum_{m=1}^n(1) \,x^n\\\\ &=\sum_{m=1}^\infty \sum_{n=m}^\infty x^n\\\\ &=\sum_{m=1}^\infty \frac{x^m}{1-x}\\\\ &=\frac{x}{(1-x)^2} \end{align}$$
METHODOLOGY 2:
Let $S=\sum_{n=1}^\infty nx^n$. Note that we can write
$$\begin{align} x S&=\sum_{n=1}^\infty nx^{n+1}\\\\ &=\color{blue}{\sum_{n=1}^\infty (n+1)x^{n+1}}-\color{red}{\sum_{n=1}^\infty x^{n+1}}\\\\ &=\color{blue}{S-x}-\color{red}{\frac{x^2}{1-x}}\\\\ (1-x)S&=x+\frac{x^2}{1-x}\\\\ S&=\frac{x}{(1-x)^2} \end{align}$$
as expected!
Mark Viola
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I would have replaced 1/2 by x to get a more general result with no additional effort. – marty cohen Nov 13 '16 at 20:51
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@martycohen Good suggestion; edited accordingly. I've also added a second pre-calculus approach. -Mark – Mark Viola Nov 13 '16 at 21:03
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The series is obviously convergent. Set $S:=\sum_{n=1}^\infty \frac{n}{2^n}$. This gives $$S=\sum_{n=1}^\infty \frac{n}{2^n}= \sum_{n=1}^\infty \frac{n-1}{2^{n-1}}+\sum_{n=1}^\infty \frac{1}{2^{n-1}} =\frac{1}{2} \sum_{n=1}^{\infty}\frac{n}{2^n}+ \sum_{n=1}^{\infty}\frac{1}{2^n}=\frac{S}{2} + \frac{1}{2}\ \frac{1}{1-\frac{1}{2}}=\frac{S}{2}+1$$ and therefore $$S=\frac{S}{2}+1$$ that is $S=2$.