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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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marked as duplicate by Workaholic, Willie Wong, Martin Sleziak, user91500, E. Joseph Nov 18 '16 at 8:06

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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.

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    $\begingroup$ Differentiating an infinite sum seems like the recipe of Coca-Cola. $\endgroup$ – the_candyman Nov 13 '16 at 20:45
  • $\begingroup$ @the_candyman quite a lot is known about swapping sum and differentiation... $\endgroup$ – Gabriel Romon Nov 13 '16 at 20:55
  • $\begingroup$ @LeGrandDODOM it was ironic $\endgroup$ – the_candyman Nov 13 '16 at 20:56
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Hint: $${\displaystyle {\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}\quad {\text{ for }}|x|<1\!}$$

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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!

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  • $\begingroup$ I would have replaced 1/2 by x to get a more general result with no additional effort. $\endgroup$ – marty cohen Nov 13 '16 at 20:51
  • $\begingroup$ @martycohen Good suggestion; edited accordingly. I've also added a second pre-calculus approach. -Mark $\endgroup$ – 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 $S=2$.

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