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My Work

I felt the best way to go about this problem was to compare it to a well known MacLaurin series. I noticed it resembled the reciprocal of the absolute value of the MacLaurin series of $\ln(1+x)$ where $x = \dfrac12$ but I had the problem of the $n$ in the numerator still. None of the well known series have an $n$ in the numerator.

The Well Known MacLaurin Series

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My Question

Can someone give me a hint as to which well known MacLaurin series this resembles? Once I have that I know the solution!

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If you want to directly proceed via MacLaurin series, try the MacLaurin series of $\dfrac{x^2}{(1-x)^2}$, which unfortunately is not there in your list. Else, proceed as follows: Let $S_m = \displaystyle \sum_{n=1}^m \dfrac{n}{2^{n+1}}$. \begin{align} S_m & = \dfrac14 + \dfrac28 + \dfrac3{16} + \dfrac4{32} + \cdots + \dfrac{m-1}{2^m} + \dfrac{m}{2^{m+1}} &\spadesuit\\ \dfrac{S_m}2 & = \,\,\,\,\,\,\,\,\,\,\,\dfrac18 + \dfrac2{16} + \dfrac3{32} + \cdots + \dfrac{m-2}{2^m} + \dfrac{m-1}{2^{m+1}} + \dfrac{m}{2^{m+2}} & \diamondsuit \end{align} $\spadesuit-\diamondsuit$ now gives us $$\dfrac{S_m}2 = \dfrac14 + \dfrac18 + \cdots + \dfrac1{2^{m+1}} - \dfrac{m}{2^{m+2}}$$ I trust you can conclude from this making use of the first MacLaurin series.

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  • $\begingroup$ Would the second solution work if the series is infinite (Which it is in my question)? We wouldn't have a last term in that case would we. $\endgroup$ – Dunka Nov 21 '14 at 1:31
  • $\begingroup$ It works if the series is infinite because that "last term" vanishes as $m$ goes to infinity. $\endgroup$ – Milo Brandt Nov 21 '14 at 1:47
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The following is valid since $|1/2|<1$, and uses a series you have on your list. $$\begin{align} \sum_{n=0}^\infty n\left(\frac12\right)^{n+1} &=\left.\sum_{n=0}^\infty nx^{n+1}\right|_{x=1/2}\\ &=x^2\left.\sum_{n=0}^\infty nx^{n-1}\right|_{x=1/2}\\ &=x^2\left.\sum_{n=0}^\infty \frac{d}{dx}x^{n}\right|_{x=1/2}\\ &=x^2\left.\frac{d}{dx}\sum_{n=0}^\infty x^{n}\right|_{x=1/2}\\ &=x^2\left.\frac{d}{dx}\frac{1}{1-x}\right|_{x=1/2}\\ &=x^2\left.\frac{1}{(1-x)^2}\right|_{x=1/2}\\ &=\left.\frac{x^2}{(1-x)^2}\right|_{x=1/2}\\ &=\left.\frac{(1/2)^2}{(1-1/2)^2}\right|_{x=1/2}\\ &=\left.\frac{(1/2)^2}{(1/2)^2}\right|_{x=1/2}\\ &=1 \end{align}$$

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See my answer here for a visual explanation.

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  • $\begingroup$ Nice! I never saw that before. $\endgroup$ – Akiva Weinberger Nov 21 '14 at 2:13
  • $\begingroup$ Very nice graphical proof! $\endgroup$ – hypergeometric Nov 21 '14 at 2:39

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