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Question as stated:

Is there a closed form solution to the series $\sum\limits_{k = 0}^n \dfrac{k}{2^k}$?

I am guessing that this series is expected to converge, since $2^k$ dominates $k$, but I am not certain.

Mikasa
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Norman
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  • Think about the derivative of a finite geometric series. – Gabriel Romon Sep 18 '17 at 04:51
  • See the accepted answer here: https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn – Hans Lundmark Sep 18 '17 at 05:14
  • Just very briefly: $$\because \forall p, \ \frac{0}{p} = 0$$ $$\implies \sum_{k = 0}^{n}\frac{k}{2^k} = \sum_{k = 1}^{n}\frac{k}{2^k} = 1 + \sum_{k = 3}^{n}\frac{k}{2^k}$$ And we can work out from here that: $$\sum_{k = 3}^{n}\frac{k}{2^k} = 1 \iff n = \infty$$ meaning... – Mr Pie Sep 18 '17 at 06:35

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We can write, $$S_n=\color{blue}{\frac{1}{2}}+\color{red}{\frac{2}{2^2}}+\frac{3}{2^3}+\cdots+\frac{n-1}{2^{n-1}}\frac{n}{2^n}$$ $$\frac{S_n}{2}=\color{red}{\frac{1}{2^2}}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots+\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}$$ $$S_n-\frac{S_n}{2}=\color{blue}{\frac{1}{2}}+\color{red}{\frac{1}{2^2}}+\cdots+\frac{1}{2^n}-\frac{n}{2^{n+1}}$$ $$S_n=2\Biggr(\frac{1}{2}\frac{\Big(\frac{1}{2}\Big)^n-1}{\frac{1}{2}-1}-\frac{n}{2^n}\Biggr)$$ $$S_n=2\Biggr(-\Big(\frac{1}{2}\Big)^n+1-\frac{n}{2^n}\Biggr)$$ $$S_n=2+\frac{n-2}{2^{n}}$$

neonpokharkar
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Let $S(x) = \sum\limits_{I=1}^n x^k$ , convergent for $|x| \lt 1$, geometric series.

Differentiate : $S'(x) = \sum\limits_{I=1}^n k x^{k-1}$.

$xS'(x) = \sum\limits_{I=1}^n kx^k$.

Cf.https://en.m.wikipedia.org/wiki/Geometric_series

Peter Szilas
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hint: relate the sum in question with the sum: $\displaystyle \sum_{k = 0}^n kx^k$ and $x = 1/2$ in your example. This near "geometric sum" is very popular and can be calculated in various ways.

DeepSea
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