$∑_{n=1}^\infty$ $n\over2^{n-1}$
or
1 + $2\over2$ + $3\over4$ + $4\over8$ + $5\over16$ + $\ldots$
How to go about evaluating the above, showing that it sums to 4?
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Just like you compute $\int xe^xdx$, integration by parts. – Pp.. Jan 13 '15 at 18:54
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1See this. – David Mitra Jan 13 '15 at 18:56
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Another very similar question is Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2?, which was closed as a duplicate of the question in David Mitra's comment. So this one should be closed as such as well. – epimorphic Jan 13 '15 at 19:48
3 Answers
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For these sorts of problems it is helpful to start with geometric series: $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$ If we differentiate this series we find $$\sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}.$$ This series is valid for all $x \in (-1,1)$.
Joel
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$\sum_{n\geq 1} \frac{n}{2^{n-1}} = f'(1)$ where $f(x) = 2 \sum_{n\geq 0} \frac{x^n}{2^n} = 2 \frac{1}{1 - \frac{x}{2}}$ so that $f'(x) = 2 \frac{\frac{1}{2}}{\left(1-\frac{x}{2}\right)^2}$ and $f'(1) = 4$.
Olórin
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prove this by induction $\sum_{i=1}^m\frac{i}{2^{i-1}}=-4\, \left( 1/2 \right) ^{m+1} \left( m+1 \right) -4\, \left( 1/2 \right) ^{m+1}+4 $
Dr. Sonnhard Graubner
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2yes again or you got the proof it is an exercise for you – Dr. Sonnhard Graubner Jan 13 '15 at 19:03
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