How do I evaluate this sum:
$$\sum_{n=1}^\infty{\frac{1}{n2^n}}$$
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2Hi. It would be nice to include your thoughts on the question. – Julien Nov 03 '13 at 15:04
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Hint: Taylor expansion of $-\ln(1-x)$ – Raymond Manzoni Nov 03 '13 at 15:23
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See also http://math.stackexchange.com/questions/1153499/show-ln2-sum-limitsn-1-infty-frac1n2n – Martin Sleziak Aug 30 '16 at 23:10
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Another way is
$$\sum_{n=1}^\infty{\frac{1}{n2^n}}=\sum_{n=1}^\infty{\int_0^{1/2}x^{n-1}\,\mathrm{d}x=\int_0^{1/2}\sum_{n=1}^\infty x^{n-1}\,\mathrm{d}x=\int_0^{1/2}{\frac{1}{1-x}}}\,\mathrm{d}x=\ln{2}$$
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This works in this case, but in general one must take care about the interchange between summation and integration. In particular, a sufficient condition is that the series $S(x) = \sum_{n=0}^\infty f_n(x)$ is uniformly convergent on the domain of integration. (being all $f_n$ continuous on the same region.) – user91126 Nov 03 '13 at 16:21
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Hint: Think of this as the result of plugging $x=\frac{1}{2}$ in to a power series.
Nick Peterson
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