How can I prove the following equality; $$\sum_{n=1}^{\infty}\frac{1}{n2^n}=\ln 2\,.$$
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2$$- \ln(1-x) = \sum_{n \ge 1} \frac{x^{n}}{n}$$ with a particular value of $x$. – Matthew Cassell Nov 30 '16 at 01:26
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Notice that
$$\frac1{1-r}=\sum_{n=0}^\infty r^n$$
Integrate both sides with respect to $r$ from $0$ to $x$ to get that
$$-\ln(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty\frac{x^n}n$$
Now proceed to use $x=1/2$.
Simply Beautiful Art
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1The right idea. In an exam I would insist that you justify, why it is ok to integrate this infinite series term-by-term. – Jyrki Lahtonen Nov 30 '16 at 05:39
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Because $\sum_{n=0}^{k}r^k$ converge uniformly to $\frac{1}{1-r}$ in $[-a,a]$ for $|a|<1$ as $k\rightarrow\infty,.$ – David Dec 05 '16 at 08:35
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@David I think you ought to check that? I think we need DCT here. – Simply Beautiful Art Dec 05 '16 at 14:25