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Evaluate $$\sum_{n=1}^\infty \frac{1}{n2^{2n}}$$

I'd be glad for guidance, because, frankly, I don't have clue here.

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  • $\begingroup$ Cannot start with $n=0$ $\endgroup$ – marwalix Feb 27 '15 at 16:26
  • $\begingroup$ Does it help any to write it as $\sum\frac1n(\frac14)^n$? Does the series $\sum\frac{x^n}{n}$ look familiar at all? $\endgroup$ – Steven Stadnicki Feb 27 '15 at 16:29
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We could write this just as well as $$ \sum_{n=1}^\infty \frac{1}{n}(1/4)^n $$ Consider the function $$ f(x) = \sum_{n=1}^\infty \frac{1}{n}x^n \quad x \in (-1,1) $$ noting that $f(0) = 0$. We evaluate $$ f'(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x} \quad x \in (-1,1) $$ It follows that $$ f(4) = f(0) + \int_0^{1/4} f'(t)\,dt = \int_0^{1/4} \frac 1{1-t}\,dt $$

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    $\begingroup$ You cannot start with $n=0$ $\endgroup$ – marwalix Feb 27 '15 at 16:33
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$$\sum_{n=1}^\infty \frac{1}{n2^{2n}}=\sum_{n=1}^\infty \frac{(1/4)^n}{n}=-\ln\left(1-\frac14\right)=-\ln\frac34$$


Since, $$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+...$$

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  • $\begingroup$ You are off by a negative and that's because the anti derivative of $\frac{1}{1-x}$ has a negative upfront by the chain rule (+1) $\endgroup$ – imranfat Feb 27 '15 at 16:31
  • $\begingroup$ @imranfat edited $\endgroup$ – RE60K Feb 27 '15 at 16:31
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    $\begingroup$ @AlonAlon If you take the geometric series and integrate (The constant turns out to be zero) and then replace x by 1/4, you get ADG's answer $\endgroup$ – imranfat Feb 27 '15 at 16:32
  • $\begingroup$ @imranfat, following your guidance: $$\sum_{n=0}^\infty \int x^n \, dx = \sum_{n=1}^\infty \frac{x^n}{n}$$. So we have $$\int \frac{1}{1-x} \, dx = -\ln(1-x) $$. Applying $x=4$ we get $-\ln(3/4)$ $\endgroup$ – AlonAlon Feb 27 '15 at 16:54
  • $\begingroup$ Now, I'm not so sure what should be the bounds of those integrals. Can you help me with that? $\endgroup$ – AlonAlon Feb 27 '15 at 16:55
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We have

$$\sum_{n = 1}^\infty \frac{1}{n2^{2n}} = \sum_{n = 1}^\infty \frac{(1/4)^n}{n} = -\log\left(1 - \frac{1}{4}\right) = \log\frac{4}{3}$$

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For any $z$ such that $|z|<1$ we have:

$$\sum_{n=1}^{+\infty}\frac{z^n}{n} = \sum_{n=1}^{+\infty}\int_{0}^{z}x^{n-1}\,dx = \int_{0}^{z}\frac{dx}{1-x} = -\log(1-z) $$ and by plugging in $z=\frac{1}{4}$ it follows that:

$$ \sum_{n\geq 1}\frac{1}{n 2^{2n}}=\color{red}{\log 4-\log 3}. $$

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  • $\begingroup$ You are able to make the interchange because it's a power-series. Right? $\endgroup$ – AlonAlon Feb 27 '15 at 17:44
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    $\begingroup$ @AlonAlon: We can exchange $\sum$ and $\int$ since we have uniform convergence due to the fact that $|z|<1$. $\endgroup$ – Jack D'Aurizio Feb 27 '15 at 17:50
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Think about $\sum_{n=1}^\infty\frac{x^n}{n}=\log{(1-x)}$. The sum of your series is therefore $\log{(4/3)}$

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