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There is an infinite sum given: $$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$$ It should be solved using integration, derivation or both. I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!

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    $\begingroup$ $Li_{2} \left( \frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{ (\ln 2)^2}{2}$. $\endgroup$ Apr 23, 2017 at 19:26
  • $\begingroup$ Thank you! But how did you get that? @DonaldSplutterwit $\endgroup$
    – Hendrra
    Apr 23, 2017 at 19:32
  • $\begingroup$ I copied it from here ... en.wikipedia.org/wiki/Spence%27s_function ... I am scribbling in my note book now ... I will get back to you when I have managed to derive it ... $\endgroup$ Apr 23, 2017 at 19:34
  • $\begingroup$ Thank you very much :) I'm really looking forward to the solution! $\endgroup$
    – Hendrra
    Apr 23, 2017 at 19:35
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    $\begingroup$ Have a look at math.stackexchange.com/a/1056111/44121 $\endgroup$ Apr 23, 2017 at 20:10

3 Answers 3

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Note that we have $\int_0^x t^{n-1}\,dt=\frac{x^n}{n}$. Then, we can write

$$\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n-1}\,dt\int_0^x s^{n-1}\,ds\\\\ &=\int_0^x\int_0^x \frac{1}{1-st}\,ds\,dt\\\\ &=-\int_0^{x} \frac{\log(1-sx)}{s}\,ds\\\\ &=-\int_0^{x^2} \frac{\log(1-s)}{s}\,ds\\\\ &=\text{Li}_2(x^2) \end{align}$$

Evaluating at $x=1/\sqrt{2}$ yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^2\,2^n}=\text{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)}$$

And we are done!


To evaluate $\text{Li}_2(1/2)$, we exploit the relationship

$$\text{Li}_2(1-x)=-\text{Li}_2\left(1-\frac1x\right)-\frac12\log^2(x)$$

Letting $x=1/2$ yields

$$\text{Li}_2(1/2)=-\text{Li}_2\left(-1\right)-\frac12\log^2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$

where we used

$$\begin{align} \text{Li}_2(-1)&=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\\\\ &=\frac{\pi^2}{12} \end{align}$$

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  • $\begingroup$ That's a nice solution! Thanks. However I must ask about $\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2}$. Why are we considering such a sum? $\endgroup$
    – Hendrra
    Apr 23, 2017 at 19:43
  • $\begingroup$ You're welcome. My pleasure. The value of that sum when $x=1/\sqrt 2$ is $\sum_{n=1}^\infty \frac{1}{n^2\,2^n}$. $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 19:45
  • $\begingroup$ The value truly is $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$! Thus I have the next question. Why did you decided to take an $x = \frac{1}{\sqrt{2}}$? Just because it works? $\endgroup$
    – Hendrra
    Apr 23, 2017 at 19:48
  • $\begingroup$ Well, yes. We took $x=1/\sqrt 2$ in order that we would have the sum of interest. $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 19:51
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    $\begingroup$ @Tyberius The first integral yields $$\int_0^x \int_0^x \frac{1}{1-st}\,dt\,ds=-\int_0^x \frac{\log(1-sx)}{s}\,ds=-\int_0^{x^2}\frac{\log(1-s)}{s}\,ds=\text{Li}_2(x^2)$$ $\endgroup$
    – Mark Viola
    Apr 23, 2017 at 20:20
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Another way to calculate it: consider $$ f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}. $$ Differentiating w.r.t. $x$ gives $$ f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad (xf'(x))'=\sum_{n=1}^{+\infty}x^{n-1}=\frac{1}{1-x}. $$ Now integrating with $f(0)=0$ $$ xf'(x)=-\ln(1-x)\quad\Rightarrow\quad f(x)=-\int_0^x\frac{\ln(1-t)}{t}\,dt. $$ Motivation for termwise differentiation for power series is straightforward.

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My answer to a duplicate question says:

$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$ Integrating from 0 to t we get $$\int_{0}^{t}\frac{1}{(1-x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$ Dividing by t and integrating $$\int_{0}^{0.5}-\frac{\ln(1-t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n-1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$ This on calculating is $$ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$

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  • $\begingroup$ This answer is identical to your answer to this question. If the questions are the same, this one should be flagged as a duplicate. If not, then it would be more efficient and less noisy to cite or quote (with attribution) from the other answer. The link between the answers would also serve as a link between two closely related questions. In this case, the questions were duplicates. $\endgroup$
    – robjohn
    May 26, 2019 at 1:31
  • $\begingroup$ @robjohn So should I change anything ??. I posted it here because the other one was marked as duplicate. Should I remove it from here?? $\endgroup$ May 26, 2019 at 3:12
  • $\begingroup$ Even though the other question was closed, your answer there is still active and getting votes. One should be removed, but you might edit one to reference the other; you might just say something like, "my answer to a duplicate question says..." $\endgroup$
    – robjohn
    May 26, 2019 at 5:28
  • $\begingroup$ Okay , I will link my answers $\endgroup$ May 26, 2019 at 9:55

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