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!

1$Li_{2} \left( \frac{1}{2} \right) = \frac{\pi^2}{12}\frac{ (\ln 2)^2}{2}$. – Donald Splutterwit Apr 23 '17 at 19:26

Thank you! But how did you get that? @DonaldSplutterwit – Hendrra Apr 23 '17 at 19:32

I copied it from here ... https://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 ... – Donald Splutterwit Apr 23 '17 at 19:34

Thank you very much :) I'm really looking forward to the solution! – Hendrra Apr 23 '17 at 19:35

1Have a look at https://math.stackexchange.com/a/1056111/44121 – Jack D'Aurizio Apr 23 '17 at 20:10
3 Answers
Note that we have $\int_0^x t^{n1}\,dt=\frac{x^n}{n}$. Then, we can write
$$\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n1}\,dt\int_0^x s^{n1}\,ds\\\\ &=\int_0^x\int_0^x \frac{1}{1st}\,ds\,dt\\\\ &=\int_0^{x} \frac{\log(1sx)}{s}\,ds\\\\ &=\int_0^{x^2} \frac{\log(1s)}{s}\,ds\\\\ &=\text{Li}_2(x^2) \end{align}$$
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
$$\text{Li}_2(1x)=\text{Li}_2\left(1\frac1x\right)\frac12\log^2(x)$$
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
$$\begin{align} \text{Li}_2(1)&=\int_0^{1}\frac{\log(1x)}{x}\,dx\\\\ &=\sum_{n=1}^\infty\frac{(1)^{n1}}{n^2}\\\\ &=\frac{\pi^2}{12} \end{align}$$
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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? – Hendrra Apr 23 '17 at 19:43

You're welcome. My pleasure. The value of that sum when $x=1/\sqrt 2$ is $\sum_{n=1}^\infty \frac{1}{n^2\,2^n}$. – Mark Viola Apr 23 '17 at 19:45

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? – Hendrra Apr 23 '17 at 19:48

Well, yes. We took $x=1/\sqrt 2$ in order that we would have the sum of interest. – Mark Viola Apr 23 '17 at 19:51

Thank you. Your comments were really helpful. I have to ask about one more thing. How did you get $\int_0^x\int_0^x \frac{1}{1st}\,ds\,dt$? – Hendrra Apr 23 '17 at 19:54

You're welcome again. And good question. We sum the geometric series $$\sum_{n=1}^\infty (st)^{n1}=\frac{1}{1st}$$ – Mark Viola Apr 23 '17 at 19:56

Thanks! And the last question  how did you integrate $\int_0^x\int_0^x \frac{1}{1st}\,ds\,dt$? – Hendrra Apr 23 '17 at 20:11

Why does the upper limit of the integrand become squared in the last step of your initial derivation. – Tyberius Apr 23 '17 at 20:19

1@Tyberius The first integral yields $$\int_0^x \int_0^x \frac{1}{1st}\,dt\,ds=\int_0^x \frac{\log(1sx)}{s}\,ds=\int_0^{x^2}\frac{\log(1s)}{s}\,ds=\text{Li}_2(x^2)$$ – Mark Viola Apr 23 '17 at 20:20
Another way to calculate it: consider $$ f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}. $$ Differentiating w.r.t. $x$ gives $$ f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad (xf'(x))'=\sum_{n=1}^{+\infty}x^{n1}=\frac{1}{1x}. $$ Now integrating with $f(0)=0$ $$ xf'(x)=\ln(1x)\quad\Rightarrow\quad f(x)=\int_0^x\frac{\ln(1t)}{t}\,dt. $$ Motivation for termwise differentiation for power series is straightforward.
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My answer to a duplicate question says:
$$\frac{1}{1x}=\sum_{n=0}^{\infty} x^{n}$$ Integrating from 0 to t we get $$\int_{0}^{t}\frac{1}{(1x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$\ln(1t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$ Dividing by t and integrating $$\int_{0}^{0.5}\frac{\ln(1t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$ This on calculating is $$ \dfrac{\pi^2}{12}\dfrac{ln^2(2)}{2},$$
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This answer is identical to your answer to [this question](https://math.stackexchange.com/questions/3238429/evaluatinganinfiniteseriessumn1inftyfrac1n22n). 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. – robjohn May 26 '19 at 01:31

@robjohn So should I change anything ??. I posted it here because the other one was marked as duplicate. Should I remove it from here?? – Mr.HiggsBoson May 26 '19 at 03:12

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](https://math.stackexchange.com/a/3238446) to a duplicate question says..." – robjohn May 26 '19 at 05:28
