4
$\begingroup$

This question already has an answer here:

I have been trying to find the sum of the series $$\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$ but I couldn't find any methods (such as a fourier series) that seem to get me anywhere.

WolframAlpha gave $ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$ but how would one get to this?

$\endgroup$

marked as duplicate by Guacho Perez, E.H.E, Parcly Taxel, max_zorn, mihaild May 24 at 22:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

I think a better way to go about this is as follows. Put $g(x)=-\int_0^x\frac{\ln(1-t)}t\text dt$. As you know from the other answers, you are trying to compute $\frac 12g(1/2)$. Now prove that $g(x)+g(1-x)=g(1)-\ln(x)\ln(1-x)$ using the definition of $g$, a substitution and integration by parts. Then expand the Taylor series for the integrand of $g(1)$ and integrate term by term to get $g(1)=\sum_{n\ge1}n^{-2}=\frac{\pi^2}6$ (this is called the Basel Problem). Finally, plug in $x=1/2$ and simplify.

$\endgroup$
5
$\begingroup$

Hint Let $f(x)=\sum_{n=1}^\infty \frac{1}{n^2}x^n$. Then $$f'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n-1} \\ xf'(x)=\sum_{n=1}^\infty \frac{1}{n}x^{n}\\ f'(x)+xf''(x)=\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$$

For simplicity let $y=f'(x)$. Then you need to solve: $$y+xy'=\frac{1}{1-x}$$ or equivalently $$(xy)'=\frac{1}{1-x}$$

You can find $y=f'$ and hence $f$ from here. Plug in $x=\frac{1}{2}$.

$\endgroup$
  • $\begingroup$ Please forgive me, because this is just a special value of the dilogarithm - see my answer. $\endgroup$ – Parcly Taxel May 24 at 17:03
  • $\begingroup$ @ParclyTaxel The issue is that using the dilogarithm is an overkill, Calculus students don't know what that is... $\endgroup$ – N. S. May 24 at 17:50
  • 2
    $\begingroup$ @ParclyTaxel If you change $2^n$ to $3^n$ the same works, you just plugin $x=\frac{1}{3}$ at the end.... Besides, if my students would write that the answer is the dilogarithm function, WITHOUT details, I would not give them points for the solution. How is that different than plugging it into WolframAlpha ? Besides check the edit for the ODE part ;) $\endgroup$ – N. S. May 24 at 18:05
  • 1
    $\begingroup$ @ParclyTaxel As for the link you added, I fail to see the relevance to this post... $\endgroup$ – N. S. May 24 at 18:08
  • 2
    $\begingroup$ But evaluating the dilogarithm is exactly as hard as computing the sum. $\endgroup$ – Guacho Perez May 24 at 18:19
1
$\begingroup$

This sum is in fact the series representation of $\operatorname{Li}_2\left(\frac12\right)$, which has a known closed form.

$\endgroup$
1
$\begingroup$

$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$ 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 $$ \dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.