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$$ \sum_{n=1}^\infty \frac{1}{n^2 3^n} $$ I tried to use the regular way to calculate the sum of a power series $(x=1/3)$ to solve it but in the end I get to an integral I can't calculate.

Thanks

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Given

$$ \sum_{n=1}^\infty \frac{1}{n^2 3^n}. \tag 1 $$

Write

$$ S(x) = \sum_{n=1}^\infty \frac{\exp( x n)}{n^2}. \tag 2 $$

So we get

$$ \sum_{n=1}^\infty \frac{1}{n^2 3^n} = S(-\ln(a)). \tag 3 $$

Note that

$$ \frac{d^2 S}{dx^2} = \sum_{n=1}^\infty \exp( x n) = \frac{\exp(x)}{1 - \exp(x)}. \tag 4 $$

Then

$$ S(x) = \int dx \int dx \frac{\exp(x)}{1 - \exp(x)} = - \int dx \ln(1 - \exp(x)) = \operatorname{Li}_2( \exp(x) ). \tag 5 $$

Thus

$$ \sum_{n=1}^\infty \frac{\exp( x n)}{n^2} = \operatorname{Li}_2( \exp(x) ). \tag 6 $$

Put in $x = -\ln(3)$ and we get

$$ \bbox[16px,border:2px solid #800000] { \sum_{n=1}^\infty \frac{1}{n^2 3^n} = \operatorname{Li}_2(1/3).} \tag 7 $$

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I suppose that you get to an integral of $\frac{ln(1-x)}{x}$.

This cannot be expressed with a finite number of elementary function. In fact, the sum is the series definition of a special function called "dilogarithm" which belongs to the family of "polylogarithms". $$ \sum_{n=1}^\infty \frac{x^n}{n^2}=Li_2(x) $$ The integral definition of dilogarithm is : $$ Li_2(x)=-\int_0^x \frac{\ln(1-t)}{t}dt $$ In case of $x=1/3$ : $$ \sum_{n=1}^\infty \frac{1}{n^2 3^n}=Li_2(1/3)=0.366213... $$ For the meaning and the use of special functions, for example see :

http://mathworld.wolfram.com/SpecialFunction.html

http://mathworld.wolfram.com/Dilogarithm.html

http://mathworld.wolfram.com/Polylogarithm.html

A more general paper :

https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

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  • $\begingroup$ May I ask how the constants of integration disappear during the manipulation (as shown in my answer)? $\endgroup$ – Kari Aug 4 '15 at 10:38
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    $\begingroup$ They don't really disappear, but they can be $=0$ while computing them in setting $x=0$. $\endgroup$ – JJacquelin Aug 4 '15 at 11:27
  • $\begingroup$ Gotcha, thanks! $\endgroup$ – Kari Aug 4 '15 at 12:32
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$$ \begin{aligned} I \ := \sum_{n=1}^{\infty} \dfrac{x^n}{n^2} & \implies \dfrac{\text{d}I}{\text{d}x} = \sum_{n=1}^{\infty} \dfrac{x^{n-1}}{n} \\ & \implies x \dfrac{\text{d}I}{\text{d}x} = \sum_{n=1}^{\infty} \dfrac{x^{n}}{n} \\ & \implies \dfrac{\text{d}}{\text{d}x} \left( x \dfrac{\text{d}I}{\text{d}x} \right) = \sum_{n=1}^{\infty} x^{n-1} = \dfrac{1}{1-x} \\ & \implies x \dfrac{\text{d}I}{\text{d}x} = \int \dfrac{1}{1-x} \text{ d}x \\ & \implies x \dfrac{\text{d}I}{\text{d}x} = \log \left( \dfrac{1}{1-x} \right) \\ & \implies I = \int \dfrac{1}{x} \ \log \left( \dfrac{1}{1-x} \right) \text{ d}x \ = \mathrm{Li}_2 (x) \end{aligned} $$

$$ \therefore \ \sum_{n=1}^{\infty} \dfrac{1}{n^2 3^n} \ := \ \mathrm{Li}_2 \left( \dfrac{1}{3} \right) $$

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  • $\begingroup$ I cannot understand why someone downnoted your answer which seems correct. May be because your integrals are undefined ? $\endgroup$ – JJacquelin Aug 4 '15 at 11:33
  • $\begingroup$ May I ask what you mean by undefined? Is it to do with the constant of integration? (Also, it seems as though I have a serial downvoter. They've been going through a lot of my posts.) $\endgroup$ – Kari Aug 4 '15 at 12:27
  • $\begingroup$ By "undefined integral", I mean "integral without the bounds". $\endgroup$ – JJacquelin Aug 4 '15 at 13:22
  • $\begingroup$ Oh, thank you. Yea, I wasn't too sure of dealing with the constants of integration, but knew that evaluating the equalities at certain values of $x$ would lead to the constants vanishing. $\endgroup$ – Kari Aug 4 '15 at 19:13

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