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How to find the following sum $$\sum\limits_{n=1}^{\infty}\dfrac{\left(\dfrac{3-\sqrt{5}}{2}\right)^n}{n^3}$$

I have tried by rationalizing. But after that I got stuck.

J. W. Tanner
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vqw7Ad
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    May be easier to consider $\sum r^n/n^3$ instead of using this specific value for $r$ – MPW Apr 29 '19 at 14:08
  • Note also that if you call this sum $s(r)$, then $rs'(r)=\sum r^n/n^2$, and likewise $r(rs'(r))'=\sum r^n/n$ etc. Taking a derivative and multiplying by $r$ will reduce the power of $n$ in the denominator. Continue until you get a geometric sum, which you can write as a closed formula. Then reverse the process by integrating. – MPW Apr 29 '19 at 14:17
  • Yes, except that the integrations aren't elementary. – Robert Israel Apr 29 '19 at 14:21
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    Almost by definition, the sum evaluates to ${\rm Li}_3\left(\frac{3-\sqrt{5}}{2}\right)$, where ${\rm Li}_n(z)$ is a polylogarithm function. According to WA, this value can be reexpressed in more common constants. $${\rm Li}_3\left(\frac{3-\sqrt{5}}{2}\right) = \frac{4 \zeta(3)}{5} + \frac{\pi^2}{15} \log\left(\frac{3-\sqrt{5}}{2}\right) - \frac{1}{12}\log^3\left(\frac{3-\sqrt{5}}{2}\right)$$ – achille hui Apr 29 '19 at 14:21

2 Answers2

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For $x<1$,by differentiation,

$$S(x):=\sum_{n=1}^\infty\frac{x^n}{n^3}$$

$$xS'(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$$

$$x(xS'(x))'=\sum_{n=1}^\infty\frac{x^n}{n}$$

$$(x(xS'(x))')'=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}.$$

Then by integration

$$(xS'(x))'=-\frac{\log(1-x)}x$$ and the closed-form stops here.

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By definition, for $|r|<1$, $$ \sum_{n=1}^\infty r^n/n^3 = \text{polylog}(3,r)$$

This is not an elementary function.

Robert Israel
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