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I need to find sum of first n terms of series $\sum_{1}^{\infty} {\frac{x^{3n}}{3n(3n-1)(3n-2)}}$. I tried but I just don't know how to transform it into any known form of power series.

EDIT: I tried partial fraction decomposition and telescoping, something cancels but not for every $n$.

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    $\begingroup$ Partial fractions + telescoping? $\endgroup$ – user296602 Apr 16 '18 at 20:10
  • $\begingroup$ It's going to be tough...I think it'd be easier to sum the whole series. $\endgroup$ – DonAntonio Apr 16 '18 at 20:10
  • $\begingroup$ I tried partial fraction decomposition but it didn't help me $\endgroup$ – Nebeski Apr 16 '18 at 20:12
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$$ \frac{1}{3n(3n-1)(3n-2)} = \frac{1}{2}\left[\frac{1}{3n-2}-\frac{2}{3n-1}+\frac{1}{3n}\right] = \frac{1}{2}\int_{0}^{1}\left(z^{3n-3}-2 z^{3n-2}+z^{3n-1}\right)\,dz $$ hence $$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}x^{3n} z^{3n-3}(1-z)^2\,dz=\frac{1}{2}\int_{0}^{1}\frac{x^3(1-z)^2}{1-x^3 z^3}\,dz. $$ On the other hand $\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x)$, hence $$ \sum_{n\geq 1}\frac{x^{3n}}{3n} = -\frac{1}{3}\log(1-x^3) $$ and by letting $\omega=\exp\left(\frac{2\pi i}{3}\right)$, $$ \sum_{n\equiv 1\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega^2 \log(1-\omega x)+\omega \log(1-\omega^2 x)\right]$$ $$ \sum_{n\equiv 2\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega \log(1-\omega x)+\omega^2 \log(1-\omega^2 x)\right]$$ due to the discrete Fourier transform. In particular $$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)}=-\frac{1}{6}\left[(x^2-2x+1)\log(1-x)+(\omega^2 x^2-2\omega x)\log(1-\omega x)+(\omega x^2-2\omega^2 x) \log(1-\omega^2 x)\right]. $$

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    $\begingroup$ @Nebeski: the key steps are just $\int_{0}^{1}x^{\text{whatever}>-1}\,dx = \frac{1}{\text{whatever}+1}$ and the exchange of $\sum$ and $\int$ which is allowed by the absolute convergence in any compact subset of $(-1,1)$. $\endgroup$ – Jack D'Aurizio Apr 16 '18 at 22:46
  • $\begingroup$ Plus the fact that if $n\in\mathbb{N}$ then $1^n+\omega^n+\omega^{2n}$ equals $3$ or $0$ according to $n$ being a multiple of $3$ or not (DFT). $\endgroup$ – Jack D'Aurizio Apr 16 '18 at 22:47
  • $\begingroup$ When $x$ is real, can the answer be simplified to an expression that doesn't involve complex constants? $\endgroup$ – Harry Richman yesterday

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