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I am wondering if the sum $$S=\sum_{k=1}^{\infty}\frac{1}{k(3k-1)}$$ has an exact expression. And when I plugged it into Wolfram Alpha it spitted out: $$S=\frac{1}{6}\Big(-\sqrt{3} π + 9 \ln(3)\Big)$$ I am wondering how is the answer obtained? Is there a simple way of not using math beyond second year university to arrive to that answer?

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  • $\begingroup$ There is a typo. k instead of x $\endgroup$ – HAMIDINE SOUMARE Apr 12 at 23:11
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$$\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k$$ $$\frac{1}{1-x}=\sum_{k=1}^{\infty}x^{k-1}$$ $$\int_{0}^{x}\frac{1}{1-x}dx=\int_{0}^{x}\sum_{k=1}^{\infty}x^{k-1}dx$$ $$-\log(1-x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k}$$ let $x\rightarrow x^3$ $$-\log(1-x^3)=\sum_{k=1}^{\infty}\frac{x^{3k}}{k}$$ $$\frac{-1}{x^2}\log(1-x^3)=\sum_{k=1}^{\infty}\frac{x^{3k-2}}{k}$$

$$\int_{0}^{1}\frac{-1}{x^2}\log(1-x^3)dx=\int_{0}^{1}\sum_{k=1}^{\infty}\frac{x^{3k-2}}{k}dx$$ $$\frac{1}{6} (- \sqrt{3} π + 9 \log(3))=\sum_{k=1}^{\infty}\frac{1}{k(3k-1)}$$

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    $\begingroup$ I was thinking of something similar, only taking an appropriate combination of $\log(1-x), \log(1-\zeta x), \log(1-\zeta^2 x)$ with $\zeta = e^{2\pi i/3}$ to get the right coefficients of $\frac{1}{3}$ for $\frac{x^{3k-1}}{3k-1}$ and $-\frac{1}{3}$ for $\frac{x^{3k}}{3k}$. $\endgroup$ – Daniel Schepler Apr 13 at 0:05
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First, lets transform the sum into

$$\sum_{k=1}^{\infty}\frac{1}{k(3k-1)}=\frac{1}{3}\sum_{k=1}^{\infty}\frac{1}{k(k-\frac{1}{3})}.$$

Using a method given here, this becomes

$$=\frac{1}{3}\left(3H\left(0\right)-3H\left(-\frac{1}{3}\right) \right)$$

(where $H(x)$ denotes the $x$th harmonic number). This simplifies down to

$$=-H\left(-\frac{1}{3}\right)$$

(as the $0$th harmonic number is $0$). Now, there is a certain relationship between the generalized harmonic numbers and the digamma function. That is

$$\psi(x)=H(x-1)-\gamma$$

(where $\gamma$ is the Euler-Mascheroni constant). Thus, our value becomes

$$=-\psi\left(\frac{2}{3}\right)-\gamma.$$

Getting to the punchline finally, we can use Gauss's Digamma Theorem to get

$$=-\gamma+\gamma+\log(2\cdot 3)+\frac{\pi}{2}\cot\left(\frac{2}{3}\pi \right)-2\cos\left(\frac{2\cdot 2}{3}\pi \right)\log\left[\sin\left(\frac{1}{3}\pi\right)\right]$$

$$=\log(2)+\log(3)-\frac{\sqrt{3}\pi}{6}+\log\left(\frac{\sqrt{3}}{2}\right)$$

$$= \log(2)+\log(3)-\frac{\sqrt{3}\pi}{6}+\frac{1}{2}\log\left(3\right)-\log(2)$$

$$= -\frac{\sqrt{3}\pi}{6}+\frac{3}{2}\log\left(3\right)=\frac{1}{6}\left(-\sqrt{3}\pi+9\log(3)\right).$$

I'm not sure if this qualifies as a simple way of obtaining the answer, but I'm can't think of anything easier.

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