Sum of the series $\sum\limits_{n=0}^\infty \frac{1}{(3n+1)^3}$

Solution 1:

Let $$S_1=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3} ,S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^3}.$$ It's easily seen that $$S_1+S_2+\sum_{n=1}^{\infty}\frac{1}{(3n)^3}=\zeta(3).$$ That is, $$S_1+S_2=\zeta(3)-\frac{1}{27}\zeta(3)=\frac{26}{27}\zeta(3)\tag{1}$$ However, $$S_1-S_2=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}=\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^3}=\frac{4\pi^3}{81\sqrt{3}}\tag{2}$$

To calculate $(2)$, consider the Weierstrass form of the sinc function $$\frac{\sin x}{x}=\prod\limits_{n=1}^\infty\left(1-\frac{x^2}{\pi^2 n^2}\right)$$ and take $\frac{d^3}{dx^3}\log(\cdot)$ of both sides. Set $x=\frac{\pi}{3}.$

Update. I found that this question has been already asked here, so I link these two: Sum related to zeta function

Also, I give an alternative way to calculate $(2)$.

Notice that $$\frac{2}{\sqrt{3}} \,\sin\left( \frac{2 \pi n}{3} \right) =\begin{cases} 0 & n\equiv 0 \\ 1 & n\equiv 1 \\ -1 & n\equiv 2 \end{cases} \pmod{3}.$$ Therefore $$S_1 - S_2 =\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty} \frac{\sin(2\pi n/3)}{n^3}. \tag{3}$$

Finally, $(2)$ follows from the Fourier expansion $$ \sum_{n=1}^{\infty} \frac{\sin(2 \pi x n)}{n^3}=\frac{\pi^3}{3}\left( \{x\}-3\{x\}^2+2\{x\}^3\right), \tag{4}$$

where $\,\{x\}$ is the fractional part of $x$.

Solution 2:

We have: $$ \int_{0}^{1} x^{3n}\log^2 x\,dx = \frac{2}{(3n+1)^3} $$ hence: $$ \sum_{n\geq 0}\frac{1}{(3n+1)^3} = \frac{1}{2}\int_{0}^{1}\frac{\log^2 x}{1-x^3}\,dx=-\frac{1}{2}\left.\frac{d^2}{ds^2}\int_{0}^{1}\frac{1-x^s}{1-x^3}\,dx\,\right|_{s=0}$$ by using the same technique of this question.

On the other hand, the last integral is related with the digamma function: $$\int_{0}^{1}\frac{1-x^s}{1-x^3}\,dx = \frac{1}{2}\log 3+\frac{\pi\sqrt{3}}{18}+ \frac{1}{3}\,H_{\frac{s-2}{3}}$$ so the original series just depends on a linear combination of $\zeta(3)$ and $\psi''\left(\frac{1}{3}\right)$.

The other case is similar.