Infinite Series $\sum\limits_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$

I want to evaluate $$\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$$ let $f(z)=\sum_{n=1}^\infty\frac{1}{2^{3n}}z^{3n}$, then $$\sum_{n=1}^\infty f\left(\frac{1}{n}\right)=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{2^{3m}}\frac{1}{n^{3m}}\leq\frac{1}{7}\zeta(3)<\infty$$ so we can switch order of summation $$\sum_{n=1}^\infty f\left(\frac{1}{n}\right)=\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$$ and now $$f(z)=\sum_{n=1}^\infty\left(\frac{z^3}{2^3}\right)^n=\frac{z^3}{8-z^3}, f\left(\frac{1}{n}\right)=\frac{1}{8n^3-1}$$ Hence, $$\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}=\sum_{n=1}^\infty\frac{1}{8n^3-1}$$ Is there an analytic method for evaluating $\sum_{n=1}^\infty\frac{1}{8n^3-1}$?

• shouldn't the summands in the first row be $\frac{1}{2^{3n}(3n)^{3m}}$? – MichalisN Dec 17 '13 at 14:04
• You can express $\sum\frac{1}{n-a}-\frac{1}{n}$ in terms of the digamma function, and then apply this to a partial fraction decomposition of $\frac{1}{8n^3-1}$. – Einar Rødland Dec 17 '13 at 15:31
• To conclude your evaluation the usual trick is to expand in partial fractions and to use the digamma function $\psi$ (and derivatives) as shown in A&S page $264$. You should obtain the same expansion that in my answer. – Raymond Manzoni Dec 17 '13 at 18:13

You may use the same method that proposed in your other thread:

Start with a generating function for the $\zeta(n)$ terms (using the digamma function as proposed by Einar Rødland) : $$\psi(1-x)=-\gamma-\sum_{n=1}^\infty \zeta(n+1)\;x^n$$ multiply by $x$ (so that $\zeta(n)$ corresponds to $x^n$ since we want the coefficients $\zeta(3n)$) $$x\,\psi(1-x)=-x\,\gamma-\sum_{n=2}^\infty \zeta(n)\;x^n$$ As for the multiplication theorem a closed form for your answer will be given by : $$\sum_{n=1}^\infty \zeta(3n)x^{3n}=-\frac {x\,\psi(1-x)+x\,e^{2\pi i/3}\,\psi\left(1-x\,e^{2\pi i/3}\right)+x\,e^{-2\pi i/3}\,\psi\left(1-x\,e^{-2\pi i/3}\right)}3$$ Setting $x=\dfrac 12$ returns the wished numerical answer $\,\approx 0.16838922476583426924744\cdots$
(I don't known a much simpler form at this point).

MORE GENERALLY (we considered only the specific case $\,f(x):=x\,\psi(1-x)\,$ and $N=3$) :

Consider $\,\displaystyle f(x):=\sum_{j=0}^\infty a_j\,x^j\,$ (this could be extended to Lambert series) then the sub-series $\,\displaystyle f_N(x):=\sum_{j=0}^\infty a_{Nj}\,x^{Nj}\,$ (i.e. keeping every $N$-th term) will be obtained with : $$f_N(x)=\frac 1N\sum_{k=0}^{N-1}f\left(x\;e^{\dfrac{2\pi i k}N}\right)$$ simply because, expanding $f_N(x)$ in powers of $x$, we get a geometric series for $N$ not dividing $j$ : $$\sum_{k=0}^{N-1}\left(e^{\dfrac{2\pi i\, k}N}\right)^j=\sum_{k=0}^{N-1}\left(e^{\dfrac{2\pi i\, j}N}\right)^k=\frac{e^{2\pi i\, j}-1}{e^{\frac{2\pi i\,j}N}-1}=0$$ while for $N$ dividing $j$ we are simply adding $N$ times $1$ : $$\sum_{k=0}^{N-1}\left(e^{\dfrac{2\pi i\, j}N}\right)^k=\sum_{k=0}^{N-1} 1=N$$

• Dear Raymond, Maybe you would please elaborate a bit how to get from the line above "As for the multiplication theorem" to the last line. Thanks and all the best, – user12802 Feb 15 '14 at 19:41
• Hi @96Tears. Let's suppose that $\,\displaystyle f(x):=\sum_{n=0}^\infty a_n,x^n$ (with $f(x)=x\,\psi(1-x)\,$ here) then $\,\displaystyle f(x)+f\left(x\,e^{2\pi i/3}\right)+f\left(x\,e^{-2\pi i/3}\right)=3\sum_{k=0}^\infty a_{3k}\,x^{3k}$ (because for $n\neq 3k$ the three contributions cancel and add only for $n= 3k$). To obtain $\displaystyle\sum_{k=0}^\infty a_{mk}\,x^{mk}$ replace the $3$-th roots of unity by the $m$-th roots. Cheers, – Raymond Manzoni Feb 15 '14 at 20:03
• Nice generalization above - thanks very much. Best, – user12802 Feb 16 '14 at 12:50