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 $\sum_{n=1}^\infty\frac{1}{8n^3-1}$?
 A: 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 $1$ :
$$\sum_{k=0}^{N-1}\left(e^{\dfrac{2\pi i\, j}N}\right)^k=\sum_{k=0}^{N-1} 1=N$$
