Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\sum_{n=1}^\infty\frac{1}{8n^3-1}$?

share|cite|improve this question
shouldn't the summands in the first row be $\frac{1}{2^{3n}(3n)^{3m}}$? – Michalis 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
up vote 13 down vote accepted

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 $1$ : $$\sum_{k=0}^{N-1}\left(e^{\dfrac{2\pi i\, j}N}\right)^k=\sum_{k=0}^{N-1} 1=N$$

share|cite|improve this answer
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, – Andrew 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, – Andrew Feb 16 '14 at 12:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.