Is it possible to simplify the below sum? or maybe find a nice closed form (or asymptotics)?

$$2\sum_{n = 0}^{m - 1}(-1)^n\zeta(4m - 2n + 1)\zeta(2n + 2)$$

I am very much aware about even values of Riemann zeta (Euler's formula in terms of Bernoulli numbers) and that would definitely give us a nice closed form for the term $\zeta(2n + 2)$, but what about the term $\zeta(4m - 2n + 1)$? This seems too complicated to simplify simply because we know very little about the odd values of Riemann zeta. However I'm not familiar with other closed forms for $\zeta(2n + 1)$, so I cannot say for sure that the sum can't be simplified more.

Out of curiosity: Can we simplify or find a closed form for the product of even zeta and odd zeta that is $\zeta(2n + 1)\zeta(2n)$? Or can we get a nice closed form for $\zeta(4m + 3)$?


EDIT $\textbf{1}$: @Claude Leibovici, in his answer claims that

If $m$ is even and large then

$$ 2\sum_{n = 0}^{m - 1}(-1)^n\,\zeta(4m - 2n + 1)\,\zeta(2n + 2) \approx \pi \coth (\pi )-2 \tag{1}\label{1}$$

and if $m$ is odd then $$2\sum_{n = 0}^{m - 1}(-1)^n\,\zeta(4m - 2n + 1)\,\zeta(2n + 2) \approx \pi \coth (\pi ) \tag{2}\label{2}$$

However, I have no idea on how to prove $\eqref{1}$ and $\eqref{2}$, so proving $\eqref{1}$ and $\eqref{2}$ (probably using some elementary tools) is sufficient for answering the question completely. Thanks.

  • 1
    $\begingroup$ Where did you encounter this sum? Perhaps some more context would help in finding a solution. $\endgroup$
    – Paramanand Singh
    Apr 18, 2021 at 6:56
  • $\begingroup$ @ParamanandSingh I encountered it while playing with the product $\zeta(2n)\zeta(2n + 1)$ $\endgroup$ Apr 18, 2021 at 6:57
  • $\begingroup$ Wow great problem. I cannot wait to see who has the best answer. $\endgroup$ Apr 26, 2021 at 21:10

3 Answers 3


As is shown in this answer $$ \sum_{k=1}^\infty\zeta(2k)x^{2k}=\frac12(1-\pi x\cot(\pi x))\tag1 $$ substituting $x\mapsto ix$, we get $$ \sum_{k=1}^\infty(-1)^{k-1}\zeta(2k)x^{2k}=\frac12(\pi x\coth(\pi x)-1)\tag2 $$ and therefore, $$ \sum_{k=1}^\infty(-1)^{k-1}(\zeta(2k)-1)x^{2k}=\frac12(\pi x\coth(\pi x)-1)-\frac1{1+x^2}\tag3 $$ Taking the limit of $(3)$ as $x\to1$, we get $$ \sum_{k=1}^\infty(-1)^{k-1}(\zeta(2k)-1)=\frac12(\pi\coth(\pi)-2)\tag4 $$ Since $n\le m-1$, we have that $4m-2n+1\ge2m+3$. This means that $\zeta(4m-2n+1)=1+O\!\left(4^{-m}\right)$ $$ \begin{align} &2\sum_{n=0}^{m-1}(-1)^n\color{#C00}{\zeta(4m-2n+1)}\color{#090}{\zeta(2n+2)}\tag{5a}\\ &=2\color{#C00}{\left(1+O\!\left(4^{-m}\right)\right)}\sum_{n=1}^m(-1)^{n-1}(\color{#090}{\zeta(2n)}-1)+2(m\bmod2)\tag{5b}\\ &=2\left(1+O\!\left(4^{-m}\right)\right)\frac12(\pi\coth(\pi)-2)+2(m\bmod2)\tag{5c}\\[6pt] &=\bbox[5px,border:2px solid #C0A000]{\pi\coth(\pi)-2+2(m\bmod2)+O\!\left(4^{-m}\right)}\tag{5d} \end{align} $$ Explanation:
$\text{(5b)}$: apply $\zeta(4m-2n+1)=1+O\!\left(4^{-m}\right)$
$\phantom{\text{(5b):}}$ then substitute $n\mapsto n-1$
$\phantom{\text{(5b):}}$ then compensate for $2\sum_{n=1}^m(-1)^n$ with $2(m\bmod2)$
$\text{(5c)}$: apply $(4)$ noting that $\zeta(2n)-1=O\!\left(4^{-n}\right)$
$\text{(5d)}$: rearrange to make things look nicer

Note that since $n\le m-1$, $$ \zeta(4m-2n+1)-\zeta(4m-2n+2)=O\!\left(4^{-m}\right)\tag6 $$ and so $\text{(5d)}$ also works for $\zeta(4m-2n+2)$ in place of $\zeta(4m-2n+1)$.

Graphical Verification

Plotting $4^m$ times the absolute error of the approximation given in $\text{(5d)}$ shows that it is bounded and appears to tend to a limit around $0.6$.

enter image description here


I suspect that this is known but I am unable to find references

$$S_m=2\sum_{n = 0}^{m - 1}(-1)^n\,\zeta(4m - 2n + 1)\,\zeta(2n + 2)$$ If $m$ is even and large, it is asymptotic to $$ \pi \coth (\pi )-2$$ and if $m$ is odd to $$ \pi \coth (\pi )$$ and the values are very quickly approached.

  • 1
    $\begingroup$ How did you conclude that If $m$ is odd then $$2\sum_{n = 0}^{m - 1}(-1)^n\,\zeta(4m - 2n + 1)\,\zeta(2n + 2) \approx \pi\coth(\pi)$$ $\endgroup$ Apr 18, 2021 at 7:22
  • $\begingroup$ @BooleanWick. I just knew it. The problem is that I don't remember where it could be proved (problem of age !). $\endgroup$ Apr 18, 2021 at 7:32
  • $\begingroup$ This is really surprising. While $\cot x$ has expansion in terms of Bernoulli numbers/zeta function, it deals with $\zeta(2n)$. Don't know how this is coming up in a sum involving both even and odd zeta values. $\endgroup$
    – Paramanand Singh
    Apr 18, 2021 at 9:03
  • $\begingroup$ @ParamanandSingh Yes, indeed. Do you have any insight on how to prove it? Thanks. $\endgroup$ Apr 18, 2021 at 9:11
  • 1
    $\begingroup$ (+1) for a good memory! :-) $\endgroup$
    – robjohn
    Apr 20, 2021 at 13:59

Not an answer, but this is too long for a comment and basically sums up my train of thoughts.

Let's rewrite the summation with $2j = 2n+2$ and $2k = 4m+3$ (an odd number), i.e., $$2\sum_{j=1}^{\frac{2k-3}{4}-1}(-1)^{j-1}\zeta(2k-2j)\zeta(2j)=-2\sum_{j=1}^{\frac{2k-3}{4}-1}(-1)^{j}\zeta(2k-2j)\zeta(2j).\quad(1)$$ Then, let's use the relation to Bernoulli's numbers: $$\zeta(2n)=\frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}.$$ and plug it in the (1): $$-2\sum_{j=1}^{\frac{2k-3}{4}-1}(-1)^j\frac{(-1)^{k-j+1}(2\pi)^{2k-2j}B_{2k-2j}}{2(2k-2j)!}\frac{(-1)^{j+1}(2\pi)^{2j}B_{2j}}{2(2j)!}=$$ $$-\frac{2(-1)^{k}(2\pi)^{2k}}{4}\sum_{j=1}^{\frac{2k-3}{4}-1} (-1)^j{{2k}\choose{2j}}B_{2k-2j}B_{2j}=\frac{(2\pi)^{2k}}{2}\sum_{j=1}^{\frac{2k-3}{4}-1} (-1)^{k-j+1}{{2k}\choose{2j}}B_{2k-2j}B_{2j}.$$ There seems to be some treatment of this sum in the literature, e.g., in Sums of Products of Bernoulli Numbers there is a reference (Section 5.2, although with a comment "A closed form for this alternating sum doesnot appear to be known.") to article "E. Grosswald, Die Werte der Riemannschen Zetafunktion an ungeraden Argumentstellen, to which unfortunately I don't have access." Maybe there is a hint to @Claude Leibovici answer. If anyone has access to that, I would be more than happy to look at the paper.

Just to add interesting relations which could be relevant to this problem (but I am sceptical)

  • $\sum_{j=1}^{k-1} {{2k}\choose{2j}}B_{2k-2j}B_{2j}=-(2k+1)B_{2k}$
  • $\coth(\pi)=\frac{1}{\pi}\sum_{j=0}^\infty\frac{B_{2j}(2\pi)^{2j}}{(2j)!}$
  • $\begingroup$ $2k = 4m + 3$? How do you justify that? $\endgroup$ Apr 18, 2021 at 13:49
  • $\begingroup$ $k$ wouldn't be an integer here..just wanted to "fit" $2k-2j$. Then looking at $k\to\infty$ it wouldn't make too much difference at the end. But I don't justify it. There is probably a better way. (It's more a wishful thinking than a rigour). $\endgroup$
    – pisoir
    Apr 18, 2021 at 15:10
  • $\begingroup$ Good to see you again (do you remember year $2013$ ?) $\endgroup$ Apr 20, 2021 at 6:53
  • $\begingroup$ @Claude Leibovici You're asking me or @pisoir? $\endgroup$ Apr 20, 2021 at 8:28
  • 1
    $\begingroup$ The formulas you cite are for integer $k$ and $n$. Furthermore, except for $n=0$, $B_{2n+1}=0$, so the formulas won’t work for the non-integer cases to which you are trying to apply them. $\endgroup$
    – robjohn
    Apr 20, 2021 at 13:51

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