I think that this identity is wrong:

$$\sum_{n=1}^\infty (\zeta(4n)-1) = \frac{1}{8}(7-2\coth \pi)$$

See http://mathworld.wolfram.com/RiemannZetaFunction.html (Identity 121 on this site)

It should be $$\sum_{n=1}^\infty (\zeta(4n)-1) = \frac{1}{8}(7-2\pi\coth \pi)$$

Do you agree or am I mistaken?

  • 1
    $\begingroup$ Do you mean $\sum_{i=1}^\infty(\zeta(4i)-1)$? $\endgroup$ – Angina Seng Jun 13 '18 at 19:16
  • $\begingroup$ Yes, otherwise it would diverge $\endgroup$ – Mister Set Jun 13 '18 at 19:18
  • $\begingroup$ I think you are correct from my own numerical check. $\endgroup$ – Cameron Williams Jun 13 '18 at 19:22
  • 2
    $\begingroup$ Your title is misleading. The sum in Wolfram Alpha is correct wolframalpha.com/input/… You mean error in Wolfram MathWorld document $\endgroup$ – James Arathoon Jun 13 '18 at 19:23
  • $\begingroup$ There are a lot of errors in mathworld.wolfram.com $\endgroup$ – Angina Seng Jun 13 '18 at 19:35

I agree there is a missing $\pi$. By the integral representation for the $\zeta$ function

$$\zeta(n)-1 = \int_{0}^{+\infty}\frac{x^{n-1}}{(n-1)!}\cdot\frac{dx}{e^x(e^x-1)} \tag{1}$$ holds for any $n>1$. In particular $$ \sum_{k\geq 1}\left(\zeta(4k)-1\right) = \int_{0}^{+\infty}\frac{dx}{e^x(e^x-1)}\sum_{k\geq 0}\frac{x^{4k+3}}{(4k+3)!}=\int_{0}^{+\infty}\frac{\sinh(x)-\sin(x)}{2e^x(e^x-1)}\,dx\tag{2} $$ where $\int_{0}^{+\infty}\frac{\sinh(x)}{2e^x(e^x-1)}\,dx=\frac{3}{8}$ and $\int_{0}^{+\infty}\frac{\sin(x)}{2e^x}\,dx=\frac{1}{4}$ are trivial and $$\int_{0}^{+\infty}\frac{\sin(x)}{e^x-1}\,dx=\sum_{m\geq 1}\frac{1}{m^2+1}\tag{3} $$ due to the Laplace transform. The last series is well-known to be equal to $\frac{\pi\coth \pi-1}{2}$, for instance through the Poisson summation formula. Rearranging proves the claim (the second one).


Yes, the exact result is $$ \sum_{i=1}^{+\infty}\left(\zeta\left(4i\right)-1\right)=\frac{1}{8}\left(7-2\pi\text{coth}\left(\pi\right)\right) $$

Check it here


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