Error on MathWorld: $\sum_{n=1}^\infty (\zeta(4n)-1) = \frac{1}{8}(7-2\coth \pi)$ 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?
 A: 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).
A: 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
