TITLE: Why does the $\sum_{n>1}(\zeta(n)-1)=1?$
QUESTION [8 upvotes]: While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$
Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$
Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$
This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?

REPLY [4 votes]: For any $s>1$ we have 
$$\zeta(s)-1=\int_{0}^{+\infty}\frac{x^{s-1}}{(s-1)!}\cdot\frac{dx}{e^x(e^x-1)}\tag{1}$$
hence:
$$ \sum_{n\geq 2}\left(\zeta(n)-1\right) = \int_{0}^{+\infty}\frac{e^x-1}{e^x(e^x-1)}\,dx = 1 \tag{2} $$
and:
$$ \sum_{n\geq 2}\left(\zeta(2n-1)-1\right) = \int_{0}^{+\infty}\frac{\cosh(x)-1}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}-e^{-2x}}{2}\,dx=\frac{1}{4}\tag{3} $$
$$ \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}+e^{-2x}}{2}\,dx=\frac{3}{4}.\tag{4} $$