Why does the $\sum_{n>1}(\zeta(n)-1)=1?$ 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?
 A: Using the series representation of the Riemann-Zeta function 
$$\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$$
gives 
$$\begin{align}
\sum_{n=2}^{\infty}\left(\sum_{k=1}^{\infty}\frac{1}{k^n}-1\right)&=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{k^n}\\\\
&=\sum_{k=2}^{\infty}\left(\frac{1/k^2}{1-1/k}\right)\\\\
&=\sum_{k=2}^{\infty}\left(\frac{1}{k-1}-\frac{1}{k}\right)\\\\
&=1
\end{align}$$
A: There is a whole series of formulae like this. The proofs are all along the lines of writing out the $\zeta$ sums, changing the order of summation (making sure that this is valid, of course!), and doing the interior sum. In this case, the double sum will be
$$ \begin{align}
\sum_{n=2}^{\infty} (\zeta(n)-1) &=  \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{k^n} \\
&= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^n} \\
&= \sum_{k=2}^{\infty} \frac{1}{k^2(1-1/k)} \\
&= \sum_{k=2}^{\infty} \frac{1}{k(k-1)} \\
&= \sum_{k=2}^{\infty} \left( \frac{1}{k} - \frac{1}{k-1} \right),
\end{align} $$
which it is easy to see telescopes down to $1$. Some other examples can be found here.
A: 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:
$$ \sum_{n\geq 2}\left(\zeta(2n-1)-1\right) = \int_{0}^{+\infty}\frac{\cosh(x)-1}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}-e^{-2x}}{2}\,dx=\frac{1}{4}\tag{3} $$
$$ \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}+e^{-2x}}{2}\,dx=\frac{3}{4}.\tag{4} $$
A: Note: $$\begin{align}\sum_{n=2}^\infty (\zeta(n)-1) &= \sum_{n=2}^\infty \sum_{k=2}^\infty\frac{1}{k^n}\\
&=\sum_{k=2}^\infty\sum_{n=2}^\infty \frac{1}{k^n}
\end{align}$$
And $$\sum_{n=2}^{\infty} \frac{1}{k^n} =\frac{1}{k^2}\frac{1}{1-\frac{1}{k}}= \frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}.$$
For $\zeta(2n)$ case:
$$ \sum_{n=1}^{\infty} \frac{1}{k^{2n}} = \frac{1}{k^2-1} = \frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)$$
More generally, if $f(z)=\sum_{n=2}^\infty a_nz^n$ has radius of convergence more than $\frac{1}2$, then:
$$\sum_{n=2}^\infty a_n(\zeta(n)-1) = \sum_{k=2}^\infty f\left(\frac1k\right)$$
This can be used to show that $$\sum_{n=2}^\infty \frac{\zeta(n)-1}{n} = 1-\gamma$$ where $\gamma$ is the Euler–Mascheroni constant. Using the standard limit for $\gamma$, we see that:
$$\lim_{N\to\infty} \left(\log N -\sum_{n=2}^N \frac{\zeta(n)}{n}\right) = 0$$

Very late comment
I just noticed that if $f(z)=\sum_{n=2}^\infty a_nz^n$ has radius of convergence greater than $1,$ we get:
$$\sum_{n=2}^\infty a_n \zeta(n) = \sum_{k=1}^\infty f\left(\frac 1k\right)$$
A: Simple laymans proof:
Start with the Harmonic series

*

*1 +1/2 +1/3 + 1/4 etc

*you can sum the powers of every individual N independently

*i.e. 1/2 +1/4 +1/8 .... = 1

*1/3 +1/9 +1/27 .... =1/2

*1/5 + 1/25 +/125 ... =1/4

*As a rule Σ 1/N^n ... = 1/(N-1)

*This leaves us with 1 +1 + 1/2 +1/4 +1/5 =1/6 +1/9 etc....

*The missing terms are those 1 less than the powers (greater than 1 of course) of the integers. Since these numbers have already been used as part of an earlier series i.e. there is no 1/4 + 1/16 ...  to make 1/3 because 1/4 has been used as part 1/2 +1/4 +1/8 .... series.

*while we have an extra 1 in the beginning of the series.

*Therefore 1 = Σ 1/3 + 1/7 +1/8 ...

Simple proof without any knowledge of how to manipulate zeta functions in any way.
