fractional part of Riemann zeta $\sum_{s=2}^\infty \{\zeta (s)\}=1$ 

$$\sum_{s=2}^\infty \{\zeta (s)\}=1$$
where $\zeta (s)$ is Riemann zeta, $\{x\}$ denotes the fractional part of the real number $x$


The problem was proposed by Michael Th. Rassias
$\{\zeta(2)\}=\frac{\pi^2}6-1,$ How to go on?
Thanks a lot!
 A: Note that $\zeta(s)=1+2^{-s}+3^{-s}+\cdots$. 
The fractional part of $\zeta(s)$ where $s\geq 2$, is $\zeta(s)-1$. 
Thus, we need to consider the following sum: 
$$
\zeta(2)-1=2^{-2}+  3^{-2} + \cdots $$ $$
\zeta(3)-1=2^{-3}+  3^{-3} + \cdots 
$$
Then add the numbers vertically. 
A: Here is how you prove it.
$$ s_n = \sum_{s=2}^{n}\left\{ \zeta(s)\right\} = \sum_{s=2}^{n} (\zeta(s) - \lfloor \zeta(s)\rfloor ) = \sum_{s = 2}^{n} (\zeta(s) - 1 ) $$
$$ \implies \lim_{n\to \infty}s_n = 1. $$
Notes:
1) 

$$\left\{ x\right\} = x - \lfloor x\rfloor, $$

where $\lfloor x\rfloor$ is the floor function.
2) 

$$ \lfloor \zeta(s)\rfloor = 1,\quad \forall s\geq 2. $$

3) $$ \sum_{s=2}^{\infty} (\zeta(s) - 1 ) =1. $$.
A: Since $ 1 < \zeta(s) <2$ for $s \ge 2$, it's equivalent to showing that $ \displaystyle\sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= 1$.
In which case,
$$ \sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= \sum_{s=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^{s}} = \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{n^{s}}$$
$$ = \sum_{n=2}^{\infty} \frac{\frac{1}{n^{2}}}{1-\frac{1}{n}} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)}$$
$$=\sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n}\right) $$
$$ = \lim_{N \to \infty} \left(1- \frac{1}{2} + \frac{1}{2} -\frac{1}{3} + \ldots + \frac{1}{N-1} - \frac{1}{N} \right)$$
$$ = 1- \lim_{N \to \infty}\frac{1}{N} = 1$$
A: Hint: Note $\{\zeta(s)\} = \zeta(s) - \lfloor \zeta(s) \rfloor$. By the definition of the zeta function, $\zeta(s) > 1$ for all $s\ge 2$. What does this tell you about $\lfloor \zeta(s) \rfloor$? Once you've found your bearings with regard to the floor of $\zeta(s)$, how do you evaluate the remaining sum? Use, again, the definition of the zeta function.
