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$$\sum_{s=2}^\infty \{\zeta (s)\}=1$$ 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!

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up vote 10 down vote accepted

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,

$$ \sum_{s=2}^{\infty} \left( \zeta(s)-1 \right)= \sum_{s=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^{s}} = \sum_{n=2}^{\infty} \sum_{s=2}^{\infty} \frac{1}{n^{s}}$$

$$ = \sum_{n=2}^{\infty} \frac{\frac{1}{n^{2}}}{1-\frac{1}{n}} = \sum_{n=2}^{\infty} \frac{1}{n(n-1)}$$

$$=\sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n}\right) $$

$$ = \lim_{N \to \infty} \left(1- \frac{1}{2} + \frac{1}{2} -\frac{1}{3} + \ldots + \frac{1}{N-1} - \frac{1}{N} \right)$$

$$ = 1- \lim_{N \to \infty}\frac{1}{N} = 1$$

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The old switcharoo. :-) – Lucian May 13 '14 at 4:41

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.

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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.

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Here is how you prove it. $$ s_n = \sum_{k=2}^{n}\left\{ \zeta(s)\right\} = \sum_{k=2}^{n} (\zeta(s) - \lfloor \zeta(s)\rfloor ) = \sum_{k=2}^{n} (\zeta(s) - 1 ) $$

$$ \implies \lim_{n\to \infty}s_n = 1. $$



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

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


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

3) $$ \sum_{k=2}^{\infty} (\zeta(s) - 1 ) =1. $$.

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