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We begin by noting that $\lfloor\zeta(n)\rfloor=1 \ , \ n \geq 2$ .

Therefore our sum is

$S=\displaystyle \sum_{n=2}^{\infty}{\zeta(2n-1)-1}$

For any $n>1$ we have

$\zeta(n)-1=\int_{0}^{\infty}{\frac{x^{n-1}}{(n-1)!} \cdot \frac{dx}{e^x(e^x-1)}}$

Hence after plugging in we obtain

$S=\sum_{n=2}^{\infty}{\int_{0}^{\infty}{\frac{x^{2n-2}}{(2n-2)!}\cdot \frac{dx}{e^x(e^x-1)}}}$

We recognize the $\cosh(x)$ Taylor Series here:

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

Therefore the sum becomes

$S=\int_{0}^{\infty}{\frac{\cosh(x)-1}{e^x(e^x-1)}dx}$

Recall that

$\cosh{x}=\frac{e^x+e^{-x}}{2}$

Hence, $S=\int_{0}^{\infty}{\frac{1+e^{-2x}-2e^{-x}}{2(e^x-1)}dx}$ $=\int_{0}^{\infty}{\frac{(e^x-1)(e^{-x}-e^{-2x})}{2(e^x-1)}dx}$ $=\int_{0}^{\infty}{\frac{e^x-e^{-x}}{2}dx}$

Which is easily evaluated to be $S=\frac{1}{4}=\boxed{.250}$

Let $S = \sum\limits_{n=2}^{\infty} \left({\zeta\left(2n-1\right)} \right)$

Note that $\{x\} = x - \left\lfloor x \right\rfloor$ and $\zeta(n) - 1 < 1$ , where $n \geq 2$ , so we can write

$\begin{aligned} S & = \sum\limits_{n=2}^{\infty} \sum\limits_{k=1}^{\infty} \left(\frac{1}{k^{2n-1}}-\left\lfloor \frac{1}{k^{2n-1}} \right\rfloor \right) \\ & = \sum\limits_{n=2}^{\infty} \sum\limits_{k=2}^{\infty} \left(\frac{1}{k^{2n-1}}\right) \end{aligned}$

Note this can be expanded and rearranged into a sum of geometric series:

$\begin{aligned} S & = \left(\frac{1}{2^{3}} + \frac{1}{2^{5}} + \cdots \right) + \left(\frac{1}{3^{3}} + \frac{1}{3^{5}} + \cdots \right) + \cdots \\ & = \sum\limits_{k=2}^{\infty} \frac{ \frac{1}{k^{3}} }{1-\frac{1}{k^{2}}} \\ & = \sum\limits_{k=2}^{\infty} \frac{1}{k\left(k^{2}-1\right)} \end{aligned}$

We can use partial fractions to decompose this and simplify:

$\begin{aligned} S & = \sum\limits_{k=2}^{\infty} \left( \frac{1}{2\left(k+1\right)} + \frac{1}{2\left(k-1\right)} - \frac{1}{k} \right) \\ & = \sum\limits_{k=3}^{\infty} \left( \frac{1}{2k} \right) + \sum\limits_{k=1}^{\infty} \left( \frac{1}{2k} \right) - \sum\limits_{k=2}^{\infty} \left( \frac{1}{k} \right) \\ & = 2\sum\limits_{k=3}^{\infty} \left( \frac{1}{2k} \right) + \frac{1}{2(1)} + \frac{1}{2(2)} - \sum\limits_{k=3}^{\infty} \left( \frac{1}{k} \right) - \frac{1}{2} \\ & = \frac{1}{4} = \boxed{0.25} \end{aligned}$

This paper discusses these types of identities (from page 15 onward) and other interesting aspects of the Riemann Zeta Function.