Tinkering With The $\zeta$ Function

$\begin{array}{lllc} \displaystyle\sum_{n=1}^{\infty} \dfrac{\zeta(2n)-1}{n} & = & \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\displaystyle\sum_{k=2}^{\infty} \dfrac{1}{ k^{2n}} & \\ & = & \displaystyle\sum_{k=2}^{\infty} \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\left(\dfrac{1}{k^2} \right)^n & \\ & = & \displaystyle\sum_{k=2}^{\infty} -\ln\left(1 - \dfrac{1}{k^2}\right) & \\ & = & \displaystyle\sum_{k=2}^{\infty} \ln\left(\dfrac{k^2}{(k-1)(k+1)}\right) & \\ & = & \displaystyle\lim_{m\to\infty} \ln\left(\displaystyle\prod_{k=2}^{m} \dfrac{k^2}{(k-1)(k+1)} \right) & \\ & = & \displaystyle\lim_{m\to\infty} \ln\left(\dfrac{2m}{m+1} \right) & \\ \displaystyle\sum_{n=1}^{\infty} \dfrac{\zeta(2n)-1}{n} & = & \ln(2) & [A] \end{array}$

On the other hand,

$\begin{array}{lllc} \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{2n} - \dfrac{1}{2n+1} \right) & = & \displaystyle\sum_{m=2}^{\infty} \dfrac{(-1)^m}{m} & \\ & = & 1 - \displaystyle\sum_{m=1}^{\infty} \dfrac{(-1)^{m+1}}{m} & \\ & = & 1 - \ln(2) & [B] \end{array}$

Therefore, adding equations $(A)$ and $(B)$ gives us:

$\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{\zeta(2n)}{n} - \dfrac{1}{2n} - \dfrac{1}{2n+1} \right) = \boxed{1}$

SInce, $\displaystyle \sum_{k=2}^{\infty} \zeta(k)x^{k-1}=-\psi(1-x)-\gamma$ so we can derive $\displaystyle \sum_{k=1}^{\infty}\zeta(2k)x^{2k-1} = \frac{1}{2}(\psi(1+x)-\psi(1-x))$ , Set $x^2\to x$ and divide throughout by $x$ ,

$\displaystyle \sum_{k=1}\zeta(2k)x^{k-1}=\frac{1}{2\sqrt{x}}(\psi(1+\sqrt{x})-\psi(1-\sqrt{x}))$

Since, $\displaystyle \frac{1}{2n}=\int_{0}^{1} x^{2n-1}dx,\frac{1}{2n+1}=\int_{0}^{1} x^{2n}dx$ and summing this the desired sum in integral representaion would be , $\displaystyle S=\int_{0}^{1} \psi(1+x)-\psi(1-x)+1-\frac{1}{1-x} dx$

Now $\displaystyle\int_{0}^{1} \psi(1+x)dx = [\ln\Gamma(1+x)]_{0}^{1} =0$ , $\displaystyle \int_{0}^{1} (\psi(1-x)+\frac{1}{1-x})dx = [\ln(\Gamma(1-x)(1-x))]_{0}^{1}=0$ as $\displaystyle \lim_{x\to 1} (1-x)\Gamma(1-x)=1$

We are only left with $\displaystyle S= \boxed{1}$