The sum of $\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$ I don't have an idea how to calculate this sum. I knew that $\sum_{k=0}^{\infty}\zeta(2k+2)-1 = \frac{3}{4}$. I also knew $\sum_{k=1}^{\infty}\frac{\zeta(2k)k-k}{{2k^2+k}}=\frac{1}{2}(3-ln(4\pi))$. Thank you very much for help. The sum to calculate is $$\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$$
Write $\zeta(2k+2) - 1 = \sum_{n=2}^\infty n^{-2k-2}$ and interchange order of the two sums.  I get
$$ \sum_{n=2}^\infty \frac{\text{arctanh}(1/n)}{n} = \sum_{n=2}^\infty \frac{\ln((n+1)/(n-1))}{2n} $$
However, I still don't have a closed form.
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$\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1}:\ {\Large ?}}$.


  Note that $\pars{~\mbox{see}\ \color{#000}{\textbf{6.3.14}}\ \mbox{in A &amp; S Table}~}$
  $\ds{\left.\vphantom{\Large A}
\Psi\pars{1 + z}\right\vert_{\ \verts{z} &lt; 1} = -\gamma +
\sum_{n = 2}^{\infty}\pars{-1}^{n}\,\zeta\pars{n}z^{n - 1}}$ where $\ds{\Psi}$ and $\ds{\gamma}$ are the Digamma Function and de Euler Constant, respectively.


Then, with $\ds{\verts{z} &lt; 1}$,
\begin{align}
\Psi\pars{1 + z} &amp; = -\gamma +
\sum_{k = 2}^{\infty}\pars{-1}^{k}\,\zeta\pars{k}z^{k - 1}
\\
\Psi\pars{1 - z} &amp; = -\gamma -
\sum_{k = 2}^{\infty}\,\zeta\pars{k}z^{k - 1}
\end{align}
which leads to
\begin{align}
\Psi\pars{1 + z} - \Psi\pars{1 - z} &amp; = 
\sum_{k = 2}^{\infty}\bracks{\pars{-1}^{k} + 1}\,\zeta\pars{k}z^{k - 1} =
\sum_{k = 0}^{\infty}2\,\zeta\pars{2k + 2}z^{2k + 1}
\\[2mm]
\mbox{and}\
{\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}} &amp; = 
\sum_{k = 0}^{\infty}\bracks{\zeta\pars{2k + 2} - 1}z^{2k}
\end{align}


  Integrating over $\ds{\left[0,1\right)}$:


\begin{align}
\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} &amp; =
\int_{0}^{1}\bracks{%
{\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] &amp; =
\int_{0}^{1}\braces{%
{\bracks{\Psi\pars{z} + 1/z} - \Psi\pars{1 - z} \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] &amp; =
\int_{0}^{1}\braces{%
-\,{\bracks{\Psi\pars{1 - z} - \Psi\pars{z}} - 1/z \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] &amp; =
\int_{0}^{1}\bracks{%
-\,{\pi\cot\pars{\pi z} - 1/z \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] &amp; =
{1 \over 2}\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {2 \over 1 - z^{2}}}\dd z
\\[5mm] &amp; =
{1 \over 2}\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z -
{1 \over 2}\int_{0}^{1}{\dd z \over 1 + z}
\end{align}

$$
\bbx{\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1} =
{1 \over 2}\ \underbrace{\int_{0}^{1}\bracks{%
{1 - \pi z\cot\pars{\pi z} \over z^{2}} - {1 \over 1 - z}}\dd z}
_{\ds{\mbox{Numerically}\ \approx 2.0463}}\ -\
{1 \over 2}\,\ln\pars{2} \approx 0.6766}}
$$