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 $$\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$$
 A: 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.
A: By the integral representation for the $\zeta$ function
$$ \zeta(m)-1 = \frac{1}{(m-1)!}\int_{0}^{+\infty}\frac{x^{m-1}}{e^x(e^x-1)}\,dx $$
hence
$$\begin{eqnarray*} \sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}&=&\int_{0}^{+\infty}\frac{\sum_{k\geq 0}\frac{x^{2k+1}}{(2k+1)!(2k+1)}}{e^x(e^x-1)}\,dx\\&\stackrel{\text{IBP}}{=}&-\int_{0}^{+\infty}\frac{\sinh x}{x}\left(e^{-x}+\log(1-e^{-x})\right)\,dx\end{eqnarray*} $$
and by exploiting the ordinary generating function for $\{\zeta(2k)\}_{k\geq 1}$ we also have
$$\sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}=\int_{0}^{1}\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\,dx $$
Still no "nice" closed form, but a simple numerical task, since the last integrand function has an approximately quadratic behaviour on $(0,1)$. Simpson's rule already gives $\text{LHS}\approx\frac{2\pi^2+29}{72}\approx 0.67693$ and by using the composite rule on $5$ points (weights $1-4-2-4-1$) we have 
$\text{LHS}\approx\frac{29269-6720 \pi +210 \pi ^2}{15120}\approx \color{green}{0.6765}9486.$
A: Starting from Jack D'Aurizio's answer
$$\sum_{k\geq 0}\frac{\zeta(2k+2)-1}{2k+1}=\int_{0}^{1}\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\,dx$$ the integrand can be approximated by a $[2,2]$ Padé approximant built around $x=0$.
Since the first and third derivatives are $0$ for $x=0$, the expression of the Padé approximant of a function $F(x)$ simplifies a lot and write
$$F(x)\approx\frac{-12 \left(F(0) F''(0)\right)+ \left(F(0) F''''(0)-6
   F''(0)^2\right) x^2} {-12 F''(0)+F''''(0)\, x^2 } $$making
$$\int F(x)\,dx=\left(F(0)-\frac{6 F''(0)^2}{F''''(0)}\right)x+\frac{12 \sqrt{3} F''(0)^{5/2}
   }{F''''(0)^{3/2}}\tanh ^{-1}\left(\frac{\sqrt{F''''(0)} }{2 
   \sqrt{3F''(0)}}x\right)$$ In the considered problem, we have
$$F(0)=\frac{\pi ^2}{6}-1 \qquad F''(0)=\frac{\pi ^4}{45}-2\qquad F''''(0)=\frac{8 \pi ^6}{315}-24$$ making
$$\frac{1-3x^2-\pi x\cot(\pi x)+\pi x^3\cot(\pi x)}{2x^2(1-x^2)}\approx \frac{\frac{\pi ^2-6}{6} +\frac{\left(3150 \pi ^2-420 \pi ^4+20 \pi ^6-\pi
   ^8\right) }{210 \left(\pi ^4-90\right)}x^2 }{1-\frac{2 \left(\pi ^6-945\right) }{21 \left(\pi
   ^4-90\right)}x^2 }$$ for which the antiderivative is
$$\frac{\pi ^2 \left(-3150+420 \pi ^2-20 \pi ^4+\pi ^6\right) x}{20 \left(\pi
   ^6-945\right)}+\frac{7 \sqrt{{7}} \left(\pi ^4-90\right)^{5/2} \tanh
   ^{-1}\left(\sqrt{\frac{2 \left(\pi ^6-945\right)}{21 \left(\pi ^4-90\right)}}
   x\right)}{20 \sqrt 6 \left(\pi ^6-945\right)^{3/2}}$$ Integrating between $0$ and $1$, we get an ugly expression which is $\approx 0.676469$ while the numerical evaluation of Jack D'Aurizio's integral gives $\approx 0.676565$.
Edit
More tedious but still workable, we could build the $[4,2]$ Padé approximant. The result is $\approx 0.676558$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\sum_{k = 0}^{\infty}{\zeta\pars{2k + 2} - 1 \over 2k + 1}:\ {\Large ?}}$.

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

Then, with $\ds{\verts{z} < 1}$,
\begin{align}
\Psi\pars{1 + z} & = -\gamma +
\sum_{k = 2}^{\infty}\pars{-1}^{k}\,\zeta\pars{k}z^{k - 1}
\\
\Psi\pars{1 - z} & = -\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} & = 
\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}} & = 
\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} & =
\int_{0}^{1}\bracks{%
{\Psi\pars{1 + z} - \Psi\pars{1 - z} \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\braces{%
{\bracks{\Psi\pars{z} + 1/z} - \Psi\pars{1 - z} \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\braces{%
-\,{\bracks{\Psi\pars{1 - z} - \Psi\pars{z}} - 1/z \over 2z} -
{1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
\int_{0}^{1}\bracks{%
-\,{\pi\cot\pars{\pi z} - 1/z \over 2z} - {1 \over 1 - z^{2}}}\dd z
\\[5mm] & =
{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] & =
{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}

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