<s>[INST] 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 $$\sum_{k=0}^{\infty}\frac{\zeta(2k+2)-1}{{2k+1}}$$
 [/INST] 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 $\approx 0.676558$.
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