How to determine that $ \sum_{k=1}^\infty \frac{1}{(2k +1)^2} = \frac{\pi^2}{8}-1$ According to Mathematica it holds
$$
\sum_{k=1}^\infty \frac{1}{(2k +1)^2} = \frac{\pi^2}{8}-1 .
$$
How can this be proved? Thanks.
 A: \begin{align}
\sum_{k=1}^\infty \frac{1}{(2k +1)^2}&=\sum_{k=1}^\infty \frac{1}{k^2}-\sum_{k=1}^\infty \frac{1}{(2k)^2}-1\\
&={\pi^2 \over6}-{\pi^2 \over24}-1\\
&={\pi^2 \over8}-1
\end{align}
A: Another method (interesting but not the fastest) assuming that we don't know $\sum_{k=1}^\infty \frac{1}{k^2}={\pi^2 \over6}$.
Note that for all $x\in[-1,1]$,
$$\arcsin(x)=\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}x^{2n+1}}{4^{n}(2n+1)}}$$
Let $x\in[0,{\pi \over2}]$, we have that
$$x=\arcsin(\sin(x))=\sum _{n=0}^{\infty }{{\binom {2n}{n}}\frac {(\sin(x))^{2n+1}}{4^{n}(2n+1)}}$$
Integration from $0$ to ${\pi \over 2}$,
\begin{align}[{x^2\over2}]_0^{\pi/2}={\pi^2\over8}&=\int_0^{\pi\over2}\sum _{n=0}^{\infty }{{\binom {2n}{n}}\frac {(\sin(x))^{2n+1}}{4^{n}(2n+1)}}dx\\
&=\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}}{4^{n}(2n+1)}}\int_0^{\pi\over2}(\sin(x))^{2n+1}dx\tag{1}\\
&=\sum_{k=0}^\infty \frac{1}{(2k +1)^2}\tag{2}\\
&=1+\sum_{k=1}^\infty \frac{1}{(2k +1)^2}
\end{align}
$(1)$ : please justify the conditions;
$(2)$ : Wallis' integrals.  
Hence,
$$\sum_{k=1}^\infty \frac{1}{(2k +1)^2} = \frac{\pi^2}{8}-1$$
