Catalan’s constant β(2), calculate to six-digit accuracy

$\beta(2)=\sum\limits_{k=0}^{\infty }(-1)^k(2k+1)^{-2}=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{6^2}+...$

For fast convergence, it can be written without first term(i.e. 1)
$\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{6^2}-\frac{1}{7^2}+...=\sum\limits_{1}^{\infty}\frac{1}{(4k-1)^2}-\frac{1}{(4k+1)^2}=\sum\limits_{1}^{\infty}\frac{16k}{(16k^2-1)^2}$

Using $\int\frac{16k}{(16k^2-1)^2}dk=\frac{1}{2}\frac{1}{16k^2-1}+C$,

bounds from integral test becomes,
$\left[\frac{1}{2}\frac{1}{16k^2-1}\right]_{N+1}^\infty\le\sum\limits_{N+1}^{\infty}\frac{16k}{(16k^2-1)^2}\le\left[\frac{1}{2}\frac{1}{16k^2-1}\right]_{N+1}^\infty+\frac{16(N+1)}{(16(N+1)^2-1)^2}$

For six significant figures, at least $\left[\frac{1}{2}\frac{1}{16k^2-1}\right]_{N+1}^\infty\le10^{-7}$, which gives $N\ge\frac{\sqrt{5\times10^6}}{4}=559.01....$ where $\frac{16(N+1)}{(16(N+1)^2-1)^2}$ is negligible.

So, six significant figures can be obtained from $1-\sum\limits_{1}^{560}\frac{16k}{(16k^2-1)^2}$.

$\beta(2)=0.91596569...$, where bounds are $\sim10^-7$

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