Equation $(1)$ implies that
$$
\frac\pi{4^{n-1}}\sum_{k=1}^n\frac{\dbinom{2n-1}{n-k}}{(2k-1)^2+\pi^2}=\int_0^\infty\cos^{2n-1}(x)\,e^{-\pi x}\,\mathrm{d}x
$$
However, Jack's and both my answers say that
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\sum_{k=1}^n\frac{\dbinom{2n-1}{n-k}}{(2k-1)^2+\pi^2}=\frac1{8\sqrt\pi}\tanh(\pi^2/2)
$$
This implies that
$$
\frac1\pi\lim_{n\to\infty}\sqrt{n}\int_0^\infty\cos^{2n-1}(x)\,e^{-\pi x}\,\mathrm{d}x=\frac1{2\sqrt\pi}\tanh(\pi^2/2)
$$