= Solution
At $x=\pi/2$, all odd-indexed <cosine> terms vanish, while $\cos(2k\pi/2)=(-1)^k$. Thus the <Fourier series> from part (c) gives
$$
e^{\pi/2}=\frac{e^\pi-1}{\pi}+\frac{2(e^\pi-1)}\pi\sum_{k=1}^{\infty}\frac{(-1)^k}{4k^2+1}.
$$
Divide by $2(e^\pi-1)/\pi$ and rearrange. The <hyperbolic sine> identity $e^\pi-1=2e^{\pi/2}\sinh(\pi/2)$ yields
$$
\boxed{\sum_{k=1}^{\infty}\frac{(-1)^k}{4k^2+1}=\frac{\pi}{4\sinh(\pi/2)}-\frac12,\qquad m=4.}
$$
The evaluation is legitimate by the convergence established in part (c).
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