$$\sum_{k=0}^{\infty} \frac{(-1)^k(2k+1)}{(2k+1)^2 + 1} = \frac{\pi}{4}\mathrm{sech}(\frac{\pi}{2})$$
I tried to consider breaking up the alternating factor of the series by decomposing it into 2 separate series - one adding the even terms and the other subtracting the odd terms, but then I couldn’t figure out where to go. My attempt, letting $S$ be the sum of interest.
$$S = \sum_{k=0}^{\infty} \frac{(2(2k)+1)}{(2(2k)+1)^2 + 1} - \frac{(2(2k+1)+1)}{(2(2k+1)+1)^2 + 1}$$
Don’t know what else to do.
