sum of two series Can
\begin{equation}
\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{k^2+a^{2}}, \ \ \sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}
\end{equation}
be summed explicitly, where $a$ is a constant real number? Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!
 A: It is well known that
$$\csc z = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2 \pi^2}.$$
Changing $z$ into $\pi z$ produces
$$\pi\csc(\pi z) = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2}, \quad z \ne n \pi, \quad n \in \mathbb{Z},$$
so
$$\frac{\pi\csc(\pi z) - 1/z}{2z} = \sum_{k = 1}^\infty \frac{(-1)^k}{z^2 - k^2}.$$
Then it can be shown, by replacing $z$ with $iz$ and using $\csc(iz) = -i\operatorname{cosech} z$, that
$$\operatorname{cosech} z = \frac{1}{z} + 2z\sum_{k = 1}^\infty \frac{(-1)^k}{z^2 + k^2}, \quad z \ne n \pi i, \quad n \in \mathbb{Z},$$
or
$$\frac{\pi\operatorname{cosech}(\pi z) - 1/z}{2z} = \sum_{k = 1}^\infty \frac{(-1)^k}{z^2 + k^2}.$$
The second series looks familiar, but I cannot remember what it equals.
A: $$\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{k^2+a^{2}}$$
This series converges to
$$\dfrac{1}{2a^2}+\dfrac{\pi\mathrm{csch}(\pi a)}{2a}$$
Now,
$$\sum_{k= 0}^{\infty}\frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}$$
This series also converges by the comparison test.
Now,
\begin{equation}
\sum_{k= 0}^{\infty}\left(\frac{\left( -1\right) ^{k}}{k^2+a^{2}}+ \frac{\left( -1\right) ^{k}}{(k-1/2)^2+a^{2}}\right)
\end{equation}
Use the comparison test to see that this is convergent.
