Closed form for $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n^2+a^2}}$ Do the convergent sum
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n^2+a^2}}$$
posses a closed form? ($a \in \mathbb{R}$)
Special case is known, for $a=0$ one recalls well known alternating harmonic series :
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln 2$$
 A: As we are considering a function of $a$, it is always useful to build a Taylor series for it.
Unfortunately, in this case it's possible only when $|a|<1$, which we are going to assume here.
Using the binomial series and some identities, we can expand the radical as:
$$\frac{1}{\sqrt{n^2+a^2}}=\frac{1}{n} \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2} \left( \frac{a}{2n} \right)^{2k}$$
Now we need to find a closed form for the following series, which is well known:
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^{2k+1}}=-\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)$$
We need to carefully separate the case with $k=0$ so we don't need to deal with divergencies. Finally we have:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)  \left( \frac{a}{2} \right)^{2k}$$

Introducing a new function:

$$g(y)=\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2} \zeta(2k+1)  y^{2k}$$

We can write:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -g \left( \frac{a}{2} \right)+g \left( \frac{a}{4} \right)$$
$$|a|<1$$


Using the integral form for the zeta function we can write:
$$\zeta(2k+1)=\frac{1}{(2k)!} \int_0^\infty \frac{x^{2k}}{e^x-1}dx$$
Now after summation the function under the integral has a closed form in terms of the Bessel function:
$$g(y)=\int_0^\infty \frac{J_0 (2 yx)-1}{e^x-1}dx$$
Which makes it possible to write the original series neatly as:

$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+a^2}}=- \log 2 -\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx$$

What's more important, this formula works for $|a|>1$ as well, which can be justified by analytic continuation of the Taylor series.
