Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent? 
If we drop out the $n=0$ term and fold the remaining sum, the question can be equivalently asked for the series
\begin{equation}
\sum_{n = 1}^{\infty} (\sqrt{(a+n)^2+b^2} + \sqrt{(a-n)^2+b^2} - 2n ).
\end{equation}

My gut reaction is to say divergent, since the asymptotic behaviour (assume $n>0$, since the series is only twice that, plus a finite number of terms) is something like 
$$
\begin{align}
\sqrt{n^2 + a^2 + 2an + b^2} - n &= n\left(\sqrt{1 + \frac{a^2}{n^2} + \frac{2a}{n} + \frac{b^2}{n^2}} - 1\right) \\
&= n\left(1 + \frac{a}{n} + o\left(\frac{1}{n}\right) - 1\right) \\
&= a + o(1)
\end{align}
$$
which seems to diverge for any $a \ne 0$.
(note that this assumes that $[\cdot]$ is just a normal set of brackets, if it's the floor function, then there are only a finite number of non-zero terms).
