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}
 A: 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).
A: $$
\begin{align}
\lim_{n \to \infty }\sqrt{a^2+2an+n^2+b^2}-n=\lim_{n \to \infty }\frac{\left ( a^2+2an+b^2 \right )}{\sqrt{a^2+2an+n^2+b^2}+n}=a
\end{align}
$$
evidently divergent for 
$$
\begin{align}
a\neq 0
\end{align}
$$
A: Your two expressions are not equivalent.  The sum is not uniformly convergent, so the order you sum the terms in can matter.  The second has an explicit order of summing while the first does not and will cancel out the constant terms.  Even the second one does not converge, however.  
We have $\sqrt {(n+a)^2+b^2}-n \sim a+\frac {b^2}{2n}+o(n^{-1})$ as $n$ gets large, so $\sqrt {(n+a)^2+b^2}+\sqrt {(n-a)^2+b^2}-2n \sim\frac {b^2}{n}+o(n^{-1})$ and the sum diverges logarithmically
