I am trying to find the limit of P(x) When I am looking for a $\lim\limits_{x \to -1} P(x)$ 
where P(x)$= \sum \limits_{n=1}^\infty \left( \arctan \frac{1}{\sqrt{n+1}} - \arctan \frac{1}{\sqrt{n+x}}\right) $ do I have to ignore a summation sign sigma and dealing with it like this $ \
\lim\limits_{x \to -1}(\arctan\sqrt{n+x} - \arctan \sqrt{n+1}  )  $ ?

We write $P(-1^+)$ as
$$\begin{align}
P(-1^+)&=\arctan(1/\sqrt{2})-\pi/2+\sum_{n=2}^{\infty}\left(\arctan\left(\frac{1}{\sqrt{n+1}}\right)-\arctan\left(\frac{1}{\sqrt{n-1}}\right)\right)\\\\
&=\sum_{n=1}^{\infty}\left(\arctan \sqrt{n-1}-\arctan \sqrt{n+1}\right)\\\\
&=\lim_{N\to \infty}\sum_{n=1}^{N}\left(\arctan \sqrt{n-1}-\arctan \sqrt{n+1}\right)\\\\
&=\lim_{N\to \infty}\left(\arctan(0)+\arctan(1)-\arctan(N-1)-\arctan(N)\right)\\\\
&=0+\frac{\pi}{4}-\frac{\pi}{2}-\frac{\pi}{2}\\\\
&=-\frac{3\pi}{4}
\end{align}$$
