I am trying to find the limit of P(x)

Question

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} ) $ ?

Answer 1

$$\sum_{n\geq 1}\left(\arctan\frac{1}{\sqrt{n+1}}-\arctan\frac{1}{\sqrt{n-1}}\right)=-\frac{\pi}{2}-\frac{\pi}{4}=\color{red}{-\frac{3\pi}{4}}$$ since the LHS is a telescopic series. Here I assumed $\arctan\frac{1}{\sqrt{0}}=\lim_{x\to 0^+}\arctan\frac{1}{\sqrt{x}}=\frac{\pi}{2}$.

We are allowed to switch the limit and the integral since for any fixed $x$ $\left|\arctan\frac{1}{\sqrt{n+1}}-\arctan\frac{1}{\sqrt{n+x}}\right|$ behaves like $\frac{1}{n^{3/2}}$ for large $n$s.

Answer 2

We write $P(-1^+)$ as

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