Determine $\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$ Determine $\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$ if it exists.


I know that it exists, but I do not know how to show it. I tried to use the series definition $$\arctan (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1},$$ but I realised that we have only defined it that way for $x \in [-1, \, 1]$ shortly after.

Thanks in advance!
Note,

$$\arctan{\sqrt{n+1}} - \arctan{\sqrt{n}}
=\arctan \frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n+1}\cdot\sqrt{n}} $$
$$=\arctan  \frac{1}{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})} $$

Thus,
$$\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$$
$$=\lim_{n \rightarrow \infty}\arctan\frac1{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})}=\arctan(0)=0$$
Why did you go in for a round about series option?

When $n$ is large $ \sqrt {n+1}\rightarrow \sqrt{ n} $ and limit is directly seen zero as difference of two equal terms. Arctan tends asymptotically to $\pi/4$ for large arguments without monotonic increase.