Determine $\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) – \arctan({\sqrt{n}})$

Question

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!

Answer 1

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, $$\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$$ $$=\lim_{n \rightarrow \infty}\arctan\frac1{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})}=\arctan(0)=0$$