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!
 A: 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,
$$\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$$
$$=\lim_{n \rightarrow \infty}\arctan\frac1{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})}=\arctan(0)=0$$
A: The $\text{arctan}$ function is the inverse function of $$\tan:\left(-\frac{\pi}2,\frac{\pi}2\right)\rightarrow\Bbb R$$
as this function is monotonically increasing, we have
$$\lim_{x\to\frac\pi 2}\tan x=+\infty\iff \lim_{x\to+\infty}\arctan x=\frac\pi2$$
therefore
$$\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}})=\frac{\pi}{2}$$
$$\lim_{n \rightarrow \infty}\arctan({\sqrt{n}})=\frac{\pi}{2}$$
so the difference is $0$.
A: 
I thought it might be instructive to present a way forward that relies on an integral representation of the arctangent function along with straightforward bounds of the ensuing integral.  To that end we proceed.


Define the arctangent function by the integral
$$\arctan(t)=\int_0^t \frac{1}{1+x^2}\,dx\tag1$$
Using $(1)$, we can write the term of interest as
$$\arctan(\sqrt{n+1})-\arctan(\sqrt n)=\int_\sqrt{n}^\sqrt{n+1} \frac{1}{1+x^2}\,dx\tag2$$
Noting that $0<\frac{1}{1+x^2}\le \frac1{n+1}$ for $x\in[\sqrt{n}, \sqrt{n+1}]$, we assert from $(2)$ that
$$0<\int_\sqrt{n}^\sqrt{n+1} \frac{1}{1+x^2}\,dx\le \frac{\sqrt{n+1}-\sqrt{n}}{1+n}\tag3$$
whence application of the squeeze theorem to $(3)$ yields the coveted limit
$$\lim_{n\to\infty}\left(\arctan(\sqrt{n+1})-\arctan(\sqrt n)\right)=0$$
And we are done!
A: A bit late answer but I was just wondering that no solution used the MVT. So, I add this here using 


*

*$(\arctan \sqrt{x})' = \frac{1}{2\sqrt{x}(1+x)}$
According to MVT there is $\xi_n \in (n,n+1)$ such that
$$\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})=\frac{1}{2\sqrt{\xi_n}(1+\xi_n)}$$
Hence,
$$\underbrace{\frac{1}{2\sqrt{n+1}(1+(n+1))}}_{\stackrel{n\to \infty}{\rightarrow}0} < \arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}}) < \underbrace{\frac{1}{2\sqrt{n}(1+n)}}_{\stackrel{n\to \infty}{\rightarrow}0}$$
A: $\lim_{n\to\infty}\arctan(\sqrt{n})=\arctan(+\infty)=\pi/2$ and, analogously, $\lim_{n\to\infty}\arctan(\sqrt{n+1})=\pi/2$.
So your limit is $0$.
A: $\arctan (\tan [\arctan \sqrt{n+1} -\arctan √n]\big )=$
$\arctan \big (\dfrac{\sqrt{n+1}-√n}{1+\sqrt{n+1}√n}\big );$
$0< f(n):=\dfrac{\sqrt{n+1}-√n}{1+\sqrt{n+1}√n}<$
$\dfrac{\sqrt{n+1}-√n}{\sqrt{n+1}√n}=$
$(n+1)^{-1/2}-n^{-1/2};$
$\lim_{n \rightarrow \infty} f(n)=0.$
Note $\arctan$ is continuous.
$\lim_{n \rightarrow \infty} \arctan (f(n))=0$.
A: 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.
