limit with $\arctan$

Question

I have to find the limit and want ask about a hint:

$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$

I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard

Answer 1

For $x>0$ we have $$\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$$ so \begin{align}(u_n=\arctan^{\frac{1}{2}}(n+1)- \arctan^{\frac{1}{2}}(n)&=\sqrt{\frac{\pi}{2}}\left[(1-\frac{2}{\pi}\arctan\frac{1}{n+1})^{1/2}-(1-\frac{2}{\pi}\arctan\frac{1}{n})^{1/2}\right]\\&=\sqrt{\frac{\pi}{2}}\left(\frac{1}{\pi}\frac{1}{n(n+1)}+O(\frac{1}{n^3})\right)\end{align} and then we find $$u_n\underset{n\rightarrow \infty}{\sim}\frac{1}{\sqrt{2\pi}}\frac{1}{n^2}$$ hence we find that your desired limit is $0$ but we find also $$\lim_{n\to\infty}n^2\left(\arctan^{\frac{1}{2}}(n+1)- \arctan^{\frac{1}{2}}(n)\right)=\frac{1}{\sqrt{2\pi}}$$ Added We have $\arctan(x)=_0 x-\frac{x^3}{3}+O(x^5)$ so we find $$\arctan(n+1)^{\frac{1}{2}}- \arctan(n^{\frac{1}{2}})=\arctan(\frac{1}{n^{1/2}})-\arctan(\frac{1}{(n+1)^{1/2}})=\frac{1}{2}n^{-3/2}+O(n^{-5/2})$$ and then we have $$\lim_{n \to \infty} (n^{\frac{3}{2}}(\arctan((n+1)^{\frac{1}{2}})- \arctan((n)^{\frac{1}{2}}))=\frac{1}{2}$$

Answer 2

Remember a trigonometric identity: $$ \arctan a + \arctan b = \arctan\frac{a+b}{1-ab}. $$ After that, you may want to use the fact that $\sqrt{n+1}-\sqrt{n}\to0$ as $n\to\infty$. And then maybe L'Hopital's rule.

The fact about limits that I asserted above can be seen by rationalizing the numerator: $$ \sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}. $$

Answer 3

Write

$$\tan \left(\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) \right) = \frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n(n+1)}} = \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}$$

Thus $$\arctan\left( (n+1)^{1\over2}\right) - \arctan\left( n^{1\over2}\right) = \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) $$

And $$ \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}} \sim \frac{n^{1\over 2} \left( 1 + \frac{1}{2n} + \mathcal{O}\left(\frac{1}{n^2} \right) -1\right)}{n} \sim \frac{1}{2n^{3\over2}}$$

So, since $\arctan x \underset{x \rightarrow 0}{\sim} x$,

$$ \arctan \left( \frac{\sqrt{n}\left(\sqrt{1+{1\over n}}-1\right)}{1+\sqrt{n(n+1)}}\right) \sim \frac{1}{2n^{3\over2}} $$

And your limit is $1 \over 2$.

Answer 4

Here is another approach, just compute the Taylor series of $\arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}$ at the point $x=\infty$

$$ \arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}}= \frac{1}{2}\, \frac{1}{x^{3/2}}-\frac{7}{8}\, \frac{1}{x^{5/2}}+O \left( \frac{1}{x^{7/2}} \right). $$

So the above expansion gives

$$ n^{3/2}(\arctan((n+1)^{\frac{1}{2}}) - \arctan((n)^{\frac{1}{2}})= \frac{1}{2}\, -\frac{7}{8}\, \frac{1}{n}+O \left( \frac{1}{n^{2}} \right)$$ $$ \implies \lim_{n\to \infty} n^{3/2}( \arctan((x+1)^{\frac{1}{2}})- \arctan((x)^{\frac{1}{2}})=\frac{1}{2}. $$