Convergence bound for $p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{1}{n})}\right)$

Question

For my engineering-related work, I ended up with the following expression

$p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{1}{n})}\right)$,

where $n\in \left[1, \infty\right)$. Now I am aware that for large $n$, the nominator and thus also the $\text{arctan}$ "quickly" converges to zero.

I can of course visualize that "quick" convergence.

What I am actually interested in is a more comprehensible statement on the order of convergence like "an exponential/quadratic convergence".

I have looked into the Taylor expansion of the $\text{arctan}$ but got lost after that.

Answer 1

$$ p(n) = \arctan(\frac{2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})}}{(2+\frac{1}{n})} ) = \arctan(\frac{(2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}{(2+\frac{1}{n})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}) $$

After using $(a-b)(a+b) = a^2 - b^2 $, we have:

$$ p(n) = \arctan(\frac{\frac{1}{n}\cdot(2+\frac{1}{n})}{(2+\frac{1}{n})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}) = \arctan(\frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}})$$

Now using the taylor expansion of $\arctan$ function (near zero) : $\arctan(x) = x - \frac{x^3}{3} + o(x^3) = x + o(x^2)$

we have: as $n\to \infty$, then $p(n) = \frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}} + o(\frac{1}{n^2}) $

And after a little thinking, as $n \to \infty $, then $\frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}}$ isn't that different from $\frac{1}{4n}$

Formally ($E$ is error)$$E(n) = \frac{1}{n} \cdot \frac{1}{2 + \sqrt{4+\frac{1}{n}\cdot(2+\frac{1}{n})}} - \frac{1}{4n} = \frac{1}{n}(\frac{2 -\sqrt{4+\frac{1}{n}(2+\frac{1}{n})} }{8+4\sqrt{4+\frac{1}{n}(2+\frac{1}{n})}})$$

So similarly (multiplying both sides by $2 +\sqrt{4+\frac{1}{n}(2+\frac{1}{n})}$, we get:

$$ E(n) = -\frac{1}{4n^2}\cdot (2 + \frac{1}{n})\cdot \frac{1}{(2 +\sqrt{4+\frac{1}{n}(2+\frac{1}{n})})^2 }, $$ so after ellaborating a little bit: $E(n) = -\frac{1}{4n^2} + o(\frac{1}{n^2}) $ ( just tackle the error function once again)

SO:

$$ p(n) = \frac{1}{4n} - \frac{1}{4n^2} + o(\frac{1}{n^2})$$