Methods to approximate $\arctan(n)$ where $n> 2$ is a natural number

Question

I would like to approximate $\arctan(n)$ where $n>2$ is a natural number :

First method

I use integral like this : $$\int_{0}^{\frac{1}{n}}\frac{(1-x^2)^{4n}(x^2)^{4n}}{1+x^2}dx=2^{4n}\arctan{\frac{1}{n}}-\frac{a}{b}$$

Where $a,b$ are positive integer .

Now we can frame the integral remarking that :$\frac{1}{2}\leq \frac{1}{x^2+1}\leq 1$ on $[0;\frac{1}{n}]$

Remains to apply the formula $$\arctan(\frac{1}{n})+\arctan(n)=\frac{\pi}{2}$$

Second method

I use power series of $ \arctan$ at $x=0$:

$$\arctan(x)=x - \frac{x^3}{3} +\frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + O(x^{12}) $$

Third Method

We can use the Shafer-Fink inequality http://files.ele-math.com/abstracts/mia-17-109-abs.pdf .

Fourth Method

Using the fact that $\arctan(x)$ for $x>0$ is concave we have by Jensen's inequality:

$$\arctan(n)\leq 2\arctan(0.5(\frac{1}{n}+n))-\arctan(\frac{1}{n})$$

Well all of this more or less classical so I was wondering if there is other methods geometric by example ?

Any helps is greatly appreciated .

Thanks a lot for your contributions.

Answer 1

The best geometric method: draw a right triangle of sides $1$ and $n$ and mesure the small angle.

Answer 2

Note that

$$ \arctan n = \text{arccot} \frac1n =\frac\pi2- \arctan \frac1n$$

Then, apply your second method to the term $\arctan \frac1n$ to obtain the approximation

$$ \arctan n =\frac\pi2- \frac1n +\frac{1}{3n^3} -\frac{1}{5n^5} + ... $$