Does $f(10^n)$ begin every time by $n$ $9$ after the comma? Trying to find a formula to give me the rank of people in a game, I found for small values that if $$f=\frac{2}{\pi} \arctan$$ then $f(10^n)$ always become by $n$ $9$ after the comma if $n \ge 1$. It surprises me, I wouldn't expect that and I'm curious to know how should I prove it ! 
Ex :
$$\begin{align}&f(10)=0.936...\\
 &f(100)=0.9936...\\
 &f(1000)=0.99936...\\\end{align}$$
Any idea or hint to prove that is always the case (or not) would be appreciate ! 
Thank you in advance
PS : I have no idea for the tags
 A: Note that $\arctan x = \frac{\pi}{2}-\arctan(1/x)$. So $\arctan 10^n = \frac{\pi}{2}-\arctan 10^{-n}$.
Finally, it helps to know the power series for $\arctan$: 
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots\text{ for }|x|<1$$
So $$\frac{2}{\pi}\arctan(10^n)=1-\frac{2}{\pi}10^{-n} +\frac{2}{\pi}\frac{10^{-3n}}{3}-\frac{2}{\pi}\frac{10^{-5n}}{5}\dots$$
So the first $n+1$ digits agree with $1-\frac{2}{\pi}10^{-n}$.
A: Your observation can be written as
$$\left|\frac{2}{\pi}\arctan(10^n) - 1 \right| < \frac{1}{10^n}$$
To prove it, let's prove the stronger fact that for all $x > 0$
$$\left|\frac{2}{\pi}\arctan(x) - 1 \right| < \frac{1}{x}$$
Using that $\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}$ for all $x > 0$, we get
$$\left|\frac{2}{\pi}\arctan(x) - 1 \right| = \left|\frac{2}{\pi}\arctan\frac{1}{x} \right|$$
And since we have $\arctan(u) < u$ for all $u > 0$ (this follows from concavity of $\arctan$ on $[0,+\infty[$, draw a picture to see it). Setting $u = \frac{1}{x} > 0$, we finally get the expected inequality
$$\left|\frac{2}{\pi}\arctan(x) - 1 \right| = \left|\frac{2}{\pi}\arctan\frac{1}{x} \right| < \frac{2}{\pi x} < \frac{1}{x}$$
