<s>[INST] $x\in [0,1]$ and $n\in N$ $\implies$ $\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$ Let $x\in [0,1]$ and $n\in N$. According to the solutions manual this is true:
$$\left| \frac{1-2xn}{n^2} \right| \leq \frac{1}{\sqrt n}$$
How can I see this is true ?
 [/INST] Since $0 \le x \le 1$,
$0 \ge 1-2xn \ge 1-2n$.
So, $|1-2xn| \le 2n-1$.
We thus want to show that
$\frac{2n-1}{n^2} \le \sqrt{n}$
or $2n-1 \le n^{2.5}$.
Looking at the $2n-1$, I think of
$(n-1)^2 = n^2-2n+1$.
Since $(n-1)^2 \ge 0$,
$0 \le (n-1)^2 = n^2-2n+1$
or $2n-1 \le n^2$
and since $n^2 \le n^{2.5}$ we are done.
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