Prove: For every $\epsilon>0$, $100 n \log_2 n \leq \epsilon n^2$ for every sufficiently large $n$.

Question

Here's what I tried: When $n$ is large enough, $n$ must be greater than 0, so both sides of the above inequality can be reduced by one $n$. The result after reduction is: $$ 100 \log_2 ⁡n \leq \epsilon n.$$ By moving $n$ on the right side of the inequality to the left, we get $$\frac{100 \log_2 ⁡n}{n} \leq \epsilon.$$ Then I didn't know how to continue.

Can you see how it's equivalent to proving $\lim_{x\to\infty} \frac{x}{2^x} =0$? — Chris Lewis, Jun 22 '24 at 08:37
To Claude Leibovici: I cannot see the relationship between Lambert function and my question. — Wei-Cheng Liu, Jun 22 '24 at 09:10

score 1 · Accepted Answer · answered Jun 22 '24 at 10:23

Let $\varepsilon \mapsto C \varepsilon $ where $C = 100 / \ln \left( 2 \right)$. See we are proving that

$$\frac {\ln \left( n \right)}{n} < \varepsilon$$

for sufficiently large $n$. To this end, define $x = \ln \left( n \right)$ and see

$$\frac {x}{{e}^{x}} = \frac {x}{\sum_{k \ge 0} {x}^{k} / k!} < \frac {x}{{x}^{2} / 2} = \frac {2}{x}.$$

Let $N = 2 / \varepsilon$. For all $n > {e}^{N}$, it is known for certain that $\ln \left( n \right) / n < \varepsilon$. $\blacksquare$

Prove: For every $\epsilon>0$, $100 n \log_2 n \leq \epsilon n^2$ for every sufficiently large $n$.

1 Answers1