Prove that, $\frac{n}{\ln(n)}<\frac{n-2}{2}$ for $n>12$.

Question

Prove that, $$\frac{n}{\ln(n)}<\frac{n-2}{2}$$ for $n>12$.
Or is this something that is really obvious?

Answer 1

We have

$$\frac{n}{\ln(n)}<\frac{n-2}{2} \iff\ln n>\frac{2n}{n-2}$$

and for $x>2$

• $f(x)=\ln x$ is increasing
• $g(x)=\frac{2x}{x-2} \implies g'(x)=-\frac4{(x-2)^2}$ is decreasing

then it suffices to check the case $n=13$.

Answer 2

If $f(x)=\frac x{\log x}$, then $f'(x)\leqslant\frac14<\frac12$ for each $x>1$. And if you$g(x)=\frac{x-2}2$, you have $g'(x)=\frac12>f'(x)$. So, since$$f(13)\simeq5.07<\frac{11}2=g(13),$$you have $x\geqslant13\implies f(x)<g(x)$.

Answer 3

If a calculator is allowed, you can easily check that when $n>12$

$\ln{n}>\ln12>2.4$, thus, we have $\frac{n}{\ln{n}}<\frac{n}{2.4}$. Now, it's easy to check $\frac{n}{2.4}<\frac{n-2}{2}$ when $n>12$

Answer 4

Consider that you look for the zero of function $$f(x)=\frac{x}{\log (x)}-\frac{x-2}{2}$$ which is decreasing.

Make a Taylor expansion around $x=e^2$ to get $$f(x)=1-\frac{1}{4} \left(x-e^2\right)+O\left(\left(x-e^2\right)^3\right)$$ Ignoring the higher order terms, then the approximation $$x =4+e^2 =11.3891 \implies n=12$$ Just to be sure, do it again around $x=e^3$ to get $$f(x)=\left(1-\frac{e^3}{6}\right)-\frac{5}{18} \left(x-e^3\right)-\frac{\left(x-e^3\right)^2}{54 e^3}+O\left(\left(x-e^3\right)^3\right)$$ giving another approximation $$x=\frac{1}{2} \left(\sqrt{216 e^3+189 e^6}-13 e^3\right)=11.3828$$

Using Newton method, the exact solution would be $x=11.3395$