How can I prove this sequence is decreasing from $n=3$ without using derivatives?

Question

Consider the sequence $\{\frac{\log n}{n}\}_{n\geq 1}$.

I tried to prove that $\frac{\log(n+1)}{n+1}<\frac{\log(n)}{n}$, but I can't find out from which number $n$ the inequality is satisfied.

Answer 1

The inequality $$\frac{\log(n+1)}{n+1}<\frac{\log(n)}{n}$$ is equivalent to $$\left(1+\frac{1}{n}\right)^n<n$$ which holds for $n\geq 3$ because $\left(1+\frac{1}{n}\right)^n<e$. Note that it does not hold for $n=1,2$.

P.S. Alternative ending.

$$\left(1+\frac 1 n\right)^n=1+1+\binom n 2\frac 1 {n^2}+\binom n 3 \frac 1 {n^3}+\dotsb+\frac 1 {n^n}\\ \leq 1+1+\frac{1}{2!}+\frac 1{3!}+\dotsb+\frac 1 {n!}<1+1+\frac 1 2 +\frac{1}{2^2} +\frac{1}{2^3}+\dots +\frac{1}{2^n}<3.$$

Answer 2

for $ n> 2 $, $ n^{n+1} > (n+1)^n$. From there on you could use log and continue from where you left.