Validate the proof that $x_n = {\ln n \over n}$ is bounded by $\ln2$ and find $\sup\{x_n\}$

Question

Let $n\in \mathbb N$ and: $$ x_n = \left\{ \ln n \over n \right\} $$ Prove $x_n$ is bounded by $\ln 2$. Find $\sup\{x_n\}$

Consider nominator and denominator, $\ln n$ is growing slower than $n$ therefore: $$ {\ln n \over n} <1 $$

Suppose:

$$ {\ln n \over n} < \ln2 \iff \ln n < n\ln2 \iff \ln n < \ln{2^n} \iff n < 2^n $$

So $n < 2^n$ which is true for any $n \in \mathbb N$

It it valid? And also for the second part how can I find supremum $x_n$ without using calculus?

Answer 1

if $n\in \mathbb N$

for some $N>0, n>N \implies \frac {\ln n}{n}> \frac {\ln (n+1)}{n+1}$

Or $n^{n+1} > (n+1)^{n}$

It is not true when $n=2$ as $2^3 = 8 < 9 = 3^2$

But it is true for $n > 2$

sup $x_n = \frac {\ln 3}{3}$