Step 1

Let $\left({a}_{n}\right)}_{n\ge 1$ be a sequence defined by

$a}_{n}=\frac{\mathrm{ln}(n+1)}{cn$

$c\in {\mathbb{R}}_{+}^{\times}$

By definition,

${a}_{n}{\sim}_{n\to +\mathrm{\infty}}ell\iff {\mathrm{\forall}}_{\u03f5}>0\mathrm{\exists}N\in \mathbb{N},n\ge N:$

$|{a}_{n}-ell|<\u03f5$

that is,

$\frac{\mathrm{ln}(n+1)}{cn}{\sim}_{n\to +\mathrm{\infty}}0\iff {\mathrm{\forall}}_{\u03f5}>0\mathrm{\exists}N\in \mathbb{N},n\ge N:$

$|\frac{\mathrm{ln}(n+1)}{cn}-0|<\u03f5$

Since that

$|\frac{\mathrm{ln}(n+1)}{cn}-0|=\left|\frac{\mathrm{ln}(n+1)}{cn}\right|\le \left|\frac{n+1}{cn}\right|=\frac{n+1}{cn}$

$<\u03f5\iff n>\frac{1}{c(\u03f5-\frac{1}{c})}$

Therefore,

Let $\u03f5>0$ and $N\in \mathbb{N}$ such that $n>\frac{1}{c(\u03f5-\frac{1}{c})}$ with $c\in {\mathbb{R}}_{+}^{\times}$. Then for all $n\in \mathbb{N}$ such that $n\ge N$, we have

$|\frac{\mathrm{ln}(n+1)}{cn}-0|=\left|\frac{\mathrm{ln}(n+1)}{cn}\right|\le \left|\frac{n+1}{cn}\right|=\frac{n+1}{cn}$

$=\frac{1}{c}+\frac{1}{cn}<\frac{1}{c}+\u03f5-\frac{1}{c}=\u03f5$

Note: We know that for all $x\in {\mathbb{R}}_{+}^{\times}$ we have that

$\mathfrak{a}ac\{x-1\}\left\{x\right\}\le \mathrm{ln}\left(x\right)\le x-1$

Reference: For all $x\in {\mathbb{R}}_{+}^{\times}$ we have $\frac{x-1}{x}\le \mathrm{ln}\left(x\right)\le x-1$