Verification of the limit of a succession

Let $(a_{n})_{n\geqslant 1}$ be a sequence defined by $$a_{n}=\frac{\ln(n+1)}{cn},\quad c\in \mathbb{R}_{+}^{*}$$ By definition, $$a_{n}\underset{n\to +\infty}{\sim }\ell \iff \forall \varepsilon>0\exists N\in \mathbb{N}, n\geqslant N:\quad |a_{n}-\ell|<\varepsilon. $$ that is, $$\frac{\ln(n+1)}{cn}\underset{n\to +\infty}{\sim }0 \iff \forall \varepsilon>0\exists N\in \mathbb{N}, n\geqslant N:\quad \left|\frac{\ln(n+1)}{cn}-0\right|<\varepsilon. $$ Since that $$\left|\frac{\ln(n+1)}{cn}-0\right|=\left|\frac{\ln(n+1)}{cn} \right|\leqslant \left|\frac{n+1}{cn}\right|=\frac{n+1}{cn}<\varepsilon \iff n>\frac{1}{c(\varepsilon-\frac{1}{c})}$$

Therefore,

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

$$\left|\frac{\ln(n+1)}{cn}-0 \right|=\left|\frac{\ln(n+1)}{cn} \right|\leqslant \left| \frac{n+1}{cn}\right|=\frac{n+1}{cn}=\frac{1}{c}+\frac{1}{cn}<\frac{1}{c}+\varepsilon-\frac{1}{c}=\varepsilon.$$

Note: We know that for all $x\in \mathbb{R}_{+}^{*}$ we have that $$\frac{x-1}{x}\leqslant \ln(x)\leqslant x-1.$$ Reference: For all $x\in \mathbb{R}_{+}^{*}$ we have $\frac{x-1}{x}\leqslant \ln(x)\leqslant x-1$.

First note $$ e^t\ge1+t+\frac12t^2, t\ge0 $$ Let $t=\ln(1+n)$ and then $n=e^t-1$. So $$ \frac{\ln(1+n)}{n}=\frac{t}{e^t-1}\le\frac{t}{t+\frac12t^2}=\frac{1}{1+\frac12t}=\frac{2}{2+\ln(1+n)}. $$ For $\forall \varepsilon>0$, letting $$ \frac{2}{2+\ln(1+n)}<\varepsilon $$ gives $$ n>e^{\frac{2(1-\varepsilon)}{\varepsilon}}-1. $$ Define $$ N=\lfloor e^{\frac{2(1-\varepsilon)}{\varepsilon}}\rfloor+1 $$ Then when $n\ge N$, one has $$ \bigg|\frac{\ln(1+n)}{n}-0\bigg|<\varepsilon $$ or $$ \lim_{n\to\infty}\frac{\ln(1+n)}{n}=0. $$