Verification of the limit of a succession limn→+∞ln⁡(n+1)2n=0 Procedure. |ln⁡(n+1)2n|<ϵ

Step 1
Let ${\displaystyle {\left(a_n\right)}_{n\ge 1}}$ be a sequence defined by
${\displaystyle a_n=\frac{\ln (n+1)}{cn}}$
${\displaystyle c\in {\mathbb{R}}_{+}^{\times }}$
By definition,
${\displaystyle a_n{\sim }_{n\to +\mathrm{\infty }}ell\iff {\mathrm{\forall }}_{ϵ}>0\mathrm{\exists }N\in \mathbb{N},n\ge N:}$
${\displaystyle |a_n-ell|<ϵ}$
that is,
${\displaystyle \frac{\ln (n+1)}{cn}{\sim }_{n\to +\mathrm{\infty }}0\iff {\mathrm{\forall }}_{ϵ}>0\mathrm{\exists }N\in \mathbb{N},n\ge N:}$
${\displaystyle |\frac{\ln (n+1)}{cn}-0|<ϵ}$
Since that
$|\frac{\ln (n+1)}{cn}-0|=\left|\frac{\ln (n+1)}{cn}\right|\le \left|\frac{n+1}{cn}\right|=\frac{n+1}{cn}$
${\displaystyle <ϵ\iff n>\frac{1}{c(ϵ-\frac{1}{c})}}$
Therefore,
Let ${\displaystyle ϵ>0}$ and ${\displaystyle N\in \mathbb{N}}$ such that ${\displaystyle n>\frac{1}{c(ϵ-\frac{1}{c})}}$ with ${\displaystyle c\in {\mathbb{R}}_{+}^{\times }}$. Then for all ${\displaystyle n\in \mathbb{N}}$ such that ${\displaystyle n\ge N}$, we have
$|\frac{\ln (n+1)}{cn}-0|=\left|\frac{\ln (n+1)}{cn}\right|\le \left|\frac{n+1}{cn}\right|=\frac{n+1}{cn}$
${\displaystyle =\frac{1}{c}+\frac{1}{cn}<\frac{1}{c}+ϵ-\frac{1}{c}=ϵ}$
Note: We know that for all ${\displaystyle x\in {\mathbb{R}}_{+}^{\times }}$ we have that
${\displaystyle \mathfrak{a}ac\{x-1\}\left\{x\right\}\le \ln \left(x\right)\le x-1}$
Reference: For all ${\displaystyle x\in {\mathbb{R}}_{+}^{\times }}$ we have ${\displaystyle \frac{x-1}{x}\le \ln \left(x\right)\le x-1}$

Step 1
First note
${\displaystyle e^t\ge 1+t+\frac{1}{2}{t}^2}$
${\displaystyle t\ge 0}$
Let ${\displaystyle t=\ln (1+n)}$ and then ${\displaystyle n=e^t-1}$
So
${\displaystyle \frac{\ln (1+n)}{n}=\frac{t}{e^t-1}\le \frac{t}{t+\frac{1}{2}{t}^2}=\frac{1}{1+\frac{1}{2}t}}$
${\displaystyle =\frac{2}{2+\ln (1+n)}}$
For ${\displaystyle {\mathrm{\forall }}_{ϵ}>0}$, letting
${\displaystyle \frac{2}{2+\ln (1+n)}<ϵ}$
gives
${\displaystyle n>e^{\frac{2(1-\kappa )}{\kappa }}-1}$
Define
${\displaystyle N=\lfloor e^{\frac{2(1-\kappa )}{\kappa }}\rfloor +1}$
Then when ${\displaystyle n\ge N}$, one has
${\displaystyle |\frac{\ln (1+n)}{n}-0|<ϵ}$
or
$\underset{n\to \mathrm{\infty }}{lim}\frac{\ln (1+n)}{n}=0$