Verification of the limit of a succession $$\lim_{n\to +\infty} \frac{\ln(n+1)}{2n} = 0$$
Procedure.
$$\big|\frac{\ln(n+1)}{2n}\big| < \epsilon$$
Due to the fact that $n\to +\infty$ the absolute value is meaningless. Now we can say
The problem is that now I am stuck. I tried with some majorisations like
$$\ln(1+n) \leq 1+n$$ or
$$\ln(1+n) \geq \ln(n)$$
But they did not help me.
Is there a sort of easy way to proceed?
 A: 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 $\frac{x-1}{x}\leqslant \ln(x)\leqslant x-1$.
A: 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
$$ \lim_{n\to\infty}\frac{\ln(1+n)}{n}=0. $$
