$$\lim_{n\to +\infty} \frac{\ln(n+1)}{2n} = 0$$
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?
Best Answer
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 $\frac{x-1}{x}\leqslant \ln(x)\leqslant x-1$.