# Verification of the limit of a succession

analysiscalculusepsilon-deltalimitsproof-explanation

$$\lim_{n\to +\infty} \frac{\ln(n+1)}{2n} = 0$$

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?

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