2022-01-13

Verification of the limit of a succession
$\underset{n\to +\mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(n+1\right)}{2n}=0$
Procedure.
$|\frac{\mathrm{ln}\left(n+1\right)}{2n}|<ϵ$

braodagxj

Expert

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

Annie Levasseur

Expert

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

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