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

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?

Heidegger
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    If you're trying to prove it through $\varepsilon$-$N$, you'll probably need to resort to the definition of logarithm, or to some known inequalities that you might have already been given in your textbook or something. Can you share this context? – Git Gud Dec 04 '21 at 18:52
  • @GitGud It's actually an exercise our professor left to us. I was indeed searching for some logarithms inequalities, yet I found nought. – Heidegger Dec 04 '21 at 19:12
  • So there are things you have covered. It's likely your professor gave you enough material to solve this. You need to try to get a sense of what that might be and share it in your question. – Git Gud Dec 04 '21 at 19:22

2 Answers2

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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 $\frac{x-1}{x}\leqslant \ln(x)\leqslant x-1$.

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

xpaul
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