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I have to find the limit : (let $k\in \mathbb{R}$)

$$\lim_{n\to \infty}n^k \left(\Big(1+\frac{1}{n+1}\Big)^{n+1}-\Big(1+\frac{1}{n}\Big)^n \right)=?$$

My Try :

$$\lim_{n\to \infty}\frac{n^k}{\Big(1+\frac{1}{n}\Big)^n} \left(\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}-1\right)$$

we know that :

$$\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}>1$$

now what do i do ?

Jack D'Aurizio
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Fricul38
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3 Answers3

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$$\left(1+\frac{1}{n}\right)^n = \exp\left[n\log\left(1+\frac{1}{n}\right)\right]=e-\frac{e}{2n}+O\left(\frac{1}{n^2}\right) $$ hence $$ \left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n = \frac{e}{2n^2}+O\left(\frac{1}{n^3}\right) $$ and for a fixed $k\in\mathbb{R}$ $$ \lim_{n\to +\infty}n^k\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n\right]$$ is non-trivial only if $k=2$. Otherwise, it is either $0$ or $+\infty$.

Jack D'Aurizio
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  • thank you very much . now how without using from big o notation ? – Fricul38 Dec 24 '17 at 16:04
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    @Fricul38: you may prove the same through the AM-GM inequality and squeezing, but it is an overcomplication from my point of view. What is your issue with Landau's notation? It is very practical. – Jack D'Aurizio Dec 24 '17 at 16:09
  • I prove with AM-GM that : $$n^k \left((1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n \right)>0$$ Now what ? – Fricul38 Dec 24 '17 at 16:22
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    @Fricul38: you need to prove a non-trivial inequality, of course. It is well known that ${(1+1/n)^n}_{n\geq 1}$ is increasing, through the AM-GM or not. You need tight bounds for $$\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n}-1=\left(1+\frac{1}{n+1}\right)\left(1-\frac{1}{(n+1)^2}\right)^n-1,$$ but again, such detour is not needed. – Jack D'Aurizio Dec 24 '17 at 17:04
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$$\lim_{n\to \infty}n^k \left((1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n \right)= \lim_{n\to \infty}n^k \left(\frac{e}{2n^2}+O((\frac{1}{n^3})) \right)$$ for n<2 limit is 0, for n=2 limit is e/2, for n>2 limit is infinity

aid78
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    You should see the other answer which explains how the equation is obtained. Just stating a complicated result is only going to confuse the asker. – Paramanand Singh Dec 25 '17 at 04:27
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Using only the Binomial Theorem and Bernoulli's Inequality: $$ \begin{align} \hspace{-1cm}\left(1+\frac1{n+1}\right)^{n+1}\!\!-\left(1+\frac1n\right)^n &=\left(\frac{n+2}{n+1}\right)^{n+1}-\left(\frac{n+1}n\right)^n\tag{1a}\\ &=\color{#C00}{\left(\frac{n+1}n-\frac1{(n+1)n}\right)^{n+1}}-\color{#090}{\left(\frac{n+1}n\right)^n}\tag{1b}\\ &=\color{#C00}{\left(\frac{n+1}n\right)^{n+1}-(n+1)\left(\frac{n+1}n\right)^n\frac1{(n+1)n}}\\ &\color{#C00}{{}+\frac{(n+1)n}2\left(\frac{n+1}n\right)^{n-1}\left(\frac1{(n+1)n}\right)^2+O\!\left(\frac1{n^3}\right)}\\ &-\color{#090}{\left(\frac{n+1}n\right)^n}\tag{1c}\\ &=\frac{(n+1)n}2\left(\frac{n+1}n\right)^{n-1}\left(\frac1{(n+1)n}\right)^2+O\!\left(\frac1{n^3}\right)\tag{1d}\\ &=\frac1{2(n+1)n}\left(\frac{n+1}n\right)^{n-1}+O\!\left(\frac1{n^3}\right)\tag{1e}\\[3pt] &=\frac{e}{2n^2}+O\!\left(\frac1{n^3}\right)\tag{1f} \end{align} $$ Explanation:
$\text{(1a)}$: combine terms
$\text{(1b)}$: add and subtract $\frac1{(n+1)n}$
$\text{(1c)}$: expand the first three terms of the Binomial expansion
$\text{(1d)}$: the sum of the first two terms of the Binomial expansion
$\phantom{\text{(1d):}}$ equals $\left(\frac{n+1}n\right)^n$
$\text{(1e)}$: simplify
$\text{(1f)}$: $\frac1{(n+1)n}=\frac1{n^2}+O\!\left(\frac1{n^3}\right)$ and $\left(\frac{n+1}n\right)^{n-1}=e+O\!\left(\frac1n\right)$
$\phantom{\text{(1f):}}$ the latter equation uses this answer to show that
$\phantom{\text{(1f):}}$ $\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$ with Bernoulli's Inequality

robjohn
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