# Does this sequence $x_{n}=(\sqrt[n]{e}-1)\cdot n$ converge?

Does the sequence defined by $$x_{n} =\left(\sqrt[n]{e}-1\right)\cdot n$$ converge.

For finding the limit one has to solve for $\displaystyle\lim_{x \to \infty} x_{n}$ which I think I can solve, but how do I prove that it converges/diverges.

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It often happens that finding the limit is in itself a proof of its existence (if you don't use tricky methods that assume convergence implicitly). If you could show us how you find $\lim x_n$, then we'd see if you have already proved convergence without realizing it. –  Dan Shved Apr 16 '13 at 4:48
@DanShved: OK. What we have is $\lim_{n\to\infty} \frac{e^{1/n}-1}{1/n} \to 1$ since this is same as $\displaystyle \lim_{h \to 0} \frac{e^{h}-1}{h} = \frac{1+h+h^{2}/2! + \cdots -1}{h} = 1$ –  hint Apr 16 '13 at 4:50
@DanShved: How did u get it means i don't understand, are u giving the hint by saying, this sequence must be decreasing/increasing and bounded above by $1$ and hence should convege to $1$. –  hint Apr 16 '13 at 4:53
Nevermind, I said that before I saw your second comment. –  Dan Shved Apr 16 '13 at 4:54
There you go: you've proved that $\lim x_n = 1$, which automatically means that $x_n$ converges. –  Dan Shved Apr 16 '13 at 4:55

Set $x=\frac{1}{n}$. Then the limit becomes $$\lim_{n\to\infty}\dfrac{\sqrt[n]{e}-1}{\frac{1}{n}}=\lim_{x\to0}\dfrac{e^x-1}{x}.$$ Can you proceed?
Hint: Derivatives.

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You have: $$x_{n} =\left(\sqrt[n]{e}-1\right)\cdot n$$ Expanding $\sqrt[n]{e}$ in series will give: $$1+\frac{1}{n}+\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ So $$\left(\sqrt n{e}-1\right)\cdot n=\left(\frac{1}{n}+\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)\right)\cdot n=1+\frac{1}{2 n}+O\left(\left(\frac{1}{n}\right)^3\right)\cdot n$$ And $$\lim_{x \to \infty} x_{n}=1$$

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HINT: $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$

Sequence converge if $\exists \lim\limits_{n \to \infty}{x_n} \quad \text{and} \quad \lim\limits_{n \to \infty}{x_n} \not=\pm\infty$

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Here is an approach purely based on the ideas of sequences . Let $x_n=\bigg(1- \dfrac1n\bigg)^n$ , then $x_{n+1}=\bigg(1- \dfrac1{n+1}\bigg)^{n+1}=\dfrac 1{\bigg(1+ \dfrac1n\bigg)^n\bigg(1+ \dfrac1n\bigg)}$ , so $(x_{n+1})$ is convergent hence $(x_n)$ is

convergent and $\lim (x_n)=\lim (x_{n+1})=\dfrac1e$. Now it is easy to prove by A.M.-G.M. inequality that

$\bigg(1- \dfrac1{n+1}\bigg)^{n+1}>\bigg(1- \dfrac1n\bigg)^n , \forall n>1$ , hence $(x_n)$ is increasing so

$\dfrac1e=\lim (x_n)=$sup{ $x_n : n \in \mathbb N$ }$≥ x_{n+1}=\dfrac 1{\bigg(1+ \dfrac1n\bigg)^{n+1}} , \forall n\in \mathbb N$ i.e. $\bigg(1+ \dfrac1n\bigg)^{n+1}≥e \space, \forall n \in \mathbb N$ .

Moreover , $e=\lim \bigg(1+\dfrac1n\bigg)^n=$sup {$\bigg(1+\dfrac1n\bigg)^n: n\in \mathbb N$ }$≥\bigg(1+ \dfrac1{n+1}\bigg)^{n+1} , \forall n \in \mathbb N$ . So we

get $\bigg(1+ \dfrac1n\bigg)^{n+1}≥e ≥\bigg(1+ \dfrac1{n+1}\bigg)^{n+1} , \forall n \in \mathbb N \implies \dfrac1n≥ \sqrt[n+1]{e}-1≥\dfrac1{n+1} , \forall n \in \mathbb N$

$\implies 1+\dfrac1n ≥ (n+1)(\sqrt[n+1]{e}-1)≥1 , \forall n \in \mathbb N$. So by squeeze theorem ,

$\lim \bigg((n+1)(\sqrt[n+1]{e}-1)\bigg)=1$ i.e. $\lim \bigg(n(\sqrt[n]{e}-1)\bigg)=1$

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Without using any information about the number $e$, it is possible to prove that if $x>0$ then the sequence $n(\sqrt[n]{x}-1)$ converges. The proof is based on algebraic inequalities and given in my post (see topic "Logarithm as a Limit"). Main idea is to show monotone and bounded nature of the sequence.

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