Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

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 $$\lim_{x \to \infty} x_{n}=1$$

share|cite|improve this answer

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

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

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.