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.
 A: 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.
A: 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 $$\lim_{x \to \infty} x_{n}=1$$
A: HINT: $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$
Sequence converge if $\exists \lim\limits_{n \to \infty}{x_n} \quad \text{and} \quad \lim\limits_{n \to \infty}{x_n} \not=\pm\infty$
A: 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. $\lim \bigg(n(\sqrt[n]{e}-1)\bigg)=1$
A: 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.
