Does the sequence converges? I am trying to prove if the sequence 
$a_n=(\root n\of e-1)\cdot n$ is convergent.  I know that the sequences $x_n=(1+1/n)^n$ and $y_n=(1+1/n)^{n+1}$ tends to the same limit which is $e$. Can anyone prove if the above sequence $a_n$ is convergent? and if so, find the limit.
My trial was to write $a_n$ as $a_n=n(e^{1/n}-1)$ and taking $1/n=m$ so that $a_n=\frac{1}{m}(e^m-1)$ and taking the limit $\lim_{x\to 0^+}\frac{e^x-1}{x}$, but I don't know how to continue.
Thanks to every one who solve this for me. 

Let $e^{1/n}-1 = x$. We then have $\dfrac1n = \log(1+x) \implies n = \dfrac1{\log(1+x)}$. Now as $n \to \infty$, we have $x \to 0$. Hence,
$$\lim_{n \to \infty}n(e^{1/n}-1) = \lim_{x \to 0} \dfrac{x}{\log(1+x)} = 1$$
