Does the sequence converges?

Question

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.

$$n(e^{\frac{1}{n}}-1)=n(1+\frac{1}{n}+O(\frac{1}{n^2})-1)=1+O(\frac{1}{n})$$ — Ethan Splaver, Apr 18 '13 at 06:58

score 2 · Answer 1 · answered Apr 18 '13 at 05:28

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, $$\lim_{n \to \infty}n(e^{1/n}-1) = \lim_{x \to 0} \dfrac{x}{\log(1+x)} = 1$$

score 1 · Answer 2 · answered Apr 18 '13 at 05:22

1

It is almost finished! Recall the definition of the derivative. Note that the limit you wrote down, apart from being one-sided, is the derivative of $e^x$ at $x=0$.

answered Apr 18 '13 at 05:22

André Nicolas

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Does the sequence converges?

2 Answers2