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Suppose $a_1>0$ and $(a_n)$ is defined by the following expression:

$$a_{n+1}=e^{a_n}-1,\forall n\in\mathbb{N}$$

Prove that $(a_n)$ either converges to a finite limit or $\lim\limits_{n\to\infty} a_n=\pm \infty$ and find $\lim\limits_{n\to\infty} a_n$

Proof: it is easy to show that $e^x>x+1$ for all $x>0$. I also proved that $a_n>0$ for all $n\in\mathbb{N}$ and that $(a_n)$ is increasing. This already means that $(a_n)$ either converges to a finite limit or $\lim\limits_{n\to\infty} a_n=\pm \infty$. Now suppose $(a_n)$ is bounded. Then $(a_n)$ converges and $L:=\lim\limits_{n\to\infty} a_n=\sup \{a_n | n\in\mathbb{N}\}$. Because $e^x$ is continuous everywhere, $\lim\limits_{n\to\infty} a_{n+1}=e^{\lim\limits_{n\to\infty}a_n}-1$ or $L=e^L-1$. It is easy to see that $L=0$ is the only solution. Thus $\sup \{a_n | n\in\mathbb{N}\}=0$ which contradicts the fact that $a_1>0$. So $(a_n)$ is not bounded, and because it is increasing $\lim\limits_{n\to\infty} a_n= \infty$.

Is my proof correct?

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  • $\begingroup$ Looks fine to me $\endgroup$ Jul 16, 2015 at 20:17

1 Answer 1

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Another way.

$a_{n+1} =e^{a_n}-1 \ge (1+a_n+a_n^2/2)-1 =a_n(1+a_n/2) $. Therefore $\frac{a_{n+1}}{a_n} > 1+a_n/2 $. Therefore, for any $k$, $\frac{a_{n+k+1}}{a_{n+k}} > 1+a_{n+k}/2 > 1+a_{n}/2 $ since $a_n$ is increasing.

Multiplying these, $\frac{a_{n+k}}{a_{n}} > (1+a_{n}/2)^k > 1+ka_n/2 $ which shows the divergence.

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