# Proving that a sequence defined by $a_{n+1}=e^{a_n}-1$ converges to finite number or infinity

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?

• Looks fine to me Jul 16, 2015 at 20:17

## 1 Answer

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.