# Calculate:$\lim_\limits{n\to \infty} a_n$ , as $a_{n+1}= e^{a_n}-1$, $a_0>0$.

Calculate:$\lim_\limits{n\to \infty} a_n$ , as $a_{n+1}= e^{a_n}-1$, $a_0>0$.

I Proved that the sequence is monotonically increasing, by depending on the behavior of $f(x)=x$ and $f(x)=e^x$ functions and the fact that $e^x \geq x.$ my intuition says that the limit is $\infty$ but first I've to prove that its not bounded from above. am I on the right direction? and if yes, can u give some suggestions on how to prove that $a_n$ is not bounded?

• The limit depends on where the sequence starts. – user296602 Dec 10 '15 at 19:39
• The value of $a_1$ (or any other term of the sequence) is needed. – ajotatxe Dec 10 '15 at 19:44
• I edited the question. its given that $a_0>0$ – F1sargyan Dec 10 '15 at 19:53
• If the sequence converges to some real vaue $\ell$, then one must have $\ell = e^\ell -1$ by continuity. – Clement C. Dec 10 '15 at 19:53
• this equality holds only when $L=0$, but can we say that its not $0$ because $a_n>0$ and its monotonically increasing? – F1sargyan Dec 10 '15 at 19:56

The sequence converges to $0$ if $a_0 \le 0$ and diverges to $+\infty$ if $a_0 > 0$.

In general, if you have a sequence $b_n$ determined recursively by a continuous strictly increasing function $g : \mathbb{R} \to \mathbb{R}$ start from some point $b$.

$$b_{n} = \begin{cases}g(b_{n-1}), & n > 0\\ b, & n = 0\end{cases}$$

Let $\mathcal{F} = \{ x \in \mathbb{R} : g(x) = x \}$ be the set of fixed points for $g$. There are three cases.

1. $g(b) = b$, then $b_n = b$ for all $n$.

2. $g(b) > b$, then $b_n$ is monotonic increasing. If $\mathcal{F} \cap (b,\infty) \ne \emptyset$, then $b_n$ converges to the smallest $x \in \mathcal{F}$ greater than $b$. The basic reason is $b_n$ will be bounded from above by this smallest $x$ and if $b_n$ converges, it need to converge to a fixed point because $g$ is continuous. If $\mathcal{F} \cap (b,\infty) = \emptyset$, then $b_n$ diverges to $+\infty$.

3. $g(b) < b$, then $b_n$ is monotonic decreasing. If $\mathcal{F} \cap (-\infty,b) \ne \emptyset$, then $b_n$ converges to the largest $x \in \mathcal{F}$ smaller than $b$. Otherwise, $b_n$ diverges to $-\infty$.

In short, $b_n$ converges to the nearest fixed point in the "right" direction. If such fixed point doesn't exists, then $b_n$ diverges.

If you apply this to $f(x) = e^x - 1$ and uses the facts $f(x) \ge x$ and equality holds when and only when $x = 0$, you can deduce the assertions I make in the beginning of this answer easily.

If $a_n>0$ for any $n\in\Bbb N$ then $$\frac{a_{n+1}}{a_n}=\frac{e^{a_n}-1}{a_n}>\frac{1+a_n-1}{a_n}=1$$ so the sequence increases.

Furthermore, $$a_{n+1}-a_n>1+a_n+\frac{a_n^2}2-1-a_n=\frac{a_n^2}2>\frac{a_0^2}2>0$$ Hence, $$a_{n}>a_0+n\frac{a_0^2}2\to\infty$$

• So it doesn't matter if $0<a_0<1$? – F1sargyan Dec 10 '15 at 20:07