let $\alpha >0$, and we define a sequence s.t $a_1 = \alpha $, and: $a_{n+1} = \sqrt{e^{a_n}-1}$, Prove that $a_n \rightarrow \infty$ 

let $\alpha >0$, and we define a sequence s.t $a_1 = \alpha $ 
and: $$ a_{n+1} = \sqrt{e^{a_n}-1}  $$ 
Prove that $a_n \rightarrow \infty$.


can I say that $e^x = 1 + x + \frac{x^2}{2!} + R_2(x) $ and: 
$e^{a_n} = 1 + a_n + \frac{a_n^2}{2!} + R_2(x) $
then : $e^{a_n} -1 > a_n + \frac{a_n^2}{2!}$
which means that $a_{n+1} > \sqrt{a_n + \frac{a_n^2}{2}}$
and then conclude from here that $a_n \rightarrow \infty$ ? 
 A: Hint: The derivative of $\sqrt{e^{x} - 1}$ is $e^x / (2\sqrt{e^x - 1})$ which is greater than or equal to $e^x/2$ if $e^x < 2$. Thus you will eventually have $e^x - 1 \geq 1$ when you iterate your sequence. Can you see how to complete the proof from there?
A: First, note that $\frac{2\sqrt{6}}{6}> \frac{3}{5}$. (Any number between   $\frac{1}{2}$ and $\frac{2\sqrt{6}}{6}$ would do here.)
Next, for all $x>0$,
$$\frac{1}{x}+\frac{x}{6} \geq \frac{2\sqrt{6}}{6} > \frac{3}{5}$$ since that function of $x$ has its minimum at $x=\sqrt{6}$. 
Now look at the ratio $\frac{a_{n+1}^2}{a_n^2}$ where $a_n >0$:
$$
\frac{a_{n+1}^2}{a_n^2} = \frac{e^{a_n}-1}{a_n^2} > \frac{1+a_n + \frac{a_n^2}{2}+ \frac{a_n^3}{6}-1}{a_n^2}$$
$$=\frac{1}{2} + \left( \frac{1}{a_n} + \frac{a_n}{6} \right) > \frac{1}{2}+\frac{3}{5} = \frac{11}{10}
$$
So each $a_{n+1}^2 > \frac{11}{10}  a_n^2$.  So for all $n$
$$
a_{n} > \alpha \left(\sqrt{\frac{11}{10}}\right)^{n-1}$$ which clearly goes to infinity for any positive value of $\alpha$.
A: Define $f(x) = \sqrt{e^{x} - 1}$, for $x > 0$. $f$ is an increasing, continuous function, since $e^{x} - 1$ and $\sqrt{x}$ both are. Hence $\{a_i\}_{i \in \mathbb{N}}$ is a monotone sequence- either always increasing or always decreasing. We are done if we can show that $f(x) > x$ for all $x > 0$- this would imply that the sequence is always increasing, and cannot be bounded above, since then it would have a limit $L$ which would be a fixed point of $f$.
Note that, for $x > 0$, $f(x) > x$ iff $e^{x} > x^{2} + 1$. Let $h(x) = e^{x} - x^{2} - 1$- we want to show $h(x) > 0$ for all $x > 0$. We have $h(0) = 0$, and $h'(x) = e^{x} - 2x$. For $x \geq 0$, $h'(x) \geq (1 + x + x^2/2) - 2x = 1 - x + x^{2}/2$, which is a nonnegative quadratic (its discriminant is $-1$), so $h'(x) > 0$. That means $h$ is an increasing function on $[0,\infty)$, which proves $h(x) > 0$ for $x > 0$.
