Bounds of a recurring sequence I have the following sequence
$x_1 > 0$ ; $x_{n+1}= \frac{x_n}{e^{x_n}}$,
and I must find if it is convergent and find its limit in that case. So far, I have proven that it is decreasing, and now I want to find its bounds. What I have at the moment is that the number $x_1=a>0$ will be its upper bound, since the sequence is decreasing. I can see that the sequence will always be between $0$ and $a$, however, I'm stuck with the proof.
Thanks in advance for any help.
 A: Well the sequence converges since it's decreasing and bounded below. Now if $L$ is its limit, then we have the following manipulations:
$$
L = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} \frac{x_n}{e^{x_n}} = \frac{L}{e^L}.
$$
Of course, the limit properties used are justified, since we know the limit of $x_n$ exists. Now, either $L=0$ or $e^L = 1$, which gives $L=0$ as well, so indeed $L=0$.
Edit: I'm not sure why this is downvoted, but I can give more details on the beginning. Induction, as other commentors have pointed out, gives that $x_n > 0$ for all $n$. Thus, $e^{x_n} > 1$ for all $n$. Hence,
$$
x_n > \frac{x_n}{e^{x_n}} = x_{n+1} ,
$$
so the sequence is decreasing. Monotone bounded sequences converge, so the sequence converges and can be said to have limit $L$.
A: The sequence is positive and decreasing. Assume by contrary that there exists some $L>0$ such that for any $n\in\Bbb N$, $x_n\ge L$. Hence
$$
\frac{x_{n+1}}{x_n}=\frac{1}{\exp(x_n)}\le \frac{1}{\exp(L)}
$$
hence
$$
\frac{x_{n+1}}{x_1}\le \exp(-nL)\implies 
x_{n+1}\le \exp(-nL)x_1
$$
implying that $x_n\to 0$ which is a contradiction. Hence $L=0$ and since $x_n$ is decreasing, we have $\lim_{n\to \infty}x_n=0$.
