# Proving a sequence converges to 0.

I am given a sequence $$x_n$$ with $$n\in\mathbb N$$, and knowing that $$x_0 > 0$$ and $$x_{n+1} = f(e^{x_n})$$, where $$f(x) = \frac{\ln{x}}{\sqrt{x}},$$ I need to show that $$x_n$$ converges to $$0$$.

My attempt:

Knowing that $$x_0 > 0$$, I showed that $$x_0 < x_1$$, for all $$x_0 > 0$$, and by induction $$x_{n+1} < x_n$$, so $$x_n$$ is strictly decreasing. Then, $$\lim\limits_{n\to \infty} x_n = \lim\limits_{n\to \infty} x_{n-1} = L,$$ and after the algebra part I end up with $$L = \frac{L}{\sqrt{e^L}}\implies \sqrt{e^L} = 1 \implies L = 0,$$ so $$x_n$$ converges to $$0$$.

Is this a legit proof, or am I missing something?

• So in short $x_{n+1}=\frac{e^{x_n/2}}{x_n}$ – Hagen von Eitzen May 3 '19 at 12:10
• You see to have a few confusions about how to properly use LaTeX/MathJax. This page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. – Brian May 3 '19 at 12:14
• How did you end up with $L = \frac{L}{\sqrt{e^L}}$? From what you wrote, $L = \frac{\sqrt{e^L}}{L}$, no? $L>0$ – Jakobian May 3 '19 at 12:15
• I wrote down the wrong function. It's actually the reciprocal of what I wrote. Sorry for the confusion. – Gigel May 3 '19 at 12:37

We have $$\tag1x_{n+1}=\frac{x_n}{e^{x_n/2}}$$ hence $$x_n>0$$ for all $$n$$. Therefore $$e^{x_n/2}>1$$ and this the sequence is strictly decreasing. Being a bounded monotonic sequence, it must have a limit $$L$$. By continuity, that limit must be a fixpoint of the transformation $$(1)$$, i.e., $$\tag2 L=\frac L{e^{L/2}}.$$ We can readily transform this to $$L(e^{L/2}-1)=0$$ and from this readthat $$L=0$$ or $$e^{L/2}=1$$ (which also implies $$L=0$$). We conclude that $$L=0$$, i.e., $$x_n\to 0$$.