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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?

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    $\begingroup$ So in short $x_{n+1}=\frac{e^{x_n/2}}{x_n}$ $\endgroup$ – Hagen von Eitzen May 3 '19 at 12:10
  • $\begingroup$ 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. $\endgroup$ – Brian May 3 '19 at 12:14
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    $\begingroup$ How did you end up with $L = \frac{L}{\sqrt{e^L}}$? From what you wrote, $L = \frac{\sqrt{e^L}}{L}$, no? $L>0$ $\endgroup$ – Jakobian May 3 '19 at 12:15
  • $\begingroup$ I wrote down the wrong function. It's actually the reciprocal of what I wrote. Sorry for the confusion. $\endgroup$ – Gigel May 3 '19 at 12:37
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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., $x_n\to 0$.

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  • $\begingroup$ My mistake, I wrote the reciprocal of the actual function. Corrected it now. $\endgroup$ – Gigel May 3 '19 at 17:10

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