Let $f : \mathbb{R} \rightarrow \mathbb{R}$ with:

$$f(x) = \frac{1}{x^2}e^{\frac{1}{x}}$$

We consider the sequence $x_n$, having $x_0 \in \left ( 0, \frac{1}{2} \right )$ and $x_{n+1} = f(\frac{1}{x_n})$, for any $n \in \mathbb{N}$.

Prove that $x_n$ is convergent and find its limit.

So far, I've only found that $x_n > 0$ $\forall n \in \mathbb{N}$. I have no idea what I should do next.

Thank you in advance.

  • $\begingroup$ Can you write out $x_{n+1}$ explicitly in terms of $x_n$ (i.e., without mentioning $f$, but only using polynomials and exponentials)? That'll let you write $x_{n+1} = g(x_n)$ for some other function $g$, whose properties you can perhaps analyze. Can you, for instance, put bounds on $x_1$? You've said it's positive. What's the largest it could possibly be? Do the same for $x_2$, and see if you start to notice anything. Also: graph $g$. $\endgroup$ – John Hughes Jul 11 '16 at 16:26


  • First, $f$ is unnecessarily confusing, because of the reciprocals. Write instead $g\colon(0,\infty)\to\mathbb{R}$ with $g(x) = x^2 e^x$ (i.e., $f(x) = g(\frac{1}{x})$: your recurrence relation is now $$ x_{n+1} = g(x_n) \qquad n\in\mathbb{N}. $$

  • We now want to prove that $(x_n)_n$ is a decreasing and positive sequence. By monotone convergence, it will converge.

    • Prove that $x\mapsto \frac{g(x)}{x} = x e^x$ is increasing on $(0,\infty)$, and bounded between $0$ and $1$ on $(0,1/2]$. Why? Because then $\frac{x_{n+1}}{x_n} = \frac{g(x_n)}{x_n}$ will be, by induction, in $(0,1)$.

    • Use the above to get that indeed $(x_n)_n$ is a decreasing and positive sequence.

    • Conclude by monotone convergence.

  • Now that you showed convergence, you know there exists $\ell\in[0,1/2]$ such that $x_n \xrightarrow[n\to\infty]{} \ell$. By continuity of the function $g$, this $\ell$ must then satisfy $$ \ell = g(\ell) $$ i.e. $\ell^2 e^\ell = \ell$. The solutions are either $0$, or values $\ell$ such that $\ell e^\ell = 1$. But by the above, $x e^x \in (0,1)$ for all $x\in (0,1/2]$, so there is no solution there... Hence, the only possible limit is $0$.

  • $\begingroup$ To experiment and get an idea of the behavior, you can try to see what the sequence looks like with different initial values. This motivates the proof -- that is, hints that indeed the sequence is decreasing. $\endgroup$ – Clement C. Jul 11 '16 at 16:57
  • $\begingroup$ Can you, please, explain a bit more on the convergence part? I don't really understand what you did. $\endgroup$ – George R. Jul 11 '16 at 17:13
  • $\begingroup$ We want to show that $(x_n)_n$ is decreasing. Since it is a sequence of positive terms, we can either show that (i) $x_{n+1}-x_n < 0$, or that (ii) $\frac{x_{n+1}}{x_n} < 1$. Both would give what we need; it's a bit simpler to show (ii) here, so that's what the above suggests. Now, by definition of $x_{n+1}$, showing (ii) amounts to showing $\frac{g(x_{n+1})}{x_n} < 1$. To do so, we show that $0<\frac{g(x)}{x} < 1$ for any $x\in(0,1/2)$: since $x_0\in (0,1/2)$, this proves that $0< x_1 < x_0 < 1/2$. But then we can perform induction, since $x_1\in (0.1/2)$: $0< x_2< x_1 < 1/2$, and so on. $\endgroup$ – Clement C. Jul 11 '16 at 17:17

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