1. Prove that if $(a_n)$ is a growing and bounded sequence, $\lim_{n\to\infty}a_n = \sup${{$a_n$}}. DONE
  2. Prove that in the interval $(0,1)$ $e^{-x} = x$ has a single solution. DONE
  3. let $c$ be the single solution from the previous proof. also, let $(a_n)$ be the sequence defined by: $a_1 = 1, a_{n+1} = e^{-a_{n}}$ for every $n$. Prove that $\lim_{n\to\infty}a_n = c$.

How can I prove the third task? The solution will probably make use of the previous proofs, yet I couldn't see how to do so..

  • 1
    $\begingroup$ Do you mean $a_{n+1}=e^{-a_n}$? $\endgroup$ – lulu Feb 20 '17 at 21:28
  • $\begingroup$ Yes, fixed it now, thanks $\endgroup$ – user371583 Feb 20 '17 at 21:29

To prove that $a_n$ is convergent, we'll see that $a_n$ is Cauchy. Let us call $f(x)=e^{-x}$, which satisfies:

  • $f$ is derivable.
  • $f([\frac1e,1])\subset [\frac1e,1]$.

Using the mean value theorem, we have, for every $a,b\in [\frac1e,1]$:

$$|f(b)-f(a)|=|f'(c)|\cdot|b-a| \leq \rho |b-a|$$

as $f'(x)=-e^{-x}$ has its minimum $\rho$ at $x=\frac1e$ and is $\rho \approx 0.7$. For $a=a_n, \ b=f(a_n)=a_{n+1}$, means:

$$|a_{n+2}-a_{n+1}|\leq \rho|a_{n+1}-a_n| \ \forall n\geq 1\qquad (1)$$

Now, it's easy to prove that $a_n$ is Cauchy, so it's convergent. Note that, from (1), we get that $$|a_3-a_2| \leq \rho |a_2-a_1|$$ $$|a_4-a_3| \leq \rho |a_3-a_2|\leq\rho^2 |a_2-a_1|$$ and in general: $$|a_{n+1}-a_{n}| \leq \rho^{n-1} |a_2-a_1| $$ To prove that $a_n$ is Cauchy, pick $n,m\in \mathbb N$. We have:

\begin{equation} \begin{split} |a_{n+m}-a_n|&=|a_{n+m}-a_{n+m-1}+a_{n+m-1}-\dots+a_{n+1}-a_n|\leq \\ &\leq |a_{n+m}-a_{n+m-1}|+\cdots+|a_{n+1}-a_n| \leq \\ &\leq (\rho^{n+m-2}+\rho^{n+m-3}+\cdots+\rho^{n-1})|a_2-a_1| \end{split} \end{equation}

and now, using that $\rho <1$ for the last inequality:

$$\rho^{n+m-2}+\rho^{n+m-3}+\cdots+\rho^{n-1} = \frac{\rho^{n-1}-\rho^{n-m-1}}{1-\rho}=\frac{\rho^{n-1}}{1-\rho}(1-\rho^{m-1}) \leq \frac{\rho^{n-1}}{1-\rho}$$

In the end, $$|a_{n+m}-a_n|\leq \frac{\rho^{n-1}}{1-\rho} |a_2-a_1|$$ and, as $n\to+\infty$, $\rho^n \to 0$ because $\rho<1$, so $|a_{n+m}-a_n|\to 0$.

Now, for the limit, as $L=\lim_n a_n$, we also have $$\lim_n a_{n+1}=L$$ because it's a subsequence of $a_n$. Now, note that $a_{n+1}= f(a_n)$, so we can write $$L=\lim_n a_n=\lim_n a_{n+1} = \lim_n f(a_n) = f(\lim_n a_n)=f(L)$$ so $L$ has to verify $L=e^{-L}$. Since you proved there's only one real number satisfying this, then $L=c$.

  • $\begingroup$ Don't we need to show first that $(a_n)$ is convergence, before marking its limit as $L$? $\endgroup$ – user371583 Feb 20 '17 at 21:49
  • $\begingroup$ It would be better if you include the proof that $a_n$ converges. $\endgroup$ – Sungjin Kim Feb 20 '17 at 21:49
  • $\begingroup$ You prove it already in what you called 1. I could try adding it to the answer if you need it. $\endgroup$ – A. Salguero-Alarcón Feb 20 '17 at 21:49
  • $\begingroup$ This was a general proof for all kind of sequences, not for that specific one. $\endgroup$ – user371583 Feb 20 '17 at 21:50
  • 1
    $\begingroup$ @flytothesurface, $(a_n)$ is not an increasing sequence. You need a little more than 1 to prove the convergence. $\endgroup$ – Sungjin Kim Feb 20 '17 at 21:51

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