# Finding the limit of recursive sequence

Assuming that the solution of $e^{-x}=x$ is $c\in (0,1)$

And give the following sequence

$$a_{n+1}=e^{-a_n}$$

$$a_1=1$$

How can i prove that the sequence converge and that the limit is $$\lim_{n \to \infty}a_n=c$$

I tried to define a new sequence $$b_n=a_n-c$$ and to prove it using the ratio test but without success

Exponential function is continuous,then

$$\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}e^{-a_n}=e^{-\lim_{n\to\infty}a_n}.$$

So the limit of this sequence is the solution of $x=e^{-x}$ which is $c$.

To see the existence, first note by induction that $1\geq a_n\geq 0$. Now we claim that subsequence with odd terms decreasing and subsequence with even terms increasing. For $k=1$, $a_{2k-1}\geq a_{2k+1}$ is immediate and to see $a_{2k}\leq a_{2k+2}$ note that $1\geq a_n$ and $e$ is monotone. Thus, $a_2=e^{-1}\leq a_4=e^{-a_3}.$

Suppose now for some $k$, $a_{2k-1}\geq a_{2k+1}$ and $a_{2k}\leq a_{2k+2}$. Then

$$e^{-a_{2k}}\geq e^{-a_{2k+2}}\implies a_{2k+1}\geq a_{2k+3},\\ e^{-a_{2k+1}}\leq e^{-a_{2k+3}}\implies a_{2k+2}\leq a_{2k+4}.$$

This tells us that both $\lim_{k\to\infty}a_{2_k}=l_0$ and $\lim_{k\to\infty}a_{2k-1}=l_1$ exist by monotone convergence theorem. Now we claim that every odd term is bigger than every even term. Trivially for $k=1$, this claim holds. Suppose now that it is true for some $k$, $a_{2k-1}\geq a_{2k}$. Then

$$a_{2k+1}=e^{-a_{2k}}\geq e^{-a_{2k-1}}=a_{2k}\implies e^{-a_2k}=a_{2k+1}\geq e^{a_{2k+1}}=a_{2k+2}$$ which shows that the claim is true for $k+1$ and induction is completed. So we show that $a_{2k}\leq a_{2k+2}\leq a_{2k+1}\leq a_{2k-1}$. Note also that $l_0$ is supremum and $l_1$ is infimum for corresponding sequences. Then $l_1\geq l_0$.

By induction one can show $a_{2k+1}=a_{2k}^{1/e}\geq a_{2k+1}^{1/e}=a_{2k+2}$. Taking limit on both sides gives us $l_0^{1/e}\geq l_1^{1/e}$ and $l_0\geq l_1$. Thus $l_0=l_1=l$ and $a_k\to l$.

• If the limit exists, it must be $c$. Showing the existence of the limit is the less easy task. Feb 13 '16 at 21:37

$f(x) = \mathrm{e}^{-x}$ is a contraction in the relevant range $(0, 1]$, thus $x_{n + 1} = f(x_n)$ converges if started inside that range.

• Can you elaborate or provide some reference? I undertand contraction is a kind of coordinate transform, how does it work here? Feb 13 '16 at 22:40
• @YuriyS Check this document. The relevant theorem is 3.6. You can also search for Banach Fixed Point Theorem. Feb 14 '16 at 12:49

Write $e^y=\exp(y)$ - it's easier to write recursive sequence:

$$a_n=\exp(-\exp(-\exp(-...\exp(-1)))))$$

Then you can write:

$$x=\lim_{n \rightarrow \infty} a_n=\exp(-x)$$

Since just one step of recursion will not change the limit of infinite sequence.

The rest is clear.

As for proving the limit exists - every term in the sequence is positive, and every term is bounded.

$$0 \leq a_n \leq 1$$

The sequence is actually not monotonic (as can bee seen by numerical experiment $a_3=0.500, a_4=0.606, a_5=0.545, a_6=0.580$ etc), so the proof of convergence is tricky

• You assume that the limit exist,but how do you prove it exist? Feb 13 '16 at 21:42
• See the edit I made. We start with 1 and you can see that the sequence is bounded. That's a start, not the whole proof of course Feb 13 '16 at 21:55