Here's my question:

Let $a_1>0$ and $a_n$ be a sequence such that: $$a_{n+1}=e^{a_n}-1$$

Prove that $\lim\limits_{n\to\infty}a_n=\infty$.

I have proven that the sequence is monotonic increasing, with the following Lemma.

Lemma $$e^x>x+1$$


With the Lemma, I can write that sequence like that:

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

Therefore the sequence is monotonic increasing.

How do I prove it's unbounded? If I do prove it's unbounded, then its limit would be $\infty$.




Using Lemma $1$, $e^x>1+x$ for $x>0$, we have

$$a_{n+1}-a_n=e^{a_n}-1-a_n>0\implies a_n\,\,\text{is increasing monotonically}$$

Assume that $a_n$ is bounded. Then, we would have $a_n\to L$ for some real number $L$. Then,

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

But, $e^L-1>L$ and we have the desired contradiction!

Therefore, $a_n$ is unbounded.

  • $\begingroup$ As my answer shows, you just need to use $e^x > 1+x+x^2/2$. $\endgroup$ – marty cohen Sep 12 '15 at 18:47
  • 1
    $\begingroup$ @martycohen That's fine. There is more than one way to skin a sequence as a wise man once told me. $\endgroup$ – Mark Viola Sep 12 '15 at 18:49
  • $\begingroup$ Nice answer @Dr.MV. $\endgroup$ – kobe Sep 12 '15 at 19:30
  • $\begingroup$ @kobe Wow! Thank you. Coming from you, that means a lot! Much appreciative. $\endgroup$ – Mark Viola Sep 12 '15 at 21:03

$a_{n+1} =e^{a_n}-1 \ge (1+a_n+a_n^2/2)-1 =a_n(1+a_n/2) $. Therefore $\frac{a_{n+1}}{a_n} > 1+a_n/2 $. Therefore, for any $k$, $\frac{a_{n+k+1}}{a_{n+k}} > 1+a_{n+k}/2 > 1+a_{n}/2 $ since $a_n$ is increasing.

Multiplying these, $\frac{a_{n+k}}{a_{n}} > (1+a_{n}/2)^k > 1+ka_n/2 $ which shows the divergence.


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