4
$\begingroup$

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 $$e^x>x+1$$

(Proven).

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 $\infty$.

Thanks,

Alan

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.