5
$\begingroup$

Does the recursive sequence $a_1 = 1, a_n = a_{n-1}+\frac{1}{a_{n-1}}$ converge?

Since the function $x+1/x$ is strictly monotonic increasing for all $x>1$, I don't think that the limit converges, but I'm not sure. Can anybody tell me whether the sequence is converging or not?

$\endgroup$
3
  • $\begingroup$ Before trying to find a formal proof, it helps to get some intuition. For that, I suggest sketching by hand the web diagram of your function. $\endgroup$ Aug 4 '14 at 18:44
  • $\begingroup$ Once you showed that it is not converging, you can find the speed by considering the differential equation $f'=1/f$ $\endgroup$
    – Bertrand R
    Aug 4 '14 at 18:46
  • $\begingroup$ No, but at large values it does become linear! $\endgroup$ Oct 19 '20 at 14:25
13
$\begingroup$

Note that $a_n^2 = \left(a_{n-1}+\dfrac{1}{a_{n-1}}\right)^2 = a_{n-1}^2 + 2 + \dfrac{1}{a_{n-1}^2} \ge a_{n-1}^2 + 2$.

Therefore, $a_n^2 \ge 2n-1$, and thus, $a_n \ge \sqrt{2n-1}$ for all $n \ge 1$. That's enough to show divergence.

$\endgroup$
1
  • 1
    $\begingroup$ How did you get that $a_n^2 \geq 2n-1$? $\endgroup$
    – Botond
    Sep 11 '18 at 18:47
11
$\begingroup$

No. If it were convergent to some $\alpha$, this value would verify $$\alpha=\alpha+\frac{1}{\alpha}.$$

$\endgroup$
8
$\begingroup$

Assume it converges, then it does so to a limit $L \ge 1$.

Then we have $L = L + \frac{1}{L}$ which is not possible.

$\endgroup$
6
$\begingroup$

It is obvious that $a_n\le n$. Thus $$a_{n+1}\ge a_n+\frac1n.$$ Since $H_n$ diverges, we can conclude that $a_n$ also diverges.

$\endgroup$
2
  • $\begingroup$ What does $H_n$ stand for? $\endgroup$
    – Vlad
    Mar 23 '17 at 1:51
  • $\begingroup$ @Vlad It is the harmonic series, defined as $H_n=\sum_{k=1}^n 1/k$. $\endgroup$ Mar 28 '17 at 11:23
0
$\begingroup$

Since the equation $x= x+\frac1x$ does not have a solution. Therefore $a_n$ does not converges. Also, since $a_1>0,$ by induction we easily have $a_n>0$ and then, $$a_{n+1} -a_n = \frac{1}{a_n}>0$$ which means $(a_n)$ is a strictly increasing and non convergent sequence. So $a_n\to\infty$. That $a_n$ is unbounded.

$\endgroup$
0
$\begingroup$

Since $a_{n+1}=a_n$ + $\frac{1}{a_n}$ and $a_n$

$\frac{1}{a_n}$ > $0$ we get that $a_{n+1}>a_n$

thus diverging

$\endgroup$

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.