If $a_1=1, a_n=a_{n-1}+\frac1{a_{n-1}}$ if $n\gt1$ then $a_{100}$ lies in the interval

Asked Feb 10 '23 at 18:04

Active Feb 10 '23 at 19:00

Viewed 93 times

If $a_1=1, a_n=a_{n-1}+\frac1{a_{n-1}}$ if $n\gt1$ then $a_{100}$ lies in the interval

A) $(14,18)$
B) $(13,20)$
C) $(14,20)$
D) $(20,26)$

(The answer given is ABC)

My Attempt:

$a_2=2, a_3=\frac52, a_4=\frac52+\frac25=\frac{29}{10}$

So, the sequence is $\frac{10}{10},\frac{20}{10}, \frac{25}{10}, \frac{29}{10}, ...$

In the numerators, the differences are $10,5,4$

Not sure what to make of it ...

EDIT:

@A.P. suggested change of variable $x_n=a_n^2$, so, $$a_n^2=a_{n-1}^2+\frac1{a_{n-1}^2}+2\\\implies x_n=x_{n-1}+\frac1{x_{n-1}}+2\\\implies x_n-2=x_{n-1}+\frac1{x_{n-1}}\\\implies x_n^2-4x_n+4=x_{n-1}^2+\frac1{x_{n-1}^2}$$

Do I need to change $x_n^2$ to $y_n$?

EDIT2: From this post, I understand that $$a_{n-1}^2+2<a_n^2\leq a_{n-1}^2+3\,.$$ But from this, how do we write the following? $$2n-1\leq a_n^2\leq 3n-2$$

edited Feb 10 '23 at 19:00

asked Feb 10 '23 at 18:04

aarbee

8,246

2

It is a classical problem. The key idea it is consider a change of variables $x_n=a_n^2$. – A. P. Feb 10 '23 at 18:11
Checked "numerically" that the value is around "$14.21 , 14.28 , 14.35$" , hence [ABC] is Correct : Proof might be using harmonic series or with the Suggestion given by @A.P. – Prem Feb 10 '23 at 18:24
Compare the sequence to $\sqrt {2n}$. – lulu Feb 10 '23 at 18:36
@A.P. I have made an edit. – aarbee Feb 10 '23 at 18:38
@Prem, thanks for confirming the answer. I have made an edit. Also, are we getting a harmonic progression here? – aarbee Feb 10 '23 at 18:39
@RezaRajaei I have made an edit explaining my doubt in the linked post. – aarbee Feb 10 '23 at 19:01
@lulu thanks but not able to follow how to think of $\sqrt{2n}$. Seems like this is the idea that's explored in the linked post. – aarbee Feb 10 '23 at 19:02
See the Answer to Duplicate Question [ here https://math.stackexchange.com/a/3599220/464087 given by @ Oscar Lanzi ] where that Answer uses Harmonic Series to get the bounds. – Prem Feb 10 '23 at 19:05
@aarbee you just iterate the recursion – Fix Feb 10 '23 at 19:09
@Fix got it, thanks. – aarbee Feb 10 '23 at 19:19

If $a_1=1, a_n=a_{n-1}+\frac1{a_{n-1}}$ if $n\gt1$ then $a_{100}$ lies in the interval

0 Answers0