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Let $(a_n)_{n\ge 1}$ be the sequence defined as the following : $$a_1=1 ,\ a_{n+1}=\dfrac{a_n}{n} + \dfrac{n}{a_n} ,\ n\ge1$$ Show that for every $n\ge4,\ \lfloor a_n^2 \rfloor = n$.
My approach to this problem was trying induction and using the function $f_n(x)=\dfrac{x}{n} + \dfrac{n}{x}$ :
Proving the base case for $n= 4$ and then by the inductive hypothesis $\lfloor a_n^2 \rfloor = n$ implies that $$\sqrt{n} \le a_n \lt \sqrt{n+1}$$
We then apply $f_n$ knowing that it is decreasing in that interval following it up with the floor function and some polishing, all leads to this inequality : $$n+1\le \lfloor a_{n+1}^2 \rfloor \le n+2$$
So I can't exactly get $n+1$ since $n+2$ is a possibility, this problem is a product of the fact that if $a\lt b$ then $\lfloor a \rfloor \le \lfloor b \rfloor$.

Any insights would be greatly appreciated! I wonder if my result is correct because it seems like the only way.

StubbornAtom
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OUCHNA
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2 Answers2

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We will prove the following result instead.

$$n+{2\over n}<{a_n}^2<n+1$$

First we check it's true for $n=4$.

Now the induction step, first notice that $x>y>1$ implies $x+{1\over x}>y+{1\over y} \equiv 1>{1\over xy}$

Therefore for RHS $${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over n}}+2=2+{n^4+n^2+4+{4\over n^2}\over n^3+2n}$$$$= 2+n-{n^2-4-{4\over n^2}\over n^3+2n}<n+2$$

For LHS,

$${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+1\over n^2}+{n^2\over n+1}+2={n^4+n^2+2n+1\over n^3+n^2}+2$$ $$=2+(n-1)+{2n^2+2n+1\over n^3+n^2}>n+1+{2\over n}>n+1+{2\over n+1}$$

cr001
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  • Nice derivation! although this doesn't really answer the question because $n+\dfrac{2}{n} \lt a_n^2 \lt n+1$ only implies $n\le \lfloor a_{n}^2 \rfloor \le n+1$ from which we can't deduce $\lfloor a_n^2 \rfloor = n$ if I'm not mistaken – OUCHNA Aug 14 '20 at 13:36
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    $n+\dfrac{2}{n} \lt a_n^2 \lt n+1$ implies ${a_n}^2$ is strictly less than $n+1$. There is no equality involved here. – cr001 Aug 14 '20 at 13:38
  • As shown in the $RHS$ part, $2+n-{n^2-4-{4\over n^2}\over n^3+2n}$ is strictly less than $n+2$ as well. – cr001 Aug 14 '20 at 13:40
  • Ohhh you are right because $n+1$ is an integer if the RHS wasn't an integer we must then have the equality case right ? – OUCHNA Aug 14 '20 at 13:41
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    Thanks a lot that is really what I was missing! – OUCHNA Aug 14 '20 at 13:42
  • Integer or not is not an issue. The $\leq$ sign is never used and only $<$ sign is involved here. Let's give a concrete example, if $1<x<2$ then $\lfloor x \rfloor$ must be $1$ and $2$ is not possible. – cr001 Aug 14 '20 at 13:43
  • I meant in the case where $5.1 \lt 5.3 \lt 5.8$ where 5.8 isn't an integer and evidently all their floors are equal – OUCHNA Aug 14 '20 at 13:55
  • Yes, if you mean the bounding values, being integer is a key factor here. – cr001 Aug 14 '20 at 13:57
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    You have an error: ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+{2\over n}}+2$ doesn't hold. what you should have is ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+{2\over n}\over n^2}+{n^2 \over n+1}+2$. Similarly ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+1\over n^2}+{n^2\over n+1}+2$ should be ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2>{n+\frac 2n\over n^2}+{n^2\over n+1}+2$. I don't think in the end it will be a problem. – fleablood Jul 01 '24 at 19:10
  • It is a problem and this solution doesn't work. The upper bound is fixable but the lower bound just won't work (you get something way too low). – fleablood Jul 02 '24 at 17:06
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The above solution doesn't seem to do the bounds properly:

It says, if we assume $n+\frac 2n < a_n^2 < n+1$ then ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{\color{red}{n+{2\over n}}\over n^2}+{n^2 \over \color{red}{n+{2\over n}}}+2=2+{n^4+n^2+4+{4\over n^2}\over n^3+2n}$

But as $n+\frac 2n < a_n^2 < n+1$, we do not have $\frac{a_n^2}{n^2} < \frac {n+\frac 2n}{n^2}$ (although we do have $\frac {n^2}{a_n^2} < \frac {n^2}{n+\frac 2n}$) and we have a similar error further down.

What we need is ${{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{\color{blue}{n+1}\over n^2}+{n^2 \over \color{red}{n+{2\over n}}}+2=$

The upper bound can be fixed:

$a_{n+1}^2 ={{a_n}^2\over n^2}+{n^2\over {a_n}^2}+2<{n+1\over n^2}+{n^2 \over n+{2\over n}}+2=2+\frac{n^4+n^2 + 2 + \frac 2n}{n^3+2n}=2+\frac {n^4 + 2n^2 -n^2 +2+\frac 2n}{n^3+2n}= 2+n -\frac {n^2-2-\frac 2n}{n^3 +2n}< n+2$

To get the lower bound cannot

$a_{n+1}^2 =\frac {a_n}{n^2}+\frac {n^2}{a_{n}^2} > \frac{n+\frac 2n}{n^2}+\frac{n^2}{n+1}=$
$\frac {(n+\frac 2n)(n+1)+n^4}{n^2(n+1)}=\frac{n^4 +n^2 +2+n+\frac 2n}{n^3+n^2}=$
$\frac{n^4 + n^3 - n^3 -n^2 +3n^2+2+n+\frac 2n}{n^3+n^2}=n-1+\frac{3n^3+2+n}{n^3+n^2} $

The official solution https://angelkiller411.wordpress.com/wp-content/uploads/2013/07/bulgarian-mathematical-olympiads-third-and-fourth-rounds-from-1995-to-2000-more-than-300-problems-with-solutions-260p.pdf page 13 problem 4.

uses that the function $f(x)= \frac xn +\frac nx$ is decreasing when $x < n$.

[If $0< a < b < n$ then $f(a)-f(b)=\frac an + \frac na -\frac bn-\frac nb=\frac{a^2b +n^2b-ab^2-an^2}{abn}=\frac{b(n^2-ab)-a(n^2-ab)}{abn}=\frac{(b-a)(n^2-ab)}{abn} > 0$]

and the induction hypothesis/bounds that $\sqrt n \le a_n \le \frac n{\sqrt {n-1}}< n$ (so $n \le a_n^2 \le \frac {n^2}{n-1}=n+\frac {n+1}{n-1} < n+1$)

As $f(x)$ is decreasing and $\sqrt n \le a_n$ then $a_{n+1} = f(a_n) \le f(\sqrt n)=\frac {\sqrt n}n + \frac n{\sqrt n}=\frac {n+1}{\sqrt n}$.

And as $a_n \le \frac {n}{\sqrt{n-1}}$ we have $a_{n+1}=f(a_n)\ge f( \frac {n}{\sqrt{n-1}})=\frac 1{\sqrt{n-1}} + \sqrt{n-1}=\frac n{\sqrt{n-1}}=\sqrt{\frac{n^2}{n-1}}=\sqrt{n +\frac n{n-1}} > \sqrt{n+1}$

So $\sqrt{n+1} < a_{n+1} < \frac{n+1}{\sqrt n}< n+1$.

Thus we've proven by induction (assuming base case $n=4$) that $\sqrt{n} \le a_n <\frac n{\sqrt{n-1}}$ and so $\lfloor a_n^2 \rfloor = n$.

fleablood
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