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.


1 Answer 1


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,

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

  • $\begingroup$ 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 $\endgroup$
    – OUCHNA
    Aug 14, 2020 at 13:36
  • 2
    $\begingroup$ $n+\dfrac{2}{n} \lt a_n^2 \lt n+1$ implies ${a_n}^2$ is strictly less than $n+1$. There is no equality involved here. $\endgroup$
    – cr001
    Aug 14, 2020 at 13:38
  • $\begingroup$ 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. $\endgroup$
    – cr001
    Aug 14, 2020 at 13:40
  • $\begingroup$ 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 ? $\endgroup$
    – OUCHNA
    Aug 14, 2020 at 13:41
  • 1
    $\begingroup$ Thanks a lot that is really what I was missing! $\endgroup$
    – OUCHNA
    Aug 14, 2020 at 13:42

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