$a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. Prove that for $n\ge4$, $\lfloor{a_n^2}\rfloor=n$

Question

Define a sequence $\left\lbrace a_{n}\right\rbrace$ by $\displaystyle{a_{1} = 1\,,\ a_{n + 1} = {n \over a_n} + {a_n \over n}.\quad}$ Prove that for $n \geq 4,\,\,\left\lfloor a_{n}^{2}\right\rfloor=n$

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The substitution $b_{n} = a_{n}^{2}$ might be helpful, but I still haven't proved the assertion yet.

Answer 1

First,we use Mathematical induction have following inequality

$$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$

we easy prove this function $$f(x)=\dfrac{x}{n}+\dfrac{n}{x} $$ is decreasing on $(0,n)$

becasue $$f'(x)=-\dfrac{n}{x^2}+\dfrac{1}{n}\le 0$$

since $a_{1}=1,a_{2}=2,a_{3}=2$,so $$\sqrt{3}\le a_{3}\le\dfrac{3}{\sqrt{2}}$$ Assmue that $$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$

then $$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}}=\dfrac{n}{\sqrt{n-1}}>\sqrt{n+1}$$ $$a_{n+1}=f(a_{n})\le f(\sqrt{n})=\dfrac{n+1}{\sqrt{n}}$$

In fact,we can prove

$$a_{n}\le \sqrt{n+1},n\ge 4$$ since $$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}})=\dfrac{n}{\sqrt{n-1}},n\ge 3$$ so $$a_{n}\ge\dfrac{n-1}{\sqrt{n-2}},n\ge4$$ and note $$a_{n+1}=f(a_{n})\le f\left(\dfrac{n-1}{\sqrt{n-2}}\right)=\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}$$

it suffces prove that $$\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}\le\sqrt{n+2},n\ge 4$$ $$\Longleftrightarrow n^3-n^2-2n+1<(n^2-n)\sqrt{n^2-4}$$ $$\Longleftrightarrow 2n^3-6n^2+4n-1=2n^2(n-3)+4n-1>0$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\tt{\large\it\mbox{Hint:}}\ \mbox{For}\ n > 1:$ $$ a_{n + 1} = {n \over a_{n}} + {a_{n} \over n}\quad\imp\quad a_{n}^{2} - \pars{na_{n + 1}}a_{n} + n^{2} = 0 $$ Since $a_{n} \in {\mathbb R}$, we'll have $\pars{na_{n + 1}}^{2} - 4n^{2} \geq 0\quad \imp\quad a_{n + 1} \geq 2$ and $$ a_{n} = \bracks{\vphantom{\LARGE A^{A^{A}}}% {a_{n + 1} \over 2} - \root{\pars{a_{n + 1} \over 2}^{2} - 1}}\,n = {n \over a_{n + 1}/2 + \root{\pars{a_{n + 1}/2}^{2} - 1}} $$ $$ \mbox{With}\quad b_{n} \equiv {a_{n} \over 2}\,,\qquad b_{1} = {1 \over 2}\,,\quad b_{n} = {n/2 \over b_{n + 1} + \root{b_{n + 1}^{2} - 1}}\,; \qquad b_{n} \geq 1\,,\quad n \geq 2 $$ We have to prove that $\floor{b_{n}^{2}} = n/4$ when $n \geq 4$. Notice that $b_{n} \leq n/2$.