$a_1=3$ and $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$. Show that it monotonically decreases and find the limit.

Question

What I've done so far: I have proved that this sequence is bounded below by 0, which is a very rough estimate. I know that the infimum is $\sqrt2$.

Anyway, the question first asks me to prove that the sequence decreases monotonically. And I've tried the following:

Suppose $a_{n+1} \le a_n$, then we would get $\frac{1}{a_n} \le \frac{a_n}{2}$. After some algebra, I ended up with $\frac{1}{a_n^2 +2}\le\frac{a_n^2 +2}{2}$. Now, $a_{n+2} - a_{n+1}=\frac{2a_n}{a_n^2 +2}-\frac{a_n^2 +2}{4a_n}$. I was looking for some similarities between these two expressions but it doesn't seem to work out.

A more general question is, if the sequence is defined recursively, what are some common strategies to find the limit? I just started studying analysis and this kind of questions kind of troubles me.

Thank you guys.

Answer 1

By AM-GM inequality: $a_{n+1} = \dfrac{a_n}{2} + \dfrac{1}{a_n} \geq 2\cdot \sqrt{\dfrac{a_n}{2}\cdot \dfrac{1}{a_n}} = \sqrt{2}$. So: $a_{n+1}^2 \geq 2$, $\forall n$. Thus we have: $a_{n+1} - a_n = \dfrac{a_n}{2} + \dfrac{1}{a_n} - a_n = \dfrac{2 - a_n^2}{2a_n} \leq 0$, proving $(a_n)$ a decreasing sequence.

From the above analysis, it is bounded below by $\sqrt{2}$, so there exists a limit $L$ of this sequence, and $L$ must satisfies the equation:$L = \dfrac{L}{2} + \dfrac{1}{L}$. Solving this equation we have: $L = \sqrt{2}$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} a_{n + 1} = {a_{n} \over 2} + {1 \over a_{n}}\quad\imp\quad a_{n}^{2} - 2a_{n + 1}a_{n} + 2 = 0\quad\imp\quad a_{n} = a_{n + 1} \pm \root{a_{n + 1}^{2} - 2} \end{align} The "$+$" sign is consistent with $a_{1} = 3$ and $\ds{a_{2} = {3 \over 2} + {1 \over 3} = {11 \over 6}}$. Then

$$ \color{#00f}{\large a_{n} = a_{n + 1} + \root{a_{n + 1}^{2} - 2} > a_{n + 1}} $$

Answer 3

I can only answer your second question.

A common way is to use let $a_n+1 = a_n = x$. This strategy derives from the banach fixed point theorem.

This problem can be solved by banach fixed point theorem.

Here is the theorem:

If mapping $K: U \to U$ where $U$ and metric $d$ formed a complete metric space and $d(U(x),U(y)) \leq k d(x,y) \forall x,y \in U \exists 0 \leq k < 1$,

the limit of $m_n = K(m_{n-1})$ is the root of equation $x = K(x)$ where $m_1 \in U$.

In this problem:

$U := [\sqrt{2},3]$, $K(x):= \frac{x}{2} + \frac{1}{x}$.

Since forall $2\leq x < y \leq 3$, $K(y)-K(x) = \frac{y}{2} + \frac{1}{y}-\frac{x}{2} - \frac{1}{x} = \frac{1}{2} (y-x)(1-\frac{2}{xy})$.

And from $x<y \leq 3$, $xy < 9 \implies 1-\frac{2}{xy}<1-\frac{2}{9}$ so take $k = \frac{7}{18}<1$. So it satifies the condition for the banach fix-point theorem.

Therefore the limit should be $\frac{x}{2}+\frac{1}{x} = x \implies x = \sqrt{2}$