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I am stuck in this problem. I could not get a way to prove the Cauchy criterion to the next exercise. I am looking for a solution which does not use the Banach fixed point theorem.

Let {$a_n$} be the sequence defined as $$a_1=2,\quad a_{n+1}=\frac {a_n}2 + \frac 1{a_n}$$ for every $n \in \Bbb N$ Prove that $\{a_n\}$ converges and find the limit.

I have already found the limit which is $\sqrt 2$. But I cannot prove it converges to that number.

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    $\begingroup$ Prove $a_n$ decreases and is bounded below by $\sqrt 2$ $\endgroup$ – saulspatz May 21 '18 at 17:57
  • $\begingroup$ Every monotone and bounded real sequence is convergent. $\endgroup$ – DiegoMath May 21 '18 at 17:59
  • $\begingroup$ Every monotone and bounded real sequence is convergent. $\endgroup$ – DiegoMath May 21 '18 at 17:59
  • $\begingroup$ @saulspatz I got your idea, how can I prove is a decreasing one? $\endgroup$ – Mike Matt May 21 '18 at 18:05
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    $\begingroup$ You might start by looking at $a_{n+1} - a_n$. When is this negative? $\endgroup$ – Xander Henderson May 21 '18 at 18:12
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1. Let's begin by studying if $(a_n)$ is increasing or decreasing.

Note that $\forall n\in\Bbb N^*$, if $a_n\ne 0$: $$a_{n+1} - a_n = \dfrac{1}{a_n} - \dfrac{a_n}2 = \dfrac{2-a_n^2}{2a_n}\;\;\;(\star)$$

Thus, we have to see if this is positive or negative. To do this, we need to see if $a_n>0$ or not, and if $2-a_n^2\ge 0$ or not (i.e. what's the position of $a_n$ with respect to $0$ and to $\pm\sqrt{2}$). Since $a_1=2>\sqrt{2}$ and $a_2 = \dfrac32>\sqrt{2}$, we may hope that $\forall n\in\Bbb N^*, a_n\ge \sqrt{2}$. Let's try to prove it by induction:

  • $a_1 = 2>\sqrt{2}$
  • if we assume that for a given $n\ge 1$, $a_n\ge \sqrt{2}$, then $$a_{n+1} - \sqrt{2} = \dfrac{1}{a_n} + \dfrac{a_n}2 - \sqrt{2} = \dfrac{2+a_n^2 - 2\sqrt{2}a_n}{2a_n} = \dfrac{(a_n - \sqrt{2})^2}{2a_n}\ge 0$$ since $a_n\ge \sqrt{2}\ge 0$.
  • thus $\forall n\in\Bbb N^*,\,a_n\ge \sqrt{2}\;\;(\triangle)$.

Thus, using $(\star)$, we can see that $\forall n\in\Bbb N^*,\,2-a_n^2\le 0$ (since $\forall n\in\Bbb N^*,\,a_n\ge \sqrt{2}$) and $a_n\ge 0$. This shows $a_{n+1}-a_n\le 0$ and $(a_n)$ is non-increasing.

2. Now, we also know that $(a_n)$ is bounded from below by $\sqrt{2}$ and non-increasing. Thus, it's a convergent sequence.

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You may use the following criterion:

Let $I\,$ be an interval, $a\in I$ and $f:I\longrightarrow I$. Then the sequence defined recursively by $$\begin{cases} x_0=a, \newline x_{n+1}=f(x_n),\end{cases}$$ is decreasing if $f(x)<x\,$ on the interval $I$, increasing if $f(x)>x\,$ on $I$.

So all you have to do is to find an interval $I$ which is stable by the function $\;f(x)=\dfrac12\Bigl( x+\dfrac 2x\Bigr)$, and on which the graph of $f$ is below the line $y=x$.

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The Arithmetic-Geometric Mean inequality tells us that

$$a_{n+1}={a_n\over2}+{1\over a_n}={1\over2}\left(a_n+{2\over a_n}\right)\ge\sqrt{a_n\cdot{2\over a_n}}=\sqrt2$$

(provided $a_n\gt0$). But if $\sqrt2\le a_n$, then

$${a_{n+1}\over a_n}={1\over2}+{1\over a_n^2}\le{1\over2}+{1\over2}=1$$

So if $a_1=2$, then $\{a_n\}$ is a decreasing sequence bounded below (by $\sqrt2$), hence has a limit $L$. The limit equation $L={L\over2}+{1\over L}$ now tells us $L^2=2$, so $L=\sqrt2$.

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