Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.
How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
 A: This is the method used in Example 3.3.5 of Introduction to Real Analysis, Bartle and Sherbert:
You could first show that $(x_n)$ is bounded below by $\sqrt2$, for $n>1$: Note that $x_n$ satisfies the quadratic equation $x_n^2-2x_n x_{n+1}+2=0$. So this equation has a real root. Its discriminant $4x_{n+1}^2-4\cdot2$ is nonnegative.
Then show $(x_n)$ is decreasing. Towards this end, use the above and write
$$
x_n-x_{n+1}= x_n-{1\over 2}\Bigl(x_n+{2\over x_n}\Bigr)={1\over 2}{(x_n^2-2)\over x_n}\ge 0.
$$
It follows that $(x_n)$ converges.  To find the limit, $L$, take the limit of both sides
of $$x_{n+1}={x_n\over 2}+{1\over x_n}$$
and solve for $L$.
Though explicitly evaluating various $x_n$ is helpful and gives intuition towards solving the problem, there is no need to do so.
A: A. $x_n>0$ , $n\in \mathbb{N}$.
B.
We will show that $\{x_2, x_3, \ldots, x_n, \ldots \}$ is monotonically non-increasing.
$$
x_n-x_{n+1} = x_n - \frac{x_n}{2} - \frac{1}{x_n}  = \frac{x_n}{2} - \frac{1}{x_n} =
\frac{x^2_n-2}{2x_n}, \qquad n>1. 
$$
But $x^2_n-2>0$, when $n>1$, because
$$
x^2_n-2 = \left( \dfrac{x_{n-1}}{2}+\dfrac{1}{x_{n-1}} \right)^2 - 2 = 
\dfrac{x^2_{n-1}}{4} + 1 + \dfrac{1}{x^2_{n-1}} - 2 = 
$$
$$
\dfrac{x^2_{n-1}}{4} - 1 + \dfrac{1}{x^2_{n-1}}  =
 \left( \dfrac{x_{n-1}}{2}-\dfrac{1}{x_{n-1}} \right)^2  \geqslant 0.
$$
So, 
$$x_n-x_{n+1} \geqslant 0.$$
Proved.
C. Since the sequense has the limit (denote $X$) $-$ as bounded and non-increasing sequence, then
$$
X = \frac{X}{2}+\frac{1}{X};
$$
$$
X^2 = \frac{X^2}{2}+1;
$$
$$
\frac{X^2}{2}=1;
$$
$$
X = \sqrt{2}.
$$
A: $
\text {Assumption: } \big(x_n\big)_{n\ge2} \text { is strictly monotonically decreasing} \tag{1} \\
$
$
\text {Assumption: } \sqrt2 \le \big(x_n\big)_{n\ge2} \tag{2}
$
Strictly monotonic & bounded $\Longrightarrow$ convergent.
$$\begin{align*}
\text {ad (1):} && x_{n+1} &< x_n \\
&& \frac{x_n}2+\frac 1{x_n} &< x_n \\
&& \frac{x_n^2}2+1 &< x_n^2 \\
&& 1 &< \frac{x_n^2}2 \\
&& 2 &< x_n^2 \\
&& \sqrt2 &< x_n
\end{align*}
$$
If we can prove $\sqrt2<x_n$, then both $(1)$ and $(2)$ hold and the convergence is proven.
$$\begin{align*}
\text {Induction by } n \\
n=2\text : & \sqrt2 < 1.5 = x_2 \\
n\to n+1\text : & \text {Assume } \sqrt2 < x_n \text{ holds.} \\
& \sqrt2 < x_{n+1} \\
& \sqrt2 < \frac {x_n}2+\frac1{x_n}=\frac{x_n^2+2}{2x_n} \\
& \text{By assumption } x_n=\sqrt2+\delta \text{ for a positive } \delta. \\
& \sqrt2 < \frac{\left( \sqrt2+\delta \right)^2+2}{2\left( \sqrt2+\delta \right)} \\
& 4+2\sqrt2\delta < \left( \sqrt2+\delta \right)^2+2 \\
& 2+2\sqrt2\delta<2+2\sqrt2\delta+\delta^2 \\
& 0 < \delta^2 & \square \\
\end{align*}
$$
A: Firstly, it is clear that $x_n>0$ for all $n \in \mathbb N$.
Secondly, the AM-GM inequality states that $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n} \geq 2 \cdot \sqrt{\frac{1}{2}}=\sqrt{2}$.
Lastly, according to the last result, for $n \geq 2$, $\frac{x_{n+1}}{x_n}=\frac{1}{2}+\frac{1}{x_n^2} \leq 1$, so that $x_{n+1} \leq x_n$.
We have shown that the sequence $(x_n)$ is bounded and eventually monotonic, thus it has a limit according to the monotone convergence theorem.
