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?

$
\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.
$$\begin{align*}
\text {Induction by } n \\
n=2\text : & \sqrt2 < 1.5 = x_2 \\
n\to n+1\text : & \text {Assume } \sqrt2 < x_n \text{ holds.} \\
& \sqrt2 < x_{n+1} \\
& \sqrt2 < \frac {x_n}2+\frac1{x_n}=\frac{x_n^2+2}{2x_n} \\
& \text{By assumption } x_n=\sqrt2+\delta \text{ for a positive } \delta. \\
& \sqrt2 < \frac{\left( \sqrt2+\delta \right)^2+2}{2\left( \sqrt2+\delta \right)} \\
& 4+2\sqrt2\delta < \left( \sqrt2+\delta \right)^2+2 \\
& 2+2\sqrt2\delta<2+2\sqrt2\delta+\delta^2 \\
& 0 < \delta^2 & \square \\
\end{align*}
$$
