limits - 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?

Answer

$\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

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