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?

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.
