$a_1=1$, $a_{n+1}=\frac{1}{2}\left(a_n+\frac{2}{a_n}\right)$. Show that the sequence is decreasing.
$a_{n+1}$ is the Arithmetic mean of $a_n$ and $\frac{2}{a_n}$. I can say that if $a_n > \frac{2}{a_n}$, then $a_{n+1}<a_n$. But thats not always true. What should be my approach?
The sequence $\left\{a_n\right\}_{n\geq 1}$ is generated by Newton's method with starting point $a_0=1$ applied to the function $f(x)=x^2-2$, because
$$ x-\frac{f(x)}{f'(x)} = x-\frac{x^2-2}{2x} = \frac{1}{2}\left(x+\frac{2}{x}\right). $$
Now $a_1=\frac{3}{2}>\sqrt{2}$ and the function $f(x)$ is positive and convex on the interval $(\sqrt{2},+\infty)$.
By the properties of Newton's method (just draw some tangents to make it clear)
$$ a_1 < \sqrt{2} <\ldots<a_4<a_3<a_2 $$ and the sequence converges to $\sqrt{2}$ quadratically. As a matter of fact, $a_n$ is a convergent of the continued fraction of $\sqrt{2}$, namely $\frac{p_{2^n}}{q_{2^n}}$.
Putting the value of $n$ as $1$ in the given relation, we get the $2^{nd}$ term equal to $\frac{3}{2}$. Again putting the value of $n$ as $2$, we get the $3^{rd}$ term equal to $\frac{17}{12}$. Similarly, the fourth term comes out to be $\frac{577}{408}$. This shows that the sequence is decreasing.
