<s>[INST] Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is  true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.
 [/INST] The sequence appears to be decreasing to $\sqrt2$. (It's the Newton-Raphson iteration for computing $\sqrt 2$ as the root of the function $f(x):=x^2-2$.) So show this in two steps:
(1) First prove by induction that $p_n\ge \sqrt2$ for every $n$. This follows from
$$p_{n+1}-\sqrt 2=\left({p_n\over2} +\frac1{p_n}\right)-\sqrt2={(p_n-\sqrt2)^2\over2p_n}.$$
(2) Then use (1) to prove by induction that $p_{n+1}\le p_n$. You should write your expression $p_n-p_{n+1}$ this way:
$$
p_n-p_{n+1}=p_n-\left({p_n\over2}+\frac1{p_n}\right)={p_n^2-2\over2p_n}
$$
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