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.
 A: 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:
$$
p_n-p_{n+1}=p_n-\left({p_n\over2}+\frac1{p_n}\right)={p_n^2-2\over2p_n}
$$
A: AM/GM. $\frac{x}{2}+\frac{1}{x}<x$ iff $x^2>2$. So it is enough to show that if $x^2>2$ then $(\frac{x}{2}+\frac{1}{x})^2>2$ or $\frac{x^2}{4}+\frac{1}{x^2}>1$. But by AM/GM $(\frac{x^2}{4}+\frac{1}{x^2})/2>\sqrt{\frac{1}{4}}$.
A: You have that $p_{n+1} - p_{n} = \frac{1}{p_n}-\frac{p_n}{2}$
So $p_n$ is decreasing if $ \forall n, \frac{1}{p_n} -\frac{p_n}{2} \leq 0$
And this is true if $\forall n, p_n^2 -2 \geq 0$, ie. $p_n > \sqrt{2}$ (as $p_n > 0$ )
You can prove this by induction. First, let's study the function $f(x) = \frac{x}{2}+ \frac{1}{x}$.
$f'(x) = \frac{1}{2}- \frac{1}{x^2}$
So $f$ is decreasing on $[\sqrt{2},+\infty[$. But $f(\sqrt{2}) = 0$, so $\forall x \geq \sqrt{2}, f(x) \geq \sqrt{2}$
Now the induction :
1) it's true for $p_1$. 
2) Suppose $p_n > \sqrt{2}$, then, by what we proved before,
$p_{n+1} = f(p_n) \geq \sqrt{2}$
Hence $\forall n, p_n \geq \sqrt{2}$, and this imply that $p_{n}-p{n+1} \geq 0$
A: A calculus approach. Let $$f(x) = \frac{x}{2}+\frac{1}{x}.$$ So $$f(x)-x =  -\frac{x}{2}+\frac{1}{x}$$ is the sum of two decreasing functions on $]0,+\infty]$. Hence $f(p_{n})-p_n<0$, the sequence is decreasing.
$f(x)$ is strictly increasing on $]\sqrt{2},2]$ (easy with the derivative), thus if $x \in ]\sqrt{2},2] $, $f(x) \in ]\sqrt{2},\frac{3}{2}] $. You begin to feel that $f$ could be a contracting map on this interval, and indeed $0<f'(x)\le \frac{1}{4}$. 
By the Mean Value Theorem For Derivatives, there exist $c_x\in[\sqrt{2},x]$ such that
$$ f'(c_x) = \frac{f(x)-\sqrt{2}}{x-\sqrt{2}}\,.$$
So:
$$  0< p_{n+1}-\sqrt{2}\le \max_{[\sqrt{2},x]} f' \times (p_n-\sqrt{2})\,$$
and 
$$  0< p_{n+1}-\sqrt{2}\le  \frac{2-\sqrt{2}}{4^n}\,$$
which provides you with some rate of convergence:

