# Cauchy Criterion.

• Jul 28th 2010, 05:53 PM
Also sprach Zarathustra
Cauchy Criterion.
Prove that the following infinite sum is converges using Cauchy Criterion.

$\displaystyle \Sigma^{\infty}_{n=1} \frac{cos{nx}-cos{(n+1)x}}{n}$

$\displaystyle x-Constant$

• Jul 29th 2010, 02:15 AM
Ackbeet
Just to clarify, you're trying to show that

$\displaystyle \displaystyle{\sum_{n=1}^{\infty} \frac{\cos(nx)-\cos((n+1)x)}{n}}}$

converges, correct? And you're required to use the Cauchy criterion?

What have you done so far?
• Jul 29th 2010, 02:47 AM
Also sprach Zarathustra
Yes.

I say the next thing:

$\displaystyle \Sigma^{\infty}_{n=1} \frac{cos{nx}-cos{(n+1)x}}{n}=2sin(\frac{1}{2}x)\Sigma^{\infty}_ {n=1}\frac{sin(\frac{2n+1}{2}x}{n}$

So, I need to prove that for every $\displaystyle \epsilon >0$, exist $\displaystyle N(\epsilon)$, for all $\displaystyle n$, $\displaystyle n>N(\epsilon)$, and for all $\displaystyle p$ natural:

$\displaystyle |S_{n+p}-S_n|<\epsilon$

I choose $\displaystyle p=n$.

$\displaystyle |S_{2n}-S_n|=|\frac{sin(\frac{2(n+1)+1}{2}x)}{n+1}+...+\fr ac{sin(\frac{4n+1}{2}x)}{2n}|<|\frac{sin(\frac{2(n +1)+1}{2}x)}{n+1}|+...+|\frac{sin(\frac{4n+1}{2}x) }{2n}|<\frac{1}{n+1}+...+\frac{1}{2n}<n\frac{1}{n+ 1}<\frac{n}{n}<1$

My $\displaystyle \epsilon$ is not containing $\displaystyle n$...
• Jul 29th 2010, 04:01 AM
Dinkydoe
Choose $\displaystyle \epsilon >0$ and write:

$\displaystyle x_n= \cos(nx)$
$\displaystyle a_n = \frac{1}{n}(x_n-x_{n+1})$

We need to show that for all $\displaystyle \epsilon>0$ we can find $\displaystyle N$ such that for all $\displaystyle n>N$ and $\displaystyle p\geq 1$ we have $\displaystyle A_{n,p}:= |a_n+\cdots +a_{n+p}|< \epsilon$

We can write: $\displaystyle A_{n,p}= |\frac{x_n}{n}-\frac{x_{n+1}}{n(n+1)}-\frac{x_{n+2}}{(n+1)(n+2)}-\cdots - \frac{x_{n+p}}{(n+p-1)(n+p)}-\frac{x_{n+p+1}}{n+p}|$

since $\displaystyle -1\leq \cos(nx)\leq 1$ for all $\displaystyle n$ we may assume (with triangle inequality)

$\displaystyle A_{n,p}\leq \frac{1}{n}+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+ \cdots + \frac{1}{(n+p-1)(n+p)}+\frac{1}{n+p}$

Now it can't be too hard to show that we can find $\displaystyle N$ such that for all $\displaystyle n>N$ and $\displaystyle p\geq 1$ we have

$\displaystyle \frac{1}{n}+ \frac{1}{n+p}< \frac{\epsilon}{2}$

and

$\displaystyle \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots + \frac{1}{(n+p-1)(n+p)} < \frac{\epsilon}{2}$