# Cauchy Criterion.

#### Also sprach Zarathustra

Prove that the following infinite sum is converges using Cauchy Criterion.

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

$$\displaystyle x-Constant$$

#### Ackbeet

MHF Hall of Honor
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?

#### 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 $$\displaystyle p=n$$.

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

My $$\displaystyle \epsilon$$ is not containing $$\displaystyle n$$...

#### 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

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

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