Cauchy Criterion.

Dec 2009
1,506
434
Russia
Prove that the following infinite sum is converges using Cauchy Criterion.

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

\(\displaystyle x-Constant\)

(Headbang)
 

Ackbeet

MHF Hall of Honor
Jun 2010
6,318
2,433
CT, USA
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?
 
Dec 2009
1,506
434
Russia
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 \(\displaystyle n\)...
 
Dec 2009
411
131
How about this then...
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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