Cauchy sequence

**SENTINEL4** · January 31st 2011, 07:54 AM

Hey guys i study for my exams at Wednesday and i would appreciate some help here.
Based on "if for every ε>0, there is $\nu_0\inN$ such that for all natural numbers $m, n > \nu_o$

"

1)Prove that the sequence $a_\nu , \nu \in N$ , $a_\nu=\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{ cos\nu}{\nu}$ is a Cauchy sequence

2) Prove that the sequence $b_\nu , \nu \in N$ , $b_\nu = 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2\nu-1}$ is not a Cauchy sequence.

Thanks in advance...

**BAdhi** · January 31st 2011, 08:18 AM

1)shouldn't it be $\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{cosv}{ v^2}$

cannot we use "if a sequence is covergent, it is a Cauchy sequence"

**BAdhi** · January 31st 2011, 08:19 AM

sorry, posts keep repeating, something wrong with my connection

**SENTINEL4** · January 31st 2011, 12:18 PM

Yes it is $a_\nu=..........+\frac{cos\nu}{\nu^2}$

I am thinking something like :
1) $|a_{\nu+k}-a_\nu|=|(\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\fr ac{cos(\nu+k)}{(\nu+k)^2})-(\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{cos\n u}{\nu^2})|=$
$|\frac{cos(\nu+1)}{(\nu+1)^2}+\frac{cos(\nu+2)}{(\ nu+2)^2}+...+\frac{cos(\nu+k)}{(\nu+k)^2}|\leq|\fr ac{cos(\nu+1)}{(\nu+1)^2}|+|\frac{cos(\nu+2)}{(\nu +2)^2}|+...+|\frac{cos(\nu+k)}{(\nu+k)^2}|\leq$
$\leq\frac{1}{(\nu+1)^2}+\frac{1}{(\nu+1)^2}+...+\f rac{1}{(\nu+1)^2}\leq\frac{1}{\nu+1}+\frac{1}{\nu+ 1}+...+\frac{1}{\nu+1}=\frac{k}{\nu+1}\rightarrow0$

2) $|b_{2\nu}-b_\nu|=|(1+\frac{1}{3}+...+\frac{1}{2(2\nu)-1})-(1+\frac{1}{3}+...+\frac{1}{2\nu-1})|=|\frac{1}{2(\nu+1)-1}+\frac{1}{2(\nu+2)-1}+...+\frac{1}{2(2\nu)-1}|\geq$
$\geq\frac{1}{4\nu-1}+\frac{1}{4\nu-1}+...+\frac{1}{4\nu-1}=\frac{\nu}{4\nu-1}=\frac{1}{4}$
so for $\epsilon=\frac{1}{4}$ , $b_\nu$ is not a Cauchy sequence.

How is my work??

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