divergence of {an} - help please

**MortenDK** · November 27th 2009, 02:58 PM

Hey people.

I have a series {an}.
I know that for every positive R > 0 exists an n so |an| >= R, where n is a natural number.

Now i have to show that {an} is divergent.

I dont exacly know how to do this. I tried to make a proof by conflict, saying that {an} was convergent and then choosing an R > epsilon + |a| where a is the limit of {an}. Then i thought i might be able to end up with something of the following:
For every R > 0 there is an n so |an| >= R. If R = epsilon + |a| -> |an| > epsilon + |a| -> |an| - |a| > epsilon..
Then i didnt know what to do, because i dont know if i can say that |an|-|a| =< |an - a| and then say that |an - a| >= |an| - |a| > epsilon -> |an - a| > epsilon.

Then it conflicts with the definition of convergence, cause it states that for any epsilon, there exists an N so |an - a| <= epsilon, which in this case isnt true.

Can i do this?
Thanks for the help.

**Abu-Khalil** · November 27th 2009, 03:19 PM

Let $\epsilon>0$ . Suppose $a_n\to A\in\mathbb{R}$ , then $\exists N:\forall n\geq N,|a_n-A|<\epsilon$ .

Let $r=\max_{1\leq n\leq N}|a_n|$ then choose $R=\max\left\{r+\epsilon,|A|+\epsilon,\right\}.$

The $r+\epsilon$ is to get a new $n$ and then, for a $n\geq N$ we have

$\epsilon>|a_n-A|\geq |a_n|-|A|\geq R+|A|+\epsilon-|A|=R+\epsilon.\Rightarrow 0>R\rightarrow\leftarrow$

**MortenDK** · November 27th 2009, 04:47 PM

Ok. I understand the last part.
I dont understand why you choose r the way you do?
Could you explain in more detail?

Thanks.

**chiph588@** · November 27th 2009, 05:09 PM

Originally Posted by Abu-Khalil

Let $\epsilon>0$ . Suppose $a_n\to A\in\mathbb{R}$ , then $\exists N:\forall n\geq N,|a_n-A|<\epsilon$ .

Let $r=\max_{1\leq n\leq N}|a_n|$ then choose $R=\max\left\{r+\epsilon,|A|+\epsilon,\right\}.$

The $r+\epsilon$ is to get a new $n$ and then, for a $n\geq N$ we have

$\epsilon>|a_n-A|\geq |a_n|-|A|\geq R+|A|+\epsilon-|A|=R+\epsilon.\Rightarrow 0>R\rightarrow\leftarrow$

In your last line, how do you justify $|a_n|\geq R+|A|+\epsilon$ ?

**Plato** · November 27th 2009, 05:13 PM

Originally Posted by MortenDK

I have a series {an}.
I know that for every positive R > 0 exists an n so |an| >= R, where n is a natural number.
Now i have to show that {an} is divergent.

It seems that you are over-thinking this problem.
The first test for divergence states: $\sum\limits_{k = 1}^\infty {a_k }$ converges only if $\left( {a_n } \right) \to 0$ .
Is that possible in this case in light of the given?

**MortenDK** · November 27th 2009, 05:32 PM

First of all, doesnt it converge if it has a limit? that limit doesnt have to be 0.
Second, yes i know its divergent, but i have to give a mathematical proof. Its a hand-in so it has to be pretty formal. I just dont know how to explain it proberly.

**Plato** · November 27th 2009, 05:39 PM

Originally Posted by MortenDK

First of all, doesnt it converge if it has a limit? that limit doesnt have to be 0.
Second, yes i know its divergent, but i have to give a mathematical proof. Its a hand-in so it has to be pretty formal. I just dont know how to explain it proberly.

Here is the proof that the series diverges.
From the given $\left( {a_n } \right) \not\to 0$ therefore $\sum\limits_{k = 1}^\infty {a_k }$ diverges.

**MortenDK** · November 27th 2009, 05:46 PM

No. A series doesnt have to go to 0 to be convergent. It only has to have a limit, meaning that when n -> infinite an -> x.
Convergent series - Wikipedia, the free encyclopedia here it can be seen as it is written in my book. It is stated around the 5th line or something like that.

Also, what is in my assignment doesnt state that it doesnt have a limit, it only states what i wrote in the beginning of my first post.

Im pretty sure that Abu-Khalil has the solution, i simply dont understand why he defines r the way he does in the middle.

**Abu-Khalil** · November 27th 2009, 05:50 PM

Originally Posted by chiph588@

In your last line, how do you justify $|a_n|\geq R+|A|+\epsilon$ ?

Oh thanks, my mistake. It was $|a_n|-|A|\geq R-|A|\geq |A|+\epsilon-|A|=\epsilon \Rightarrow \epsilon >\epsilon.$

I choose that R so I can assure that it will be a new $n$ and no a $n<N$ .

But, i have a confussion i supossed that $a_n=\sum_{k=1}^nx_k$ and given any $R>0$ we may find $n\in\mathbb{N}:|a_n|>R$ or it was $|x_k|\geq R?$

**MortenDK** · November 27th 2009, 05:56 PM

Hmm.. dunno if im just daft or something, but i dont understand it.
Could you rewrite the whole thing, editing your typo?

Sorry for the inconvinience.

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