Determining series convergence. Help! Watch

1210596
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Hey guys, i've been having some trouble with a particular series convergence question. wodnering if any of you fine people would assist me.
The specific question i have been tasked with is question (6,7)



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tombayes
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(Original post by 1210596)
Hey guys, i've been having some trouble with a particular series convergence question. wodnering if any of you fine people would assist me.
i have attached a picture of the question.



note: if the series converges than the above statement is true. if the series diverges then the above statement is false.
thanks
Well
1) either try to find a counter example or
2) prove it converges (e.g. by bounding the sum or from the definition or comparison, etc)

It is kind of pointless if I tell you the answer. If still stuck let me know.
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1210596
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hey tom,
ive tried answering it. and reached the conclusion that the series converges.
i got this by doing a double direct comparison.
let me know if there are any key points i'm missing or if its just plain wrong lol.
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tombayes
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sorry just wait a sec
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tombayes
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how does sum of a^2n converge?

Consider the following:

 \exists n such that
0 \le | \frac{a_n}{n^{2/3}}| \le |\frac{a_n}{n}|=\sqrt{\frac{a_n^  2}{n^2} }\le \frac{a_n^2}{n^2} \le \frac{1}{n^2}

It is true? recall sum a_n^2 is convergent hence a_n^2 \rightarrow 0 so \exists n: a_n^2\le 1
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1210596
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(Original post by tombayes)
how does sum of a^2n converge?

Consider the following:

 \exists n such that
0 \le | \frac{a_n}{n^{2/3}}| \le |\frac{a_n}{n}|=\sqrt{\frac{a_n^  2}{n^2} }\le \frac{a_n^2}{n^2} \le \frac{1}{n^2}

It is true?
it is given in the question that a^2.n converges.
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tombayes
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(Original post by 1210596)
it is given in the question that a^2.n converges.
by a^2.n do you mean a_n^2 because that was unclear.
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DFranklin
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(Original post by 1210596)
..
Please check carefully what you've posted says what you mean it to say, because as it stands it's pretty unintelligible to me.

I also strongly suspect that you haven't posted the entire question (because you define b_n but never use it). If you post the entire question, I suspect there is some extra context that will be useful.

[To be clear, I know how to solve this, but the way I have in mind ties in with the b_n sequence...]
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1210596
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Sorry if theres been confusion guys. i will attach the ENTIRE question now. i am not sure if that will make a difference though. i have been tasked with ONLY answering question (6,7) from the image attached.



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1210596
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(Original post by DFranklin)
Please check carefully what you've posted says what you mean it to say, because as it stands it's pretty unintelligible to me.

I also strongly suspect that you haven't posted the entire question (because you define b_n but never use it). If you post the entire question, I suspect there is some extra context that will be useful.

[To be clear, I know how to solve this, but the way I have in mind ties in with the b_n sequence...]
sorry if there's been confusion. i have uploaded the entire question now on the thread. any hints or tips as to how to go about solving the question would be appreciated.
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DFranklin
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(Original post by 1210596)
sorry if there's been confusion. i have uploaded the entire question now on the thread. any hints or tips as to how to go about solving the question would be appreciated.
OK.

Firstly, as I understand your argument, you're claiming that  0 \leq a_n \leq a_n^2.

But if \sum a_n^2 converges, we know a_n \to 0, and once a_n < 1, your inequality doesn't hold.

Now how to actually solve this?

Firstly you need to decide whether it's true than \sum \dfrac{a_n}{n^{2/3}} must converge. A good test case to consider is a_n = 1/n^\lambda (where \lambda is a constant). For the condition of the question to hold, we need \sum a_n^2 converges, and this gives us a condition on lambda.

But we also know that \sum \dfrac{a_n}{n^{2/3}} = \sum 1/n^{(\lambda+2/3)}, and this converges if \lambda+2/3 < 1, diverges if \lambda+2/3 >1, So compare this with the condition on lambda, and you can see that the series does in fact converge for all allowable values of lambda.

So, at this point I'd be pretty sure it converges.

To prove it converges, I would use set b_n = 1/n^{2/3} and use the Cauchy-Schwarz inequality:

\left(\sum_1^N a_n b_n \right)^2 \leq \sum_1^N a_n^2 \sum_1^N b_n^2.

I confess I'm not seeing an easy way of proving it without using that inequality so I hope you know it (you can google it).

(My guess was that another part of the question would hint at this approach, but it turns out my guess was wrong).
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ghostwalker
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(Original post by DFranklin)
and this converges if \lambda+2/3 < 1, diverges if \lambda+2/3 >1,
We know you meant the inequalities to be the other way around.
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DFranklin
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(Original post by ghostwalker)
We know you meant the inequalities to be the other way around.
Whoops. Yes. (It was rather late...)

PRSOM...
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