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May 6th, 2013, 11:55 PM   #1
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Convergence and divergence

Hallo all!

I am struggling with these questions, so a solution would be very helpful for my revision!
(E is the sigma symbol)
A)

Infinity
E = [(-6)^(m+1)]/[4^(5m)]
n =0

B)

Infinity
E = [cos^2 (pi*n/]/[n sqrt(n)]
n=0

C)

infinity
E = [4+cos(n)]/4^n
n=0



please and thank you!
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May 7th, 2013, 12:58 AM   #2
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Re: Convergence and divergence

Is the question whether the series diverge or converge?

If so the first one can be shown to converge using the ratio test:



We need to find the limit:



As this is less than 1 we can say the series converges.
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May 7th, 2013, 01:38 AM   #3
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Re: Convergence and divergence

For the second one the term is undefined at n=0.

Did you mean:

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May 7th, 2013, 02:03 AM   #4
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Re: Convergence and divergence

[color=#000000]1. so the series converges.

2. so the series converges.

3. so the series converges.[/color]
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May 7th, 2013, 03:24 PM   #5
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Re: Convergence and divergence

thank you zardoz!
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May 9th, 2013, 09:24 PM   #6
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Re: Convergence and divergence

Zardoz is a beast.


It is worth pointing out however, that is a nontrivial fact.

If I were your grader, I would not give full credit. lol.
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May 10th, 2013, 12:37 AM   #7
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Re: Convergence and divergence

Quote:
Originally Posted by guynamedluis
Zardoz is a beast.


It is worth pointing out however, that is a nontrivial fact.

If I were your grader, I would not give full credit. lol.
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May 10th, 2013, 12:59 AM   #8
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Re: Convergence and divergence

Quote:
Originally Posted by guynamedluis
It is worth pointing out however, that is a nontrivial fact.
I feel the contrary : this is a trivial fact. As n increases, 1/n shrinks to 0 and the exponent too becomes 0. Hence, the expression tends 1.
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May 10th, 2013, 11:30 PM   #9
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Re: Convergence and divergence

Ooooh. It is indeed.

I thought it was the sequence .

Of course, everything is trivial if you use logarithms and L'Hopital's rule.
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