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InvisibleBacchus
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Rusty on my calculus.
    #15168976 - 10/02/11 08:37 PM (12 years, 7 months ago)

I'm being a hypocrite in the fact that I get annoyed when I see homework questions here, but now I'm going to post one because I'm taking calculus III after a long time off from math. I need to determine if the sequence Formula: 0 converges or diverges as n approaches infinity. If it converges, what is the limit?

I can reason that as n goes to infinity, the numerator will swing between -1 and 1 and the denominator will increase without bound, so the expression converges to zero.

How do I formalize the logic, though? sin(2n) has no limit as n goes to infinity, so I can't take the limit of the top and bottom, can I? L'Hosptital's rule doesn't apply. Where do I go? Is it enough to say Formula: 1?

A second problem that I'm stuck on is Formula: 2
I multiply by Formula: 3 to get Formula: 4 I've played with it from here but haven't made any progress. I've used L'Hospital's rule twice, pulled out the constant and rewrote the fraction as Formula: 5

From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere. In the end, I'm still left with either infinity times infinity or infinity times zero indeterminate forms.

I tried putting this one into Wolfram|Alpha so I could expand it out to see the steps, but all I got was this: :facepalm:

Edited by Bacchus (10/02/11 09:55 PM)

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Invisiblejohnm214
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Re: Rusty on my calculus. [Re: Bacchus]
    #15170466 - 10/03/11 06:51 AM (12 years, 7 months ago)

The first case would converge on 0, as you say.

The numerator will vacillate between positive and negative numbers including magnitude 1 through 0, and the denominator gets large, so approaching infinity you have a finite number over an essentially infinite number which aproaches 0.

You can show this to be the case a number of ways, but basically if your function has f1(x)/f2(x) if the degree of f1 is less than f2 it will approach 0 at infinity.  If your prof covered some other way you can use that way, but this kind of problem is intuitive.



Fort the second one, you can use the same rule:  The numerator has degree of 1 and the denominator has degree of 2, so it goes to 0 at infinity.

Intuitively, you know 1 to any power is 1, so at infinity the magnitude of the numerator is n and the denominator is nn.



Been a while since I've done this stuff as well, but seems pretty straight forward to me.  Check out wikipedia for more discussion if you want:  http://en.wikipedia.org/wiki/Limit_of_a_function#Properties

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InvisibleDieCommie

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Re: Rusty on my calculus. [Re: Bacchus]
    #15171492 - 10/03/11 12:07 PM (12 years, 7 months ago)

Quote:

From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere.




You sure about that?  Looks like you can apply it a couple more times to me.

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Offlineindraclee
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Re: Rusty on my calculus. [Re: Bacchus]
    #15172569 - 10/03/11 04:27 PM (12 years, 7 months ago)

All I think I got these figured out. Tell me if you agree cuz been a year or two since I've done this kinda stuff.

First off I think that both of these converge. The first I used the limit comparison test and the second I used the alternating series test.

1) You know that sin(2n) is always less than or equal to one. Therefore you can say that |a_n| <= 1/(1+n^2).

  Now because 1/(1+n^2) definitely converges as n->inf and because a_n is always less than or equal to 1/(1+n^2), a_n must also converge :smile:

2) If you write out a few terms of this one (0,1/2,-2/5,3/10), you can see it is monotone decreasing. Taking the absolute value you are left with n/(n^2+1) for a_n.  This definitely converges as n->inf. These are the only requirements for the alternating series test.

Tell me if this helps. I can try to be more descriptive.


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InvisibleBacchus
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Re: Rusty on my calculus. [Re: indraclee]
    #15173827 - 10/03/11 08:12 PM (12 years, 7 months ago)

Quote:

DieCommie said:
Quote:

From here, I can keep applying L'Hospital's rule, but it doesn't get me anywhere.




You sure about that?  Looks like you can apply it a couple more times to me.




Yeah, that's why I rewrote the fraction with n^3 in the numerator. I wanted to bring it down to 1 so that I no longer had an indeterminate form, but you end up with Formula: 0 Which is nonreal. So apply L'Hospital's rule one more time to bring the denominator back to (-1)^(-n), and I'm right back at 0*infinity.

indraclee, you totally nailed it on both of them. I was scratching my head at how to actually use the squeeze and absolute value theorems. Your explanation makes perfect sense. Thanks.

Edited by Bacchus (10/03/11 08:18 PM)

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InvisibleBacchus
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Re: Rusty on my calculus. [Re: DieCommie]
    #15173839 - 10/03/11 08:15 PM (12 years, 7 months ago)

deletemepleasethanks


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Edited by Bacchus (10/03/11 08:17 PM)

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