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Math Quiz 7
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Determine whether this series converges or diverges using any one or a combination of the following tests/rules/theorems: Comparison Test Show your reasoning. |
Re: Math Quiz 7
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Using the Divergence Test and l'Hopital's Rule (and the rule that lets me take the exponent outside the limit), I got that the series diverged: http://goo.gl/KVYEKZ
Edited to add: Graph of the series (to confirm my working): Attachment 18720 After thinking about this problem a bit more, I was wondering if this logic would work: as x→∞, the sum will resemble Σ1^(2x), which is just Σ1^(x), which diverges by the Geometric Series test (since it's just 1+1+1+1 infinite times). I get the same result, but I feel like I'm messing up something by applying l'Hopital's rule like that... |
Re: Math Quiz 7
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sum((x/(x+3))^(x^2)) By your reasoning, as x goes to infinity, this sum will also resemble sum(1^(x^2))=sum(1^x), so it should also diverge. However, this series does not diverge. |
Re: Math Quiz 7
It doesn't converge because the limit as k goes to infinity of your function is equal to e^-6. This is the divergence test.
Here's my work. The -6 I got as my limit is the exponent for e^-6, which means that the function is always changing, and the sum cannot converge. ![]() |
Re: Math Quiz 7
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Nice work everybody ! Quote:
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I'm not familiar with your root test method Ether. When I used the root test, it was inconclusive. ie the limit for the kth root of (k/(k+3))^2k = 1
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In the post you are referring to, I quoted and was referring to the series that Caleb mentioned in post#3: Quote:
The Root Test, however, is inconclusive for the series I posted in post#1. |
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For all k>2, k[sup]2 > 2k Therefore, for all k>2, (k/(k+3))k2 < (k/(k+3))2k. Already proven: Sk=1OO (k/(k+3))2k converges Therefore, Sk=1OO (k/(k+3))k2 converges by the Direct Comparison Test. |
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How about this one:
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I'm perhaps a little late, but it's not actually necessary to use l'Hospital's: we can just use the limit definition of the exponential function.
We have that (1 - a/n)n converges to e-a. Thus, the sequence a_k = (1-3/(k+3))2k+6 = ((1-3/(k+3))k+3)2 will converge to (e-3)2 = e-6. The sequence of terms we have is b_k= a_k / (1-3/(k+3))6 and (1-3/(k+3))6 goes to 1 as k becomes large, so in the limit b_k converges to e-6/1 = e-6. |
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See for example http://mathworld.wolfram.com/ExponentialFunction.html But let's call it a definition. Then since it's a definition you can use it even though it's not in the list of allowable list of test/rules/theorems specified in the original problem statement I like your answer, nice work. |
Re: Math Quiz 7
I went into a really intricate solution, and got a quadratic: the limit as k goes to infinity of (k/(k+3))^k = 0, 1. Very helpful. So now I know that the series either converges or diverges. My previous attempts at a solution have had logical errors, unfortunately.
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