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#1
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infinite series
I'm having trouble determining which way to test the convergence/divergence of a series that has the same degree in both numberator and denominator.
For Example: n/(3n+1), I know it is divergent but how do I provide proof (which test)? thanks |
#2
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Re: infinite series
Well, what's the limit of each term in the infinite series as n -> infinity?
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#3
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Re: infinite series
1/3...this proves divergence based on the test for divergence. Is that how you would have done it?
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#4
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Re: infinite series
Rather than apply the names of some formulas, let's just think about what it means that that limit is 1/3. When you get far enough out, each new term that you add is going to contribute roughly 1/3 to the value of your sum. Given that, is it possible for the series to converge?
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#5
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Re: infinite series
I don't understand. I don't see why it's not possible, I'm not doubting you but I don't get it. If the limit of the series->1/3 as n->infinity than what's the prob? please explain
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#6
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Re: infinite series
no, because it's not approaching any particular number but continues on until infinity, thus divergent. Interesting. I don't understand what you mean "each new term that you add is going to contribute roughly 1/3 to the value of your sum" I don't understand this, but I do understand conceptually what you are saying.
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#7
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Re: infinite series
Conceptually is the best way to put it, because I don't think you would find it particularly illuminating for me to put it technically. The idea basically is this: as n gets huge, each additional term in the series is 1/3. So if X is the sum of the first 1000000 terms, X + 1 will be very close to the sum of the first 1000003 terms. Notice that no matter how many terms I pick, something like this will be true. So there's no limit to how big the sum gets, and so it doesn't converge.
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