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anonymous

  • 5 years ago

Question about series (taking the summation): So I really don't know how to start these problems. Do I always start by seeing if it converge or diverge. and if it does converge.. i should try making them like the theorem/series types? like geometric or harmonic? or is there any easier way to do this?

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  1. anonymous
    • 5 years ago
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    it actually depends on the series :) -If it's harmonic, then directly say it diverges since harmonic series always diverge, you can take values to check. - If t's geometric series, then you have the following rule :

  2. anonymous
    • 5 years ago
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    \[\frac{ar^k}{1-r}\] ; this equation is used if and only if the series is diverging to find the summation, and you only use it when |r|< 1 ( converge) but if the |r| > 1 , then it's diverging.

  3. anonymous
    • 5 years ago
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    i'm sorry, if and only if the series is converging*

  4. anonymous
    • 5 years ago
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    example :\[\sum_{k=0}^{\infty} 2(\frac{-1}{2})^k\] discuss the divergence and convergence of the following series. If it's convergent, then what's the sum? Solution: First find r: R( is always the one that has the power k) = -1/2 now let's check if |r| < 1: |-1/2| = 1/2 < 1 , so the series is converging. Now since it's converging we have to find the sum using the following: \[\frac{ar^k}{1-r}\] where the numerator is the first term. Substitute K with 0 and you'll getthe first term which is = 2 so : \[\frac{ar^k}{1-r} = \frac{2}{1+\frac{1}{2}} = \frac{4}{3}\] so the sum of the series is 4/3 I hope this example clarified some points ^_^

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