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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it 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 :

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ar^k}{1r}\] ; 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm sorry, if and only if the series is converging*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0example :\[\sum_{k=0}^{\infty} 2(\frac{1}{2})^k\] 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: \[\frac{ar^k}{1r}\] where the numerator is the first term. Substitute K with 0 and you'll getthe first term which is = 2 so : \[\frac{ar^k}{1r} = \frac{2}{1+\frac{1}{2}} = \frac{4}{3}\] so the sum of the series is 4/3 I hope this example clarified some points ^_^
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