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#1

$Old$ August 17th, 2009, 06:23 PM

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Determine wether the series converges or diverges

(2+3^n) / (4^n) the summation goes from n=1 to n= infinity

I understand that you must split the problem up so you get (2/4^n) + (3/4)^n

I know how to get the second part as it is a gemoretic series. But i am stuck on the first half of the problem.

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$Old$ August 17th, 2009, 06:27 PM

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Well $\sum_{n=1}^\infty \frac{2}{4^n}=2\sum_{n=1}^\infty \frac{1}{4^n}$ which is also a geometric series...

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$Old$ August 17th, 2009, 06:34 PM

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Well $\sum_{n=1}^\infty \frac{2}{4^n}=2\sum_{n=1}^\infty \frac{1}{4^n}$ which is also a geometric series...

so does that mean that in that portion of the problem (1/4^n)

a=1/4 r=1/4 and you would use the formula a/1-r in order to find the sum

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$Old$ August 17th, 2009, 07:05 PM

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a= 2 r = 1/4

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