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anonymous
 one year ago
an=2^n cosn pi
anonymous
 one year ago
an=2^n cosn pi

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so? solve for n? or what

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0determine whether the sequence converges or diverges, if it converges give the limits

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a_n=\frac{\cos(n\pi)}{2^n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the denominator goes to zero lickety split the numerator is stuck between 1 and 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i said it wrong the denominator goes to INFINITY lickety split, not zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the limit is zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and of course that means it converges correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes converges to zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks sooooooo much

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0If you need some way to justify it, you could use Squeeze Theorem :)\[\large\rm 1\le \cos(n \pi)\le 1\]Divide each side by 2^n,\[\large\rm \frac{1}{2^n}\le \frac{\cos(n \pi)}{2^n}\le \frac{1}{2^n}\]And let n approach infinity on the left and right most sides of this inequality,\[\large\rm 0\le \frac{\cos(n \pi)}{2^n}\le 0\]
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