anonymous
  • anonymous
an=2^-n cosn pi
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
so? solve for n? or what
anonymous
  • anonymous
determine whether the sequence converges or diverges, if it converges give the limits
anonymous
  • anonymous
ohh

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anonymous
  • anonymous
\[a_n=\frac{\cos(n\pi)}{2^n}\]
anonymous
  • anonymous
naw dude
anonymous
  • anonymous
the denominator goes to zero lickety split the numerator is stuck between -1 and 1
anonymous
  • anonymous
so the answer is 0?
anonymous
  • anonymous
yes of course
anonymous
  • anonymous
oh i said it wrong the denominator goes to INFINITY lickety split, not zero
anonymous
  • anonymous
so the limit is zero
anonymous
  • anonymous
and of course that means it converges correct?
anonymous
  • anonymous
yes converges to zero
anonymous
  • anonymous
thanks sooooooo much
zepdrix
  • zepdrix
If 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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