anonymous
  • anonymous
sum n=1 to Infinity (9)^n-1/2^n, converges or diverges
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\sum_{n=1}^{\infty} (9)^(n-1)/2^n\]
anonymous
  • anonymous
Assuming that's 9^(n-1)/2^n it diverges. The numerator far outpaces the denominator as n increases.
anonymous
  • anonymous
what kind of method or test should I use for this question?

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anonymous
  • anonymous
For most fractional problems like this you can do what's called the ratio test where you take the limit of the series function at infinity, and if it's greater than zero it diverges, less than zero it converges and equal to zero it is inconclusive. L'Hopital's rule can be used in most situations where you end up with ∞/∞ or 0/0. So for this one, take the limit at infinity and you end up with 9^(∞-1)/2^∞. That's essentially 9^∞/2^∞ since the -1 isn't really making a dent in infinity. you can safely take the ∞th-root of the entire expression without changing its sign, so that's 9/2 and since 9/2 > 1, it's divergent by the ratio test.
anonymous
  • anonymous
Slight correction to my previous note: We're not worried about changing the limit's sign, but in making it cross between being greater or less than 1. Taking a root is still safe.

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