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Jackywu23Best ResponseYou've already chosen the best response.0
\[\sum_{n=1}^{\infty} (9)^(n1)/2^n\]
 2 years ago

3pwoodBest ResponseYou've already chosen the best response.0
Assuming that's 9^(n1)/2^n it diverges. The numerator far outpaces the denominator as n increases.
 2 years ago

Jackywu23Best ResponseYou've already chosen the best response.0
what kind of method or test should I use for this question?
 2 years ago

3pwoodBest ResponseYou've already chosen the best response.0
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 ∞throot 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.
 2 years ago

3pwoodBest ResponseYou've already chosen the best response.0
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.
 2 years ago
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