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anonymous
 5 years ago
does the series e^n.sin(4^n) converges as n=1 to infinity
anonymous
 5 years ago
does the series e^n.sin(4^n) converges as n=1 to infinity

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I would assume that it converges to zero... but that is just an uneducated guess :) take sin(4^n).... that is sin(1/(4^n)) as the "n" gets bigger and bigger, we get a bigger and bigger number in the bottom of the fraction; large bottom fractions tend to go to zero. for example: 1/10000...000000 = .000000...00001 which is a very small number. the sin(0) = 0 which whould make the rest of the stuff = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should arrange the product as such:\[\frac{\sin(\frac{1}{4^n})}{e^{n}}\]Then in the limit as n goes to infinity, you have the indeterminate form, 0/0, so you can use L'Hopital's rule. Doing this yields\[\frac{4^{x}\log 4 \cos \frac{1}{4^x}}{e^{x}}=\frac{e^x}{4^x}\log 4 \cos \frac{1}{4^x}=\left( \frac{e}{e^{\log 4}} \right)^x \log 4 \cos (\frac{1}{4^x})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As x tends to infinity, cos(1/4^x) tends to 1, and since e^(log 4)>e, \[\left( \frac{e}{e^{\log 4}} \right)^x \rightarrow 0\]as x tends to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your expression tends to 0 as x tends to infinity.
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