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anonymous

  • 5 years ago

does the series e^n.sin(4^-n) converges as n=1 to infinity

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  1. amistre64
    • 5 years ago
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    I 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

  2. anonymous
    • 5 years ago
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    You 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\[\frac{-4^{-x}\log 4 \cos \frac{1}{4^x}}{-e^{-x}}=\frac{e^x}{4^x}\log 4 \cos \frac{1}{4^x}=\left( \frac{e}{e^{\log 4}} \right)^x \log 4 \cos (\frac{1}{4^x})\]

  3. anonymous
    • 5 years ago
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    As 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.

  4. anonymous
    • 5 years ago
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    So your expression tends to 0 as x tends to infinity.

  5. anonymous
    • 5 years ago
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    thanks a lot ! :)

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