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anonymous

  • 5 years ago

a_n= cosn/e^n, converge/diverge?

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  1. anonymous
    • 5 years ago
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    converge

  2. anonymous
    • 5 years ago
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    can u explain how u got that?

  3. anonymous
    • 5 years ago
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    intuitively you can think of cos(n) as a bounded term. So lim n-> + inf |cos(n)|/e^n less than or equal to lim n-> + inf 1/e^n Then use squeeze theorem

  4. anonymous
    • 5 years ago
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    cos(n) is bounded between 1 and -1

  5. anonymous
    • 5 years ago
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    Is there any way to use latex on this site, I have a feeling that looks confusing

  6. anonymous
    • 5 years ago
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    Perform the integral test on the expression. If the integral from 1 to infinity of the same expression converges, then the sum of discrete values converges as well. When you integrate, you get \[1/2*e^{-n}*(\sin(n)-\cos(n))\] and you can evaluate it from 1 to infinity.

  7. anonymous
    • 5 years ago
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    @zaighum: It's basically a simplified version of latex with limitations; not nearly as comprehensive but it's similar.

  8. anonymous
    • 5 years ago
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    \[\lim_{n->\inf}\]

  9. anonymous
    • 5 years ago
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    \[\lim_{n->\infty} \frac{|cosn|}{e^n} \leq \frac{1}{e^n} \]

  10. anonymous
    • 5 years ago
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    \[\rightarrow \lim_{n->\infty} \frac{-1}{e^n} \leq \lim_{n->\infty} \frac{cosn}{e^n} \leq \lim_{n->\infty} \frac{1}{e^n}\]

  11. anonymous
    • 5 years ago
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    Both side goes to 0 as so middle term goes to 0 to be squeeze theorem

  12. anonymous
    • 5 years ago
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    so bc it goes to 0, it converges?

  13. anonymous
    • 5 years ago
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    Because 1/e^n and -1/e^n converge and the our term is bounded by both, it converges

  14. anonymous
    • 5 years ago
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    oh ok, so then did we have to use the integral test? or is the sandwich theorem enough

  15. anonymous
    • 5 years ago
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    Both are equally valid here. :P

  16. anonymous
    • 5 years ago
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    ok :), thanks ya'll

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