anonymous
  • anonymous
a_n= cosn/e^n, converge/diverge?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
converge
anonymous
  • anonymous
can u explain how u got that?
anonymous
  • anonymous
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

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anonymous
  • anonymous
cos(n) is bounded between 1 and -1
anonymous
  • anonymous
Is there any way to use latex on this site, I have a feeling that looks confusing
anonymous
  • anonymous
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.
anonymous
  • anonymous
@zaighum: It's basically a simplified version of latex with limitations; not nearly as comprehensive but it's similar.
anonymous
  • anonymous
\[\lim_{n->\inf}\]
anonymous
  • anonymous
\[\lim_{n->\infty} \frac{|cosn|}{e^n} \leq \frac{1}{e^n} \]
anonymous
  • anonymous
\[\rightarrow \lim_{n->\infty} \frac{-1}{e^n} \leq \lim_{n->\infty} \frac{cosn}{e^n} \leq \lim_{n->\infty} \frac{1}{e^n}\]
anonymous
  • anonymous
Both side goes to 0 as so middle term goes to 0 to be squeeze theorem
anonymous
  • anonymous
so bc it goes to 0, it converges?
anonymous
  • anonymous
Because 1/e^n and -1/e^n converge and the our term is bounded by both, it converges
anonymous
  • anonymous
oh ok, so then did we have to use the integral test? or is the sandwich theorem enough
anonymous
  • anonymous
Both are equally valid here. :P
anonymous
  • anonymous
ok :), thanks ya'll

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