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anonymous
 5 years ago
a_n= cosn/e^n, converge/diverge?
anonymous
 5 years ago
a_n= cosn/e^n, converge/diverge?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can u explain how u got that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0intuitively 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(n) is bounded between 1 and 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is there any way to use latex on this site, I have a feeling that looks confusing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perform 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
 5 years ago
Best ResponseYou've already chosen the best response.0@zaighum: It's basically a simplified version of latex with limitations; not nearly as comprehensive but it's similar.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n>\infty} \frac{cosn}{e^n} \leq \frac{1}{e^n} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\rightarrow \lim_{n>\infty} \frac{1}{e^n} \leq \lim_{n>\infty} \frac{cosn}{e^n} \leq \lim_{n>\infty} \frac{1}{e^n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Both side goes to 0 as so middle term goes to 0 to be squeeze theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so bc it goes to 0, it converges?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because 1/e^n and 1/e^n converge and the our term is bounded by both, it converges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, so then did we have to use the integral test? or is the sandwich theorem enough

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Both are equally valid here. :P
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