Question

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

Answer 1

converge

Answer 2

can u explain how u got that?

Answer 3

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

Answer 4

cos(n) is bounded between 1 and -1

Answer 5

Is there any way to use latex on this site, I have a feeling that looks confusing

Answer 6

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.

Answer 7

@zaighum: It's basically a simplified version of latex with limitations; not nearly as comprehensive but it's similar.

Answer 8

\[\lim_{n->\inf}\]

Answer 9

\[\lim_{n->\infty} \frac{|cosn|}{e^n} \leq \frac{1}{e^n} \]

Answer 10

\[\rightarrow \lim_{n->\infty} \frac{-1}{e^n} \leq \lim_{n->\infty} \frac{cosn}{e^n} \leq \lim_{n->\infty} \frac{1}{e^n}\]

Answer 11

Both side goes to 0 as so middle term goes to 0 to be squeeze theorem

Answer 12

so bc it goes to 0, it converges?

Answer 13

Because 1/e^n and -1/e^n converge
and the our term is bounded by both, it converges

Answer 14

oh ok, so then did we have to use the integral test? or is the sandwich theorem enough

Answer 15

Both are equally valid here. :P

Answer 16

ok :), thanks ya'll