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

1. anonymous

converge

2. anonymous

can u explain how u got that?

3. 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

4. anonymous

cos(n) is bounded between 1 and -1

5. anonymous

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

6. 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.

7. anonymous

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

8. anonymous

$\lim_{n->\inf}$

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

so bc it goes to 0, it converges?

13. anonymous

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

14. anonymous

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

15. anonymous

Both are equally valid here. :P

16. anonymous

ok :), thanks ya'll