## anonymous one year ago an=2^-n cosn pi

1. anonymous

so? solve for n? or what

2. anonymous

determine whether the sequence converges or diverges, if it converges give the limits

3. anonymous

ohh

4. anonymous

$a_n=\frac{\cos(n\pi)}{2^n}$

5. anonymous

naw dude

6. anonymous

the denominator goes to zero lickety split the numerator is stuck between -1 and 1

7. anonymous

8. anonymous

yes of course

9. anonymous

oh i said it wrong the denominator goes to INFINITY lickety split, not zero

10. anonymous

so the limit is zero

11. anonymous

and of course that means it converges correct?

12. anonymous

yes converges to zero

13. anonymous

thanks sooooooo much

14. zepdrix

If you need some way to justify it, you could use Squeeze Theorem :)$\large\rm -1\le \cos(n \pi)\le 1$Divide each side by 2^n,$\large\rm -\frac{1}{2^n}\le \frac{\cos(n \pi)}{2^n}\le \frac{1}{2^n}$And let n approach infinity on the left and right most sides of this inequality,$\large\rm -0\le \frac{\cos(n \pi)}{2^n}\le 0$