Question

an=2^-n cosn pi

Answer 1

so? solve for n? or what

Answer 2

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

Answer 3

ohh

Answer 4

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

Answer 5

naw dude

Answer 6

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

Answer 7

so the answer is 0?

Answer 8

yes of course

Answer 9

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

Answer 10

so the limit is zero

Answer 11

and of course that means it converges correct?

Answer 12

yes
converges to zero

Answer 13

thanks sooooooo much

Answer 14

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,\[\large\rm -0\le \frac{\cos(n \pi)}{2^n}\le 0\]