Convergent or divergent sequence?

Question

A_n = 2^n / 3^(n+1).

I'm not sure how to find the limit.

Answer 1

A[n] = 2^n / 3^(n + 1)

To solve this, we need to use some algebraic manipulation.
Note that 3^(n + 1) is the same as 3*(3^n), so we have

A[n] = 2^n / ( 3^n * 3 )

Factoring out (1/3) gives us

A[n] = (1/3) (2^n)/(3^n)

Which we can make into a single term;

A[n] = (1/3) (2/3)^n

Taking the limit as n approaches infinity,

lim (1/3) (2/3)^n
n -> infinity

Note that, for r^n, if r < 1, then the limit as n approaches infinity will be 0. r = 2/3 in this case, so we evaluate the limit, giving us

(1/3) (0)

0

The sequence converges to 0.

Answer 2

2^n/(3^(n+1)) = 2^n/(3*3^n) =1/3*( 2/3 )^n

OK, thought I'd add one last thing since all the answerers below me are taking the limit as n-->infinity, which is not what you asked. The geometric series of (2/3)^n from n=0,1,..infinity is 3, making the complete answer, 1.

Answer 3

2^n/3^(n+1) = [(2/3)^n]*1/3
lim( 2/3 )^n = 0 => lim( [(2/3)^n]*1/3 = 0