Can you answer this Calculus question for me?

Question

Stumped again, please show me how to solve this:

Determine if the following summation converges or diverges. If it converges find its sum.

SIGMA[k=0, infinity] 3^(k+2) / 4^(k+1)

To be clear K=0 at the bottom of the sigma and infinity at the top. The k+2 is a superscript as is the k+1.

Thanks!

Answer 1

Using logs , we can convert the numerator to base 4.  It would be 4^(0.477/0.602)(k+2) which is about 4^(4/5k+8/5). Then the term being summed is about  1/4^(0.2k+3/5).  You can show that by comparison to the series

1/4+1/5+1/6...... that the problem series would have terms

(starting at k=2_of 1/4+1/4^1.2+1/4^1.4........) that are less.

This is sufficient to show the series converges.  

Answer 2

We can use the Ratio Test to test for convergence.

Consider the ratio of the (n+1)st term over the nth term.

lim k -> inf [3^(k+3) / 4^(k+2)][ 4^(k+1)/3^k+2]

=lim k-> infinity (3/4) < 1

The series converges.

Answer 3

Σ (k=0 to ∞) 3^(k+2) / 4^(k+1)

= (9/4) Σ (k=0 to ∞) (3/4)^k

= (9/4) * 1 / (1 - 3/4)

= 9.

Since it has a finite sum, it converges.