We are learning series and I really need help with this. I can't find any help with these, could someone please help? Thanks!

Question

Answer 1

As Paul M. said, you can use the ratio test, which states that

a necessary condition for a series Σa_k to converge is that

|a_k+1/a_k| < 1 for sufficiently large k.

In your case this condition becomes

(|23-5x|^k+1/4(k+1)×3^k+1) × (4k×3^k/|23-5x|^k) < 1

After simplification

|23-5x|k/3(k+1) < 1

This can be replaced by

|23-5x|/3 < 1 (1)

because

k/(k+1) < 1 and has the limit of 1.

Condition (1) is equivalent to

-3 < 23 - 5x < 3

4 < x < 26/5

The endpoints must be checked separately:

If x = 4 then the sum becomes

Σ(23-20)^k/(4k×3^k) = Σ1/4k which does not converge.

If x = 26/5 then the sum becomes

Σ(23-26)^k/(4k×3^k) = Σ(-1)^k/4k which does converge.

So the interval of convergence is (4, 26/5] and the radius is half the interval, that is 3/5.

Answer 2

The3 series converges by the ratio test within the interval x<|3/20|

To determine endpoint behavior, plug in x=3/20 to whether it converges there.