Power Series?

Question

Find the radius of convergence and interval of convergence of the series
a)sum(n^3(x-5)^n) with n=0 to infinity
b)sum((-2)^n/(sqrt(n)))(x+3)^n with n=1 to infinity

Answer 1

In each case use the ratio test for absolute convergence and then test the endpoints:

For (a),

\lim_{n \to \infty} | ((n+1)^3(x-5)^(n+1))/(n^3(x-5)^n)|=
=|x-5|

The series converges absolutely if

|x-5|<1 or -1 sum(n^3(-1)^n) which diverges by the divergence test. If x=6, the series becomes
sum(n^3), which also diverges by the divergence test. Hence, the interval of convergenc is (4,6).

(b) is similar. Using the ratio test for absolute convergence and taking the limit gives us |2(x+3)|. Thus, the series converges absolutely if |2(x+3)|<1 or |x+3|<1/2 or
-1/2 sum((-1)^n/(sqrt(n))), which converges by the alternating series test. If x=-7/2, the series becomes sum(1/sqrt(n))), which diverges because it is a p-series with p=1/2<1. Thus, the interval of convergence is (-7/2,5/2].

Check my signs carefully to make sure I didn't make an error at the endpoints! :)

Answer 2

To converge, the term(s) have to go to 0 as n goes to infinity.

For the first problem, by inspection, |x-5| < 1 so the interval of convergence would be 4 < x < 6.

For the 2'nd problem, by the same logic, |x+3| < 1 so that
-4 < x < -2 would converge.

If you need the radius of convergence, the conditions become | |z| -5 | < 1 and | |z| +3 | < 1.


Doug

Answer 3

This EASY. The diameter of infinity is infinity. Halve of infinity is still infinity. 0 times any thing is 0. 1 times anything is that number. infinity can be divided but the answer will become infinity. It can be multiply add subtracted. The answer becomes infinity because no matter subtracted multiplied divided, or even add its still infinity. Its a number that can make up for what ever goes against it.

BASIC
infinity+ 1 = infinity
infinity- 1 = infinity
infinity x 0 = infinity
infinity / 1 = infinity

TO infinity and BEYOND (literally)

oh a.) 0 to infinity
b.) 1 to infinity