Question

Use the ratio test to find the radius of convergence and interval of convergence of the power series.

Answer 1

\[\sum_{n=0}^{\infty} n^3 (x+3)^n\]

Answer 2

R=

the interval is [?] < x < -2

Answer 3

I love the ratio test. It's the best test to use. Usually works for a lot of series!
Anyways, the ratio test is: \[L = \lim_{n \rightarrow \infty }|\frac{ a_{n+1} }{ a_n }|\]
if L < 1 it's absolutely convergent (hence convergent)
L > 1 divergent
L = 1 test fails. find another test.

Answer 4

how do you find the radius ?

Answer 5

Find L.

Answer 6

\[|\frac{ (n+1)^3(x+3)^{n+1} }{ n^3(x+3)^n }|\]

Answer 7

NOTE: \[(x+3)^n (x+3)^1\]

Answer 8

Cancel out some terms. and find the limit.

Answer 9

do you find the limit of n^3(x+3)^n

Answer 10

No. The limit of the function you get AFTER you cross cancel some things in the fraction i gave up there.

Answer 11

\[\frac{ (n+1)^3(x+3) }{ n^3 }\]

Answer 12

right?

Answer 13

yes.

Answer 14

not take the limit as n approaches infinity.

Answer 15

now*

Answer 16

so the limit is 3+x

Answer 17

how do i find the radius after the limit

Answer 18

It means
|x+3|<1
, and from this equ, you will find a range of x value that satisfy this equation. That is gonna be ur Radius of Convergence.