Question

What is the radius and interval of convergence of the following series:
(3x-4)^(n)
--------- n=1 and is going to infinity
(4n)^(n)

Answer 1

\[\sum_{n=1}^{\infty} (3x-4)^n / (4n)^n\]

Answer 2

Usually the main method is to perform the ratio test on the series, and when you have the limit in terms of x, and use the geometric series boundaries -- that the absolute value of the ratio must be less than or equal to one -- and apply it to that limit (which is the ratio).

Answer 3

@:Canyoudothis... I like he way how you drew the series lol

Answer 4

...is my help a bit vague? xD

Answer 5

thanks yuki

Answer 6

@quantummodulus yeah its kinda vague

Answer 7

first you want to perform the ratio test

Answer 8

Okay, first perform the ratio test and get the final value of the limit in terms of x.

Answer 9

help me

Answer 10

I dont know the answer to the ratio test :(

Answer 11

Do you know how to perform it?

Answer 12

the nth term is \[(3x-4)^n/(4n)^n\]

Answer 13

is it a(n+1)/ a(n)

Answer 14

Okay, now apply that to each n in your original series, dividing by the original series itself.

Answer 15

...it might take a while (it's not a very clean process, usually) but once you do it should be in terms of x.

Answer 16

so would i get (4n)^(n) / (3x-4)^(n) ?

Answer 17

oh no, hold on

Answer 18

(3x-4)^(n+1) (4n)^(n)
------------ * ---------- ?
(4n+1)^(n+1) (3x-4)^(n)

Answer 19

exactly

Answer 20

soooooo? um now what haha

Answer 21

(3x-4)^n+1 can be represented as

(3x-4)*(3x-4)^n

Answer 22

so the one on the top and the other one would cancel?

Answer 23

yep

Answer 24

(3x-4)(4n)^(n)
------------
(4n+1)^(n+1)

Answer 25

now that can be written in the form\[\lim_{n \rightarrow \infty}(3x-4) * (4n)^2/(4n+1)^2\]

Answer 26

okay thanks for the help :)
ttyl

Answer 27

since you will get |3x-4| < 1, all you have to do is to
solve for the inequality.

however, the endpoints are always something you have to check
by hand, so you need to plug in the two endpoints to the original
series to see if it converges or not, ok?

let me know if you need more help.