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

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

Mathematics
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\[\sum_{n=1}^{\infty} (3x-4)^n / (4n)^n\]
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).
@:Canyoudothis... I like he way how you drew the series lol

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...is my help a bit vague? xD
thanks yuki
@quantummodulus yeah its kinda vague
first you want to perform the ratio test
Okay, first perform the ratio test and get the final value of the limit in terms of x.
help me
I dont know the answer to the ratio test :(
Do you know how to perform it?
the nth term is \[(3x-4)^n/(4n)^n\]
is it a(n+1)/ a(n)
Okay, now apply that to each n in your original series, dividing by the original series itself.
...it might take a while (it's not a very clean process, usually) but once you do it should be in terms of x.
so would i get (4n)^(n) / (3x-4)^(n) ?
oh no, hold on
(3x-4)^(n+1) (4n)^(n) ------------ * ---------- ? (4n+1)^(n+1) (3x-4)^(n)
exactly
soooooo? um now what haha
(3x-4)^n+1 can be represented as (3x-4)*(3x-4)^n
so the one on the top and the other one would cancel?
yep
(3x-4)(4n)^(n) ------------ (4n+1)^(n+1)
now that can be written in the form\[\lim_{n \rightarrow \infty}(3x-4) * (4n)^2/(4n+1)^2\]
okay thanks for the help :) ttyl
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.

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