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

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

...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.

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