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anonymous
 5 years ago
What is the radius and interval of convergence of the following series:
(3x4)^(n)
 n=1 and is going to infinity
(4n)^(n)
anonymous
 5 years ago
What is the radius and interval of convergence of the following series: (3x4)^(n)  n=1 and is going to infinity (4n)^(n)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} (3x4)^n / (4n)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Usually 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).

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0@:Canyoudothis... I like he way how you drew the series lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...is my help a bit vague? xD

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@quantummodulus yeah its kinda vague

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0first you want to perform the ratio test

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, first perform the ratio test and get the final value of the limit in terms of x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know the answer to the ratio test :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to perform it?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0the nth term is \[(3x4)^n/(4n)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, now apply that to each n in your original series, dividing by the original series itself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...it might take a while (it's not a very clean process, usually) but once you do it should be in terms of x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so would i get (4n)^(n) / (3x4)^(n) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(3x4)^(n+1) (4n)^(n)  *  ? (4n+1)^(n+1) (3x4)^(n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0soooooo? um now what haha

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0(3x4)^n+1 can be represented as (3x4)*(3x4)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the one on the top and the other one would cancel?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(3x4)(4n)^(n)  (4n+1)^(n+1)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0now that can be written in the form\[\lim_{n \rightarrow \infty}(3x4) * (4n)^2/(4n+1)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thanks for the help :) ttyl

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0since you will get 3x4 < 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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