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

1. anonymous

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

2. anonymous

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).

3. anonymous

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

4. anonymous

...is my help a bit vague? xD

5. anonymous

thanks yuki

6. anonymous

@quantummodulus yeah its kinda vague

7. anonymous

first you want to perform the ratio test

8. anonymous

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

9. anonymous

help me

10. anonymous

I dont know the answer to the ratio test :(

11. anonymous

Do you know how to perform it?

12. anonymous

the nth term is $(3x-4)^n/(4n)^n$

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

oh no, hold on

18. anonymous

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

19. anonymous

exactly

20. anonymous

soooooo? um now what haha

21. anonymous

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

22. anonymous

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

23. anonymous

yep

24. anonymous

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

25. anonymous

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

26. anonymous

okay thanks for the help :) ttyl

27. anonymous

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.