Question

Power series problem:
n!(2x-1)^n
from 1 to infinity

Answer 1

are you being asked to determine whether the power series converges or not?

Answer 2

yes

Answer 3

when simplified i got absolute value of (2x-1) Lim n--->00 n+1

Answer 4

use ratio test. works about 75-80% of the time for most power series. and best test to use when you have factorials.

Answer 5

i did that, but not sure if i am right..

Answer 6

\[L = \lim_{n \rightarrow \infty}\left| \frac{ a_{n+1} }{ a_n } \right|\]
if L <1 absolutely convergent, and thus convergent
L > 1 is divergent
L = 1 use different test. but it SHOULD work.

Answer 7

\[L = \lim_{n \rightarrow \infty}\left| \frac{ (n+1)! }{ (2x-1)^{n+1} } \times \frac{(2x-1)^n}{n!} \right| = \lim_{n \rightarrow \infty }\left| \frac{ n(n+1)}{ (2x-1)(2x-1)^n } \ \times \frac{ (2x-1)^n }{ n } \right|\]

Answer 8

I'm sure you can finish it from here.

Answer 9

so when simplified should be (n+1)(2x-1) right?

Answer 10

so i pull the 2x-1 out and im left with the Lim n--->00 of n+1 ?

Answer 11

yep.

Answer 12

thanks abbot!!

Answer 13

Your series diverges everywhere except for x =1/2.
You can do it by the divergence test, the nth terms does not go to zero, then the series diverges

Answer 14

hey Elias, how did you figure that out?

Answer 15

It is a power series. if \( x\ne \frac 1 2\), then n!(2x-1)^n goes to infinity with n, so the nth-term does not go to zero, so the series is divergent.