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

1. abb0t

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

2. Requiem

yes

3. Requiem

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

4. abb0t

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

5. Requiem

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

6. abb0t

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

7. abb0t

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

8. abb0t

I'm sure you can finish it from here.

9. Requiem

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

10. Requiem

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

11. abb0t

yep.

12. Requiem

thanks abbot!!

13. eliassaab

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

14. Requiem

hey Elias, how did you figure that out?

15. eliassaab

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.