## Requiem Group Title Power series problem: n!(2x-1)^n from 1 to infinity one year ago one year ago

1. abb0t Group Title

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

2. Requiem Group Title

yes

3. Requiem Group Title

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

4. abb0t Group Title

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 Group Title

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

6. abb0t Group Title

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

7. abb0t Group Title

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

8. abb0t Group Title

I'm sure you can finish it from here.

9. Requiem Group Title

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

10. Requiem Group Title

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

11. abb0t Group Title

yep.

12. Requiem Group Title

thanks abbot!!

13. eliassaab Group Title

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 Group Title

hey Elias, how did you figure that out?

15. eliassaab Group Title

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.