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

Calculus1
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are you being asked to determine whether the power series converges or not?
yes
when simplified i got absolute value of (2x-1) Lim n--->00 n+1

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Other answers:

use ratio test. works about 75-80% of the time for most power series. and best test to use when you have factorials.
i did that, but not sure if i am right..
\[L = \lim_{n \rightarrow \infty}\left| \frac{ a_{n+1} }{ a_n } \right|\] if L <1 absolutely convergent, and thus convergent L > 1 is divergent L = 1 use different test. but it SHOULD work.
\[L = \lim_{n \rightarrow \infty}\left| \frac{ (n+1)! }{ (2x-1)^{n+1} } \times \frac{(2x-1)^n}{n!} \right| = \lim_{n \rightarrow \infty }\left| \frac{ n(n+1)}{ (2x-1)(2x-1)^n } \ \times \frac{ (2x-1)^n }{ n } \right|\]
I'm sure you can finish it from here.
so when simplified should be (n+1)(2x-1) right?
so i pull the 2x-1 out and im left with the Lim n--->00 of n+1 ?
yep.
thanks abbot!!
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
hey Elias, how did you figure that out?
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.

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