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Requiem

  • 2 years ago

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

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  1. abb0t
    • 2 years ago
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    are you being asked to determine whether the power series converges or not?

  2. Requiem
    • 2 years ago
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    yes

  3. Requiem
    • 2 years ago
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    when simplified i got absolute value of (2x-1) Lim n--->00 n+1

  4. abb0t
    • 2 years ago
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    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
    • 2 years ago
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    i did that, but not sure if i am right..

  6. abb0t
    • 2 years ago
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    \[L = \lim_{n \rightarrow \infty}\left| \frac{ a_{n+1} }{ a_n } \right|\] if L <1 absolutely convergent, and thus convergent L > 1 is divergent L = 1 use different test. but it SHOULD work.

  7. abb0t
    • 2 years ago
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    \[L = \lim_{n \rightarrow \infty}\left| \frac{ (n+1)! }{ (2x-1)^{n+1} } \times \frac{(2x-1)^n}{n!} \right| = \lim_{n \rightarrow \infty }\left| \frac{ n(n+1)}{ (2x-1)(2x-1)^n } \ \times \frac{ (2x-1)^n }{ n } \right|\]

  8. abb0t
    • 2 years ago
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    I'm sure you can finish it from here.

  9. Requiem
    • 2 years ago
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    so when simplified should be (n+1)(2x-1) right?

  10. Requiem
    • 2 years ago
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    so i pull the 2x-1 out and im left with the Lim n--->00 of n+1 ?

  11. abb0t
    • 2 years ago
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    yep.

  12. Requiem
    • 2 years ago
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    thanks abbot!!

  13. eliassaab
    • 2 years ago
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    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
    • 2 years ago
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    hey Elias, how did you figure that out?

  15. eliassaab
    • 2 years ago
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    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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