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Finding interval of convergence

  1. Apr 22, 2013 #1
    1. Determine a power series, centered at zero for the function ∫f(x)dx. Identify the interval of convergence.

    f(x) = ln(x+1) = ∫[itex]\frac{1}{x+1}[/itex]

    2.


    3. i found the power series, which is :

    Ʃ ((-1)^(n))(x^(n+1)) / (n+1)


    Im okay with that, but i need help on finding the interval of convergence...


    to find it i do ratio test.. and i get


    -1< x < 1


    I need help on checking for the endpoints

    the answer in the book is (-1,1]


    how do i know that 1 converges? do i just plug in 1 into the function?
     
  2. jcsd
  3. Apr 22, 2013 #2

    Dick

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    No, plug 1 into the series. See if the series converges. Do the same thing at x=(-1).
     
  4. Apr 22, 2013 #3
    okay so when i plug in 1 into the power series that i found..

    i get

    Ʃ (-1^n)(1^(n+1)) / (n+1)

    how do i know if this converges or diverges?

    do i use one of the series test?
     
  5. Apr 22, 2013 #4

    Dick

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    Sure, use a series test. 1^(n+1)=1. Looks like the alternating series test might be a good one.
     
  6. Apr 22, 2013 #5
    hmm when i use the alternating series test


    is my a_n = (1^(n+1)) / (n+1)

    because when i use that for a_n

    i get (1^(infinity)) / (infinity)

    or do we ignore (1^(n+1)) because it will always equal to 1?
     
  7. Apr 22, 2013 #6

    Dick

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    That is NOT the alternating series test. State the alternating series test.
     
  8. Apr 22, 2013 #7
    isnt it the lim as n -> infinity of a_n = 0

    and

    0 < a_n+1 ≤ a_n
     
  9. Apr 22, 2013 #8

    Dick

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    That's part of the premises. State the rest including the conclusion.
     
  10. Apr 22, 2013 #9
    conclusion?
     
  11. Apr 22, 2013 #10

    Dick

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    The conclusion is that if a series satisfies certain conditions then it converges. Does your series satisfy all of the conditions.
     
  12. Apr 22, 2013 #11
    yes the series does satisfy the conditions for 1.
     
  13. Apr 22, 2013 #12

    Dick

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    Alright. So it converges at x=1. Now try x=(-1).
     
  14. Apr 22, 2013 #13
    the lim as n -> infinity of (-1^(n+1)) / (n+1)

    im i using the alternate series test correct for this one?
     
  15. Apr 22, 2013 #14

    Dick

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    No, you left out the (-1)^n factor. Write out the first few terms of the series. Does it really alternate?
     
  16. Apr 22, 2013 #15
    okay. so it does not alternate it just stays negative
     
  17. Apr 22, 2013 #16

    Dick

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    Ok, then you need another test. Does it look like it converges to you?
     
  18. Apr 22, 2013 #17
    No it does not look like it will converge
     
  19. Apr 22, 2013 #18

    Dick

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    If you can give a good reason why then then you are done. It looks pretty similar to the harmonic series 1/n to me.
     
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