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Amount of Terms Required to Find a Sum at an Indicated Accuracy

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

    [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}[/tex]
    (|error| < 0.0001)

    2. Relevant equations
    Alternating Series Test
    General knowledge of adding series up...

    3. The attempt at a solution
    Alternating Series Test
    1.) Limit x->infinity
    [tex]\frac{(-1)^{n}}{n*9^{n}}[/tex] = 0
    2.) Decreasing eventually for n (really immediately)
    [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{(n+1)*9^{n+1}} < \sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}[/tex]

    By the Alternating Series Test, this series ([tex]\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}[/tex]) is convergent.

    Well, I originally choose n = 4 terms in order to find the indicated accuracy. However, that was wrong.

    So, I'm sorta scratching my head as to what I did wrong...

    My approach:
    [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}[/tex]

    I start listing out an's terms -
    |an| < 0.0001
    a1 = 0.11111111111
    a2 = 0.006172839506
    a3 = 0.000457247370
    a4 = 0.000038103947
    a5 = 0.000003387017
    a6 = 0.000000313612
    a7 = 0.000000029867

    a4 < 0.0001, so n should equal four terms.... However, it doesn't for some odd reason?

    Any help/suggestions as to where I went wrong are appreciated and will be thanked!

    Sincerely,

    NastyAccident
     
    Last edited: Oct 14, 2009
  2. jcsd
  3. Oct 14, 2009 #2

    Mark44

    Staff: Mentor

    You have an alternating series, so every other an (starting from aa) should be negative. You show all of them as positive in your list.
     
  4. Oct 14, 2009 #3
    So, in essence, I shouldn't of taken the absolute value of each of the an's and should of left it with the +/- signs?

    If that is the case, then the problem should be okay for n = 3?

    Since:
    [tex]\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n*9^{n}}[/tex]
    is within
    [tex]\sum^{\infty}_{n=1}\frac{(1)^{n+1}}{(n+1)*9^{(n+1)}}[/tex]
    which gives me:

    an < 0.0001
    a1 = 0.061728395
    a2 = 0.0004572473708
    a3 = 0.000038103947
    a4 = 0.000003387017
    a5 = 0.000000313612
    a6 = 0.000000029867

    So, n = 3 would be correct...
     
    Last edited: Oct 14, 2009
  5. Oct 14, 2009 #4

    Mark44

    Staff: Mentor

    Right. You don't want the abs. value of your terms. Just add them up. Your estimate with a1 + a2 + a3 should be within .0001 of the actual value of the infinite sum. You did add up the three terms, right?
     
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