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Power Series

  1. Mar 20, 2006 #1
    am trying to find the intervals of convergence for the summation, first deritive, and intergral of problems like this:

    the sum of [(-1)^n+1(x-5)^n]/[n5^n] from n=1 to infinity

    i know it is an alternating series and thus i am attempting to use that test to find convergence/diverigence
    lim as n gose to infinty=0 and asub.n is less than or equal to a

    when i try to find the limit, both the top and bottom go to infinity and then i'm trying to use l'hopital rule for an indeterminat form, but thats not wokring too well

    maybe i need to use the raito test, but that doesn't seem to work well either

    please point me in the right direction
     
  2. jcsd
  3. Mar 20, 2006 #2
    The ratio test should work perfectly for this series.
     
  4. Mar 20, 2006 #3
    when i do the raito test...the lim as n goes to infinty is not 0...the series diverges ..... how dose that work out...am i wrong?
     
  5. Mar 20, 2006 #4
    I think you're wrong... Can you show your work using the ratio test?
     
  6. Mar 20, 2006 #5
    well, for the raito test one condition has to be met inorder for the series to converge, that is the abs(an-10/an)<1

    so for the series we have [(-1)^n+2(x-5)^n+1]/[(n+1)5^n+1] all divided by [(-1)^n+1(x-5)^n]/[n5^n]

    i simplify and get [-(x-5)n]/[5(n+1)]<1

    i am stuck here
     
  7. Mar 20, 2006 #6

    Well you need to remember that you're taking a limit, and that there is an absolute value so the whole (-1)^k stuff goes away.
     
  8. Mar 20, 2006 #7
    ok so i get [(x-5)n]/[5(n+1)]<1 and i need to get a interval of convergence from this

    i have no idea
     
  9. Mar 20, 2006 #8
    Did you read everything I said in my last post? You need to take the limit as n goes to infinity.
     
  10. Mar 20, 2006 #9
    so [xn-5n]/[5n+5] as n goes to infinity would be [x-5]/5?
     
  11. Mar 20, 2006 #10
    Yes, but in order for the series to converge what are the conditions on that expression.
     
  12. Mar 21, 2006 #11

    HallsofIvy

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    I don't know why you multiplied that n into x- 5, it's much simpler as
    (|x-5|/5)(n/(n+1)). What is the limit of n/(n+1) as n goes to infinity?


    (That's simpler to read in LaTex:
    [tex]\frac{|x-5|}{5}\frac{n}{n+1}[/tex]
    What is the limit of
    [tex]\frac{n}{n+1}[/tex]
    as n goes to infinity?)
     
  13. Mar 21, 2006 #12
    so [x-5]/5<1

    and the interval is
    -1<[x-5]/5<1
    0<x<10
     
  14. Mar 21, 2006 #13

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Surely, there's a | key on your keyboard?
     
  15. Mar 22, 2006 #14
    thank you for all your help
     
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