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Calc 2: power series

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the radius of convergence and interval of convergence of the series:
    [itex]\sum[/itex] (-1)n * ( x^(2n) / (2n)! )
    for n=0 to infinity


    2. Relevant equations
    well, any of the divergence tests: ratio, limit comparison, etc.


    3. The attempt at a solution
    I dont get power series at all ... well, series at all. All I know is you have to use one of the divergence tests for this .. but I dont know really how to apply them to problems .. any help at all would be appreciated

    but, i was thinking to use the ratio test. not sure if thats the best way to go.. can someone also explain how you know which test to use??

    So anyways, if you do the ratio test, I ended up with, [ (-1)*(x^2)*(2n)! ] / (2n+2)!
    and now I dont know what else to do with this. i see that x doesn't depend on n so pull that out of the limit .. and thats as far as i get
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 20, 2012 #2

    Mark44

    Staff: Mentor

    In the ratio test, you're dealing with the absolute values of terms, so you can ignore the (-1)n factor.

    That leaves you with |x|2 * (2n)! / (2n + 2)!

    (2n + 2)! = (2n + 2) * (2n + 1) * (2n)!, right?

    Under what conditions do you get convergence with the Ratio Test?
     
  4. Feb 20, 2012 #3
    yea that makes sense. so you always ignore the (-1)^n if using ratio test??

    so, if the limit L < 1 it is convergent and if it is > 1 then it is divergent. how do we know if it is greater or less than 1?
     
  5. Feb 20, 2012 #4
    doesnt that just leave with like x = + or - 1 ? or am i going about that wrong
     
  6. Feb 20, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The ratio test says that a power series converges (absolutely) if that ratio is less than one. Your condition is not that x be equal to 1 or -1 but that it lie between -1 and 1. Now, what is the interval of convergence and radius of convergence?
     
  7. Feb 20, 2012 #6
    oops yea thats what i meant, x is between -1 and 1. so the radius of convergence is just 1? and the interval of convergence is (-1,1) ? ugh i dont know
     
  8. Feb 20, 2012 #7

    Mark44

    Staff: Mentor

    I think you need to back up a minute. You're looking at this limit:
    [tex]\lim_{n \to \infty} \frac{|x|^2}{(2n + 2)(2n + 1)}[/tex]

    For convergence, this limit needs to be < 1. For what x will that be true?
     
  9. Feb 20, 2012 #8
    as long as x > 1 ?
     
  10. Feb 20, 2012 #9

    Mark44

    Staff: Mentor

    I think you're guessing. Since x doesn't have anything to do with n, you can move it outside the limit. If you do that, what do you get for the limit?
     
  11. Feb 20, 2012 #10
    youd have 1/(2n+2)(2n+1) and when n goes to infinity, that'd be 1/infinity so isnt that just like 0 ? ugh calc 3 was so much easier than this =/ that answer doesnt make sense
     
  12. Feb 20, 2012 #11
    wait no .... when n goes to 0 ... that'd be 1/2 right? so x would have to be less than + or - sqrt(2) ?
     
    Last edited: Feb 20, 2012
  13. Feb 21, 2012 #12
    Ok so I just went through this problem again and I get stuck with this:

    x*(lim as n approaches infinity of 1/((2n+2)*(2n+1)))

    So doesn't that limit just go to 0? If that's true, where do I go from there to find the interval of convergence and radius of convergence ?
     
    Last edited: Feb 21, 2012
  14. Feb 21, 2012 #13

    Mark44

    Staff: Mentor

    The limit is NOT as n goes to 0, so that shouldn't even be considered.
     
  15. Feb 21, 2012 #14
    That's why I posted again, I realized that. So is the post after that in which it goes to infinity correct?
     
  16. Feb 21, 2012 #15

    Mark44

    Staff: Mentor

    Actually, it's
    [tex]|x| \lim_{n \to \infty}\frac{1}{(2n + 2)(2n + 1)}[/tex]

    And yes, this limit is 0.
    Since the limit above is zero, no matter what x is, what does that say about any restrictions on x? That is related to what the interval of and radius of convergence are.
     
  17. Feb 21, 2012 #16
    Isn't that what I have I just didn't have the absolute value signs? I'm on my phone I keep forgetting them. But yea so for all x, it goes to zero so R= Infinity right? So that's the radius of convergence and the Interval is just (-infinity, infinity) ???
     
  18. Feb 21, 2012 #17

    Mark44

    Staff: Mentor

    Don't forget them - they're important.
    Right.
     
  19. Feb 22, 2012 #18
    Yayyy thanks so much!
     
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