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Calculating limits

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data

    I have to find the radius of convergence and convergence interval. So for what x's the series converge.
    The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
    So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

    2. Relevant equations

    CodeCogsEqn-2.gif


    CodeCogsEqn.gif (?)


    3. The attempt at a solution

    I tried using CodeCogsEqn-4.gif with an = CodeCogsEqn-3.gif

    but i don't seem to get anywhere....
    Is there another way of showing that the lim = 0 ?
     
  2. jcsd
  3. Dec 7, 2014 #2

    Mark44

    Staff: Mentor

    You need to take the limit of the entire term in the summation, including the (x + 1)n factor.
    Show us the work that you did in taking the limit of an+1/an. It's easy to get wrong with those factorials.

    BTW, this is NOT a precalculus problem, so I moved it to the calculus section.
     
  4. Dec 7, 2014 #3
    Ah ok I'm sorry.
    Yes indeed. But you would get: CodeCogsEqn-5.gif
    So you might as well just use the 'n-part', for the (x-c) is just a number (?)

    I ended up with:

    CodeCogsEqn-6.gif It's not possible to simplify this, right?
     
  5. Dec 7, 2014 #4

    Dick

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    Science Advisor
    Homework Helper

    Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.
     
  6. Dec 7, 2014 #5

    Mark44

    Staff: Mentor

    Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.
    Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
    .
    It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.
     
  7. Dec 7, 2014 #6
    Ah of course!
    (n+1)!/ n! = n+1.

    I still don't know these rules. But I used :
    n=3
    4*3*2*1/3*2*1 = 4
    = n+1
     
  8. Dec 7, 2014 #7
    Hmm I see.
    You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
    Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

    Weird that my calculus book leaves the |x +c| out then...
     
  9. Dec 7, 2014 #8

    Dick

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    You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.
     
  10. Dec 7, 2014 #9
    I've no idea how to do that...
     
  11. Dec 7, 2014 #10

    Ray Vickson

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    Isn't it the case that ##(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!##? Just keep using the definition of factorial.
     
  12. Dec 7, 2014 #11
    Oh well... I blame it on my lack of sleep, ha.
    So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)
     
  13. Dec 7, 2014 #12

    Dick

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    Ok, so do you know how to find its limit?
     
  14. Dec 8, 2014 #13
    Yes, I do. Thanks for the help all!
     
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