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How simplifying this ratio test for series?

  1. Oct 31, 2007 #1
    [SOLVED] How simplifying this ratio test for series?

    Basically, we are asked if (n!n^2)/(2n)! converges absolutely...

    I got to the point where

    lim as n --> infinity of

    [(n+1)!(n+1)^2]/(2(n+1))! X (2n)!/n!n^2

    by the ratio test.

    But I don't know how to manipulate the factorials from here on...could anyone help?


    Thank you.
     
  2. jcsd
  3. Oct 31, 2007 #2

    rock.freak667

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    (n+1)!=(n+1)n! which should cancel with the n!
    2(n+1))! =(2n+2)!=(2n+2)(2n-1)!
    (2n)!=2n(2n-1)!
     
  4. Nov 1, 2007 #3

    Gib Z

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    You could replace the factorials with their respective Stirling approximations, which is asymptotic to the factorial function, ie as n grows large the quotient of the approximation and the factorial approaches 1.
     
  5. Nov 1, 2007 #4

    Curious3141

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    (My earlier idea was wrong because I compared with a divergent series, but adapted slightly and if you're allowed to assume the convergence of zeta(2), the following might do the trick simply).

    For large n, the denominator (2n)! = (n+n)! >> (n+4)! = n!(n+1)(n+2)(n+3)(n+4) > (n!)n^4

    Cancel off the numerator and you're left with terms that are strictly less than 1/n^2 for large n. By the limit comparison test with zeta(2), this converges.
     
    Last edited: Nov 1, 2007
  6. Nov 1, 2007 #5
    I see, does anyone know a place I can look up these formulas?
     
  7. Nov 1, 2007 #6
    rock.freak,

    for the (2n+2)!, shouldn't it be (2n+2)(2n+1)! ... because you keep subtracting 1??


    And can anyone please evaluate this series...I am getting stuck at the end with

    lim as n --> infinity of (n+1)^2/n.


    Thank you.
     
  8. Nov 1, 2007 #7

    D H

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    I assume you are look at the convergence of

    [tex]\sum_n \frac {n!n^2}{(2n)!}[/tex]

    Applying the ratio test, you want to determine

    [tex]\lim_{n\to\infty}\frac {(n+1)!(n+1)^2}{(2(n+1))!}\frac {(2n)!}{n!n^2}[/tex]

    First expand the factorials: [itex](n+1)!=(n+1)n![/tex] and [itex](2(n+1))!=(2n+2)(2n+1)(2n)![/tex]. Inserting these in the limit expression yields

    [tex]\lim_{n\to\infty}\frac {(n+1)n!(n+1)^2}{(2n+2)(2n+1)(2n)!}\frac {(2n)!}{n!n^2}[/tex]

    Simplifying,

    [tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]

    Most of the remaining terms cancel in the limit. For example,

    [tex]\lim_{n\to\infty}\frac{n+1}{n} = 1+\lim_{n\to\infty}\frac{1}{n} = 1[/tex]
     
  9. Nov 1, 2007 #8

    rock.freak667

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    Yes it should...I made a mistake
     
  10. Nov 1, 2007 #9
    I still don't see how you evaluate that last (n+1)^3 limit, though.

    DH, did you expand the terms or cancel something out?

    Thank you everyone for the great help.
     
  11. Nov 1, 2007 #10

    D H

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    The only terms I canceled were [itex]n!/n![/itex] and [itex](2n)!/(2n)![/tex]. I left the evaluation of

    [tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]

    up to you, with a very strong hint in

    [tex]\lim_{n\to\infty}\frac{n+1}{n} = 1[/tex]

    I guess the hint wasn't quite enough. Just drop the +1/+2 terms: change [itex](n+1)^3[/itex], [itex]2n+2[/itex], and [itex]2n+1[/itex] to [itex]n^3[/itex], [itex]2n[/itex], and [itex]2n[/itex], respectively.
     
  12. Nov 1, 2007 #11
    I see, so I would get [tex]\lim_{n\to\infty}\frac{n^3}{2n.2n.n^2} [/tex]

    SO, here I cancel the n^3 with the n^2, correct, getting n/(2n)(2n) -- Don't I get zero??

    A little more effort on your part and I am on my way to grasping this. Thanks.
     
  13. Nov 1, 2007 #12

    D H

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    Yes, the ratio test limit is [itex]\lim_{n\to\infty}1/(4n)=0[/itex]. The series converges.
     
  14. Nov 1, 2007 #13
    but, how/why did you get 1 in your previous post?
     
  15. Nov 1, 2007 #14

    D H

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    That was not the same limit. That was

    [tex]\lim_{n\to\infty}\frac{n+1}{n}[/tex]

    which is indeed one but is also quite different from

    [tex]\lim_{n\to\infty}\frac {(n+1)^3}{(2n+2)(2n+1)n^2}[/tex]
     
  16. Nov 1, 2007 #15
    I see you were just providing an example? that actually confused me a lot because i thought it was the next sterp.

    Thanks for the help!

    I guess what i get is 1/4n, which goes to zero.
     
    Last edited: Nov 1, 2007
  17. Nov 1, 2007 #16

    D H

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    That's why I said "Most of the remaining terms cancel in the limit. For example,"

    You're welcome. I'm marking this as solved.
     
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