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I need help with finding the radius of convergence of a complex series

  1. May 3, 2012 #1
    Radius of convergence of the series n^2(x^n)/(3n!) I am stumped

    the question is: find the radius and interval of convergence of the following series {sum_(n=1)^(Infinity)}((n^2)(x^n))/(3*6*9***3n)

    I'm assuming that equal to ((n^2)(x^n))/(3n)!

    then lim_(n->infinity) of (((n+1)^2(x^(n+1))/(3n+3)!)*((3n)!/((n^2)(x^n))

    = lim_(n->infinity) of ((n+1)^2(x))/((n^2)(3n+3)(3n+2)(3n+1))
    =x*lim_(n->infinity) of ((n+1)^2)/(27n^5+54n^4+33n^3+6n^2)
    lim_(n->infinity) of (n^2)/(n^5) --> 0
    =x*0 = 0<1

    so now what? i don't know how to find the radius of convergence in this situation

    does it help to know that the {sum_(n=1)^(Infinity)}(x^n)/n! converges to e^x?

    my book scarcely mentions factorials in series, is there a good place to review them?
     
    Last edited: May 3, 2012
  2. jcsd
  3. May 3, 2012 #2
    nevermind the denominator is n!*3, silly me
     
    Last edited: May 3, 2012
  4. May 3, 2012 #3

    Dick

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    Think again. Isn't 3^n*n! more like it?
     
  5. May 3, 2012 #4
    wouldn't 3^n*n! give
    N Value
    1 3
    2 18
    3 162
    giving (3*18*163***3^3n*3n)

    are those the values you should get, my book lacks anything explaining what 3*6*9***3n would even mean.

    I thought it meant 3(n!) because the first two are right
    N Value
    1 3
    2 6 but then
    3 18
    3n 9n
    giving (3*6*18***9n)

    Is there are pattern for these, I tried to find them in my book and on Wikipedia but i don't know what this expression is called
     
    Last edited: May 3, 2012
  6. May 3, 2012 #5

    Dick

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    E.g. 3*6*9=3*1*3*2*3*3=(3*3*3)*(1*2*3)=3^3*3!. You seem to be confusing the individual terms with the whole product.
     
  7. May 3, 2012 #6
    Wow, it makes so much more sense now, thank you
    I just didn't know what it all meant. you are my hero for the day
     
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