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Homework Help: Convergence of Infinite Series

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    1+ [tex]\frac{\alpha\beta}{\gamma}[/tex] x + [tex]\frac{\alpha (\alpha+1)\beta(\beta+1)}{1.2.\gamma(\gamma+1)}[/tex][tex]x^{2}[/tex]+.....


    2. Relevant equations



    3. The attempt at a solution
    Using D'Alembert's ratio test, I get [tex]lim_{n\rightarrow\infty}[/tex][tex]\frac{U_{n+1}}{U_{n}}[/tex]=x
    so, x>1 diverging series
    x<1 converging series
    when x=1, Using Raabe's test I get
    [tex]lim_{n\rightarrow\infty}[/tex][tex]n[\frac{U_{n}}{U_{n+1}}-1][/tex]=[tex]\gamma-\alpha-\beta[/tex]
    so, Series Converges if [tex]\gamma-\alpha-\beta[/tex]>1
    and diverges if [tex]\gamma-\alpha-\beta[/tex]<1

    However the book has given the answer to be
    coverges if [tex]\gamma-\alpha-\beta[/tex]>0
    and diverges if [tex]\gamma-\alpha-\beta[/tex]<0

    Can anyone point out my mistake, please?
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2

    Gib Z

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    Homework Helper

  4. Sep 9, 2007 #3

    Avodyne

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    For Raabe's test wtih x=1, I get
    [tex]\lim_{n\rightarrow\infty}n\!\left({\textstyle{U_{n}\over U_{n+1}}-1\right)=\gamma-\alpha-\beta+1[/tex]
     
  5. Sep 9, 2007 #4
    My expression for [tex]U_{n}[/tex] = [tex]\frac{1.\alpha...(\alpha + n-1) 1.\beta...(\beta+n-1)}{1.2...(n-1) 1.\gamma...(\gamma+n-1)}[/tex]

    Using this I get [tex]\gamma-\alpha-\beta[/tex] from Raabe's test
     
    Last edited: Sep 9, 2007
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