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Direct comparison test for convergence

  1. May 17, 2005 #1
    I'm supposed to compare the series

    [tex]\sum_{n=0}^{\infty}\frac{1}{n!}[/tex]

    to some other series to see if the one above converges or diverges. I have no idea of what to compare it to.

    I know by the ratio test that the above series converges, that is if i'm doing the ratio test correctly.

    [tex]\lim_{n\rightarrow\infty}\frac{1}{(n+1)!}n!\\=\frac{1}{n(n+1)}(1*n)=\\\frac{1}{(n+1)}[/tex]

    since this limit is zero which is less than one the series converges
     
  2. jcsd
  3. May 17, 2005 #2
    I'm thinking about majoring in mathematics. How important are sequences and series to a math major? I hate them BTW.
     
  4. May 17, 2005 #3

    StatusX

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    Homework Helper

    Compare it to the sum of 1/2^n. If you can show this converges, and show that every term in this series is greater than the corresponding term in the original series, that one must also converge.
     
  5. May 17, 2005 #4

    HallsofIvy

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    Or even, and simpler, just compare it to [tex]\frac{1}{n^2}[/tex]. As soon as n> 2,
    [tex]\frac{1}{n^n}< \frac{1}{n^2}[/tex]. Now, what do you know about the convergence of [tex]\Sigma_1^\infty\frac{1}{n^2}[/tex]?
     
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