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Basic comparison test

  1. Aug 13, 2009 #1
    sum(1/n!,n,1,inf)
    the only thing i can think of is
    1/n^2>=1/n! , 1/n^2 being a convergent p series since p >1
    thus /n! is convergent
     
    Last edited: Aug 13, 2009
  2. jcsd
  3. Aug 13, 2009 #2

    Cyosis

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    nameVoid could you start putting some effort into your posts perhaps? In particular use the template and state your question plus your work clearly.
     
  4. Aug 13, 2009 #3
    whats the syntax for definite integral
     
  5. Aug 13, 2009 #4

    Cyosis

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    You should split off the first few terms to make the comparison with 1/n^2 more accurate.

    The syntax for a definite integral is [ tex] \int_{lowerlimit}^{upperlimit} [ /tex].

    This is not true for all n, take n=2 for example.
     
  6. Aug 13, 2009 #5

    statdad

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    If you are investigating the series

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

    then you can use

    [tex]
    \sum_{n=1}^\infty \frac 1 {n^2}
    [/tex]

    and compare terms (and the second series IS a convergent p-series: I'm not sure what Cyosis was getting at).
     
  7. Aug 13, 2009 #6

    Cyosis

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    I was getting at the fact that for the first few terms the inequality doesn't hold. So concluding right away that one converges may bit a bit of a jump hence I suggested to write out the first few terms until the inequality indeed holds.
     
  8. Aug 13, 2009 #7
    [tex]\sum_{n=1}^{\infty}1/n![/tex]
    [tex]\frac{1}{n!}\leq\frac{1}{n^2} , n\geq 4[/tex]

    [tex]\sum_{n=1}^{\infty}1/n^2:convergent [/tex]
    [tex]b_{n}=\sum_{n=4}^{\infty}1/n^2\geq\sum_{n=4}^{\infty}1/n!=a_{n}[/tex]
    i believe there is a property which states that if
    [tex]\sum_{n=c}^{\infty}a_{n}[/tex]
    is convergent then the series converges for all n>=1 in any case here are the first few terms
    [tex]\sum_{n=1}^{\infty}1/n^2=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}....[/tex]
    [tex]\sum_{n=1}^{\infty}1/n!=1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}....[/tex]
    this seems to confirm things
     
    Last edited: Aug 13, 2009
  9. Aug 13, 2009 #8
    Looks fine to me. Good work. But you might want to prove that 1/n! <= 1/n^2 for n>= 4.
     
  10. Aug 13, 2009 #9

    Mark44

    Staff: Mentor

    Just as a side note, a series almost identical to yours--
    [tex]\sum_{n = 0}^{\infty}\frac{1}{n!}[/tex]
    -- converges to a number that is familiar to all mathematicians. Note that 0!, by definition, is 1.
     
  11. Aug 13, 2009 #10
    Does the question state explicitly that you need to use the Comparison Test? If not, then you may want to try using the Ratio Test, which I think works equally well in this case and you don't need to do induction to check what you're comparing this against actually works.

    Denote [tex]a_n = 1/n![/tex]. Then see that [tex]\lim_{n \to \infty} |a_{n+1}/a_n| = \lim \dfrac{1}{(n+1)!}\cdot n! = \lim \dfrac{n!}{(n+1)n!} = \lim \dfrac{1}{n+1} = 0 = \limsup |a_{n+1} / a_n| < 1 [/tex]. Thus, by the Ratio Test, this converges.
     
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