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Limit convergence

  1. Sep 18, 2009 #1
    1. The problem statement, all variables and given/known data

    How does r affect the convergence of:

    [tex]\displaystyle\sum_{n=r}^{\infty} \frac{(n-r)!}{n!}[/tex]


    3. The attempt at a solution

    [tex]\displaystyle\sum_{n=r}^{\infty} \frac{(n-r)!}{n!}=\displaystyle\sum_{n=r}^{\infty} \frac{(n-1)!}{(r+n)!}=\displaystyle\sum_{n=r}^{\infty} u_n[/tex]

    I thought that DAlembert's ratio test was appropriate maybe?

    [tex]\displaystyle\lim_{n\to \infty}\left(\frac{u_{n+1}}{u_n}\right)=\displaystyle\lim_{n\to \infty}\left(\frac{n!(r+n)!}{(r+n+1)!(n-1)!}\right)=\displaystyle\lim_{n\to \infty}\left(\frac{n}{(r+n+1)}\right)=\rho[/tex]

    I think I must have made a mistake, the first summation equation looks suspicious to me but as far as I remember it holds due to the summation starting at r thus they cancel and you simply have an (n-1)! numerator. Maybe the wrong method?
     
    Last edited: Sep 18, 2009
  2. jcsd
  3. Sep 18, 2009 #2
    I made a mistake with [tex]u, u_{n+1} [/tex] the limit should actually be:

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

    The question also states that r is an integer, it seems that for any r, the limit is 1? So this method is not useful in this situation.
     
  4. Sep 18, 2009 #3

    Dick

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    Think about doing a comparison test with a p-series. Or should I say 'r-series', hint.
     
  5. Sep 18, 2009 #4
    Do you mean to compare it with r = 1, which is the sum of 1/n which diverges. Then for any integer smaller than 1 clearly the numerator is dominant hence divergent. So r>1. r=2 gives

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

    So then, for [tex]r\ge 2[/tex] I will be able to work out that since a greater series converged that these ones do also?
     
  6. Sep 18, 2009 #5

    Dick

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    Yeah, basically. Take the r=2 case. 1/(n^2-n)<1/(n^2-n^2/2)=2/n^2 as long as n^2/2>n (which is true for sufficiently large n) since you've made the denominator smaller. Can you extend this to the general case of r>1?
     
  7. Sep 18, 2009 #6
    Well r is an integer that is >=2 so is >1
     
  8. Sep 18, 2009 #7

    Dick

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    I MEANT can you prove it converges for r=3,4,5... In a way perhaps similar to what I outlined for r=2? Or don't you believe the r=2 argument?
     
  9. Sep 19, 2009 #8
    [tex]\displaystyle \sum_{n=2}^{\infty} \left(\frac{2}{n^2}=\frac{1}{n^2-\frac{1}{2}n^2}\right)[/tex] Converges, [tex] |\int^{\infty} \frac{2}{n^2}| < \infty [/tex].

    [tex] \displaystyle\sum_{n=r}^{\infty}\left(\frac{(n-r)!}{n!}= u_{n,r}\right) [/tex]

    [tex] u_{n,2} = \frac{1}{n^2-n} < \frac{1}{n^2-\frac{1}{2}n^2} \iff n<\frac{1}{2}n^2[/tex]

    So, it can be show to be true for r=2 easily, since r=2 implies n starts at 2 and therefore that all terms in the series are less than the one that converges hence it must converge, but that could have been done with the method of differences, you find the limit to be 1 in the case of r=2. I don't feel like I have access to r=3,4,5. from what I've done here though although all I need to show is that [tex] u_{n,r>2} < u_{n,2}[/tex]

    I think this is shown when considering r>2,

    [tex] u_{n,r}=\frac{(n-r)!}{n!} \le u_{n,2} [/tex] if [tex] (n-r)! \le ( n-2)! \Rightarrow r \ge 2 [/tex]

    Since [tex]u_{n,r>1}\le u_{n,2}[/tex] it must converge.
     
  10. Sep 19, 2009 #9

    Dick

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    I'm not sure what you are talking about. Try r=3. u_n=1/(n*(n-1)*(n-2)). Can you show u_n<C/n^3 from some constant C and n sufficiently large? How large is 'sufficiently large'?
     
  11. Sep 19, 2009 #10
    I really don't know
     
  12. Sep 19, 2009 #11

    Dick

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    1/(n*(n-1)*(n-2))=(1/n^3)*(1/(1*(1-1/n)*(1-2/n))). What happens to the second factor as n gets large?
     
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