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Convergence of a series with factorial

  1. Jan 23, 2008 #1
    1. The problem statement, all variables and given/known data

    My task is to determine whether the following series is convergent or not.

    [tex]\sum_{n=1}^{\infty}\left(\frac{n}{e}\right)^{n}\frac{1}{n!}[/tex]

    3. The attempt at a solution

    The limit of the terms is zero, so the series might be convergent. I tried the ratio test, but the limit is equal to 1, so that didn't help. Now I'm trying Raabe's test, but the limit is quite difficult to determine...

    Do you have any ideas, please? Any help would be greatly appreciated...
     
  2. jcsd
  3. Jan 23, 2008 #2
    Well, one step closer... Now I know it's divergent - I used the comparison test with

    [tex]\frac{1}{2\pi n}[/tex]

    which yields the second limit from here - http://en.wikipedia.org/wiki/Stirling's_approximation - which is equal to 1. This means that the series is divergent (because 1/sqrt(2*pi*n) is divergent). But I would like to do a Stirling-free determination of convergence.

    The Raabe's test yields this limit:

    [tex]\lim_{n \rightarrow \infty} n \left(\frac{en^{n}}{{\left(n+1 \right)}^{n}} - 1 \right) [/tex]

    Could you please post any hints as to how to solve this limit? Thanks in advance!
     
    Last edited: Jan 23, 2008
  4. Jan 23, 2008 #3
    L'Hopitals rule will work, but the result will be infinity i.e. >1 so converges. Just curious if you tried the root test.
     
  5. Jan 23, 2008 #4
    hey, i used the the root test, but that yields one, so it's not of much help... - i used mathematica to compute that, the limit is following:

    [tex]\lim_{n \rightarrow \infty}\frac{n}{e\sqrt[n]{n!}}[/tex]

    btw, i can't see how l'hospital rule would help - i think it justs complicates the expression...
     
  6. Jan 23, 2008 #5

    Dick

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    The Raabe's test limit is 1/2. To show this use as many terms as you need from the taylor series expansions of log(1+x) and e^x.
     
  7. Jan 23, 2008 #6
    I put the expression in Maple and get infinity as a result for the limit.
     
  8. Jan 23, 2008 #7

    Dick

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    Interesting. I put it into maxima and got .4958745375797102 at n=50. Did you even try to work it out without a machine?
     
    Last edited: Jan 23, 2008
  9. Jan 24, 2008 #8
    Could you please elaborate a little bit on that? Do you mean to rewrite this part:

    [tex](\left n+1 \right)^{n}[/tex]

    as

    [tex]e^{n\log{(n+1)}}[/tex]

    and then to do a Taylor expansion of this expression?
     
  10. Jan 24, 2008 #9

    Dick

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    Actually I would combine the nth powers and expand (1+1/n)^n. By expanding n*log(1+1/n). Then use the expansion of e^x to express everything as powers of n. Remember the expansion parameter should be a SMALL number.
     
    Last edited: Jan 24, 2008
  11. Jan 24, 2008 #10
    Yes, I did. I think maybe I got the expressions wrong. Is there a power on the 'e' in the expression?
     
  12. Jan 24, 2008 #11

    Dick

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    No power on the 'e'. Exactly whats written in post 2.
     
  13. Jan 24, 2008 #12
    but by combining the nth powers (i.e. (n^n)/(n+1)^n), wouldn't you get rather (1-1/(n+1))^n than (1+1/n)^n?
     
  14. Jan 24, 2008 #13

    Dick

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    It's easier the other way around. Write it as e/(1+1/n)^n=e*(1+1/n)^(-n).
     
  15. Jan 24, 2008 #14
    thanks a lot...!

    and btw, by expansion parameter do you mean the degree to which I expand the Taylor polynomial?
     
    Last edited: Jan 24, 2008
  16. Jan 24, 2008 #15

    Dick

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    I mean for example that the taylor expansion of log(1+x) is x-x^2/2+x^3/3... I can't use that to expand log(1+n) since the terms are getting larger and larger (it's divergent). I CAN use it to expand log(1+1/n). So the x in the taylor series needs to be small, that's what I'm referring to as 'expansion parameter'.
     
  17. Jan 24, 2008 #16
    Could you just check if this is correct? I'm not quite sure how to work with the small oh notation:

    First I get this:
    [tex]e^{1}e^{(-n)\log{(1+\frac{1}{n}})[/tex]

    Then I express the logarithm as a Taylor series:

    [tex]log(1+\frac{1}{n})= \frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2})[/tex]

    This is multiplied with (-n) and then I add 1 from the first e to get:

    [tex]e^{\frac{1}{2n}-no(\frac{1}{n^2})[/tex]

    Then I use Taylor expansion to rewrite it as following:

    [tex]1+(\frac{1}{2n}) - no(\frac{1}{n^2}) + o(\frac{1}{n})[/tex]

    - I'm not quite sure about the last o().. don't know what to write there exactly. Should I write the whole expression (see below) there?

    [tex]o\left((\frac{1}{2n}) - no(\frac{1}{n^2})\right)[/tex]

    Then I subtract 1 and multiply the rest by n, as the limit says. I get:

    [tex]\lim_{n \rightarrow \infty}\left( \frac{1}{2}- \frac{o(\frac{1}{n^2})}{\frac{1}{n^2}}+ \frac{o(\frac{1}{n})}{\frac{1}{n}}\right)[/tex]

    which is equal to 1/2... Are the steps ok?
     
    Last edited: Jan 24, 2008
  18. Jan 24, 2008 #17

    Dick

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    I think so, aside from some typos, like log(1+1/n)=1/n-1/(2n^2)+o(1/n^3). You can do things like n*o(1/n^3)=o(1/n^2). The final thing you are taking the limit of should come out to 1/2+o(1/n). Then you can ignore the o(1/n). But that's basically it.
     
    Last edited: Jan 24, 2008
  19. Jan 24, 2008 #18
    Ok, thanks you very much, it helped me a lot...

    ad "o(1/n^3)".. my book says it should be o(1/n^2) since we're using the small o notation, not the big o, so that's may be the difference.
     
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