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Whats the name

  1. Mar 11, 2005 #1
    I just wonder whats the name of the serries 1/1!+2/2!+3/3!... I know it equals e but I just whant to know how it's called.

    PS. Titels of good books abot serries would allso be welcome.
  2. jcsd
  3. Mar 11, 2005 #2


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    [tex]\displaystyle e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}[/tex], if I remember correctly.
    What are you interested in learning about series?
    Most series, as far as I know, don't have names.
    Last edited: Mar 11, 2005
  4. Mar 11, 2005 #3


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    Exercise: prove the series written in the first two posts are the same!

    I don't think this particular series has a name. However, it is the evaluation of the Taylor series for e^x at x = 1, or more specifically, the MacLauren series.
  5. Mar 11, 2005 #4


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    Hurkyl, they are not the same. The series in the first post was
    [tex]\frac{1}{1!}+ \frac{2}{2!}+ \frac{3}{3!}+...= 1+ 1+ \frac{1}{2!}+ ...[/tex]
    and so is e+ 1, not e.
  6. Mar 11, 2005 #5


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    They're both e.


    [tex]\displaystyle e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}\ldots=1+1+\frac{1}{2!}+\ldots[/tex]
  7. Mar 11, 2005 #6
    They realy are both e. I tried to prove thath and I think I maneged to prove that they are equal for very large n, where 1/n is almost equal to 0. It's actualy qouit easy to do it with the use of the binomical expresion.

    PS. But I still don't get it why it's so much fester to do it with a series. When you get to the 13'th element (13/13!) it's allready excet to 10 digits. But if you do it as (1+1/n) on n, you have to use a very large n to get such an excet figure. Why is that?

    PPS. Thanks for the info Hurkley.
  8. Mar 11, 2005 #7


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    What? I'm wrong? Moi?? Oh, blast, I started my series with n=1 instead of n= 0!
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