The Mystery of the e Series: Uncovering Its Name

[LENIN]

I just wonder what's 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.

 

[nd]

\displaystyle e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}, if I remember correctly.
What are you interested in learning about series?
Most series, as far as I know, don't have names.

 

[Hurkyl]

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.

 

[HallsofIvy]

Hurkyl, they are not the same. The series in the first post was
\frac{1}{1!}+ \frac{2}{2!}+ \frac{3}{3!}+...= 1+ 1+ \frac{1}{2!}+ ...
and so is e+ 1, not e.

 

[shmoe]

They're both e.

e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}

\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

 

[LENIN]

They really 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.

 

[HallsofIvy]

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