Prove limit of improper Integral is 1

1. Mar 16, 2008

sinClair

1. The problem statement, all variables and given/known data
Show $$\mathop{\lim}\limits_{n \to \infty}(\frac{1}{n!}\int_{1}^{\infty}x^n\frac{1}{e^x} dx )=1$$

2. Relevant equations
The hint is that $$e=\mathop{\lim}\limits_{n \to \infty}\sum_{k=0}^{n}1/k!$$

3. The attempt at a solution
First I wrote out the improper integral as limit of a proper integral. Then I tried to integrate by parts with u=x^n dv=e^-xdx...which eventually gets that proper integral to be $$1/e-b^n/e^b+n\int_{1}^{b}\frac{1}{e^x}x^(n-1)dx$$

But even when I take limit b->infinity of this remaining integral either it's going to exist and be finite in which case when i take n->infinity it will be infinite or else it dosn't exist or is not finite in which case the original integral is not improperly integral. I tried using u=e^-x and dv=x^n with little success as well--did I make a mistake somewhere here?

Thanks.

2. Mar 16, 2008

tiny-tim

Hi sinClair!

You haven't included the 1/n! :
$$\frac{1/e-b^n/e^b}{n!}\,+\,\int_{1}^{b}\frac{x^{n-1}\,e^{-x}}{(n-1)!}dx$$​

… and then you can keep doing it again …

3. Mar 17, 2008

sinClair

Are you thinking that you just keep on integrating by parts and then that will yield a sum that will be e? I'm having trouble summing up the uv part of the integration... I mean when i do it again I get -b^n*e^-b+e^-1+n(-e^-b*b^n-1+e^-1+integral.....) this sums up...?

4. Mar 17, 2008

tiny-tim

Hi sinClair!

I haven't actually cheked it, but from:
don't you eventually get something like
$$e^{-1}\sum\frac{1}{n!}\,-\,e^{-b}\sum\frac{b^n}{n!}$$​
?