Prove limit of improper Integral is 1

[sinClair]

Homework Statement

Show \mathop{\lim}\limits_{n \to \infty}(\frac{1}{n!}\int_{1}^{\infty}x^n\frac{1}{e^x} dx )=1

Homework Equations

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

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.

 

[tiny-tim]

1/e-b^n/e^b+n\int_{1}^{b}\frac{1}{e^x}x^(n-1)dx.

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 …

 

[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...?

 

[tiny-tim]

Hi sinClair!

I haven't actually cheked it, but from:

\frac{1/e-b^n/e^b}{n!}\,+\,\int_{1}^{b}\frac{x^{n-1}\,e^{-x}}{(n-1)!}dx​

don't you eventually get something like

e^{-1}\sum\frac{1}{n!}\,-\,e^{-b}\sum\frac{b^n}{n!}​

?