Question

How could I show this is true/false?

Answer 1

6 viewing. Oh boy this is exciting!

Answer 2

The suspense is killing me..

Answer 3

Me to man

Answer 4

\[\int\limits x^{-1}e^{ax}dx\] can be integrated two ways as far as I know. \[\int\limits \sum_{n=0}^{\infty} \frac{a^n x^{n-1}}{n!}dx=\sum_{n=0}^{\infty}\frac{(ax)^n}{n*n!}\] The other way is to do integration by parts to get \[\int\limits x^{-1}e^{ax}dx= \frac{e^{ax}}{ax}+\frac{1}{a}\int\limits x^{-1}e^{ax}dx=...\]Continually repeating will get you this infinite sum \[e^{ax}\sum_{n=0}^{\infty}\frac{n!}{(ax)^n}=\sum_{n=0}^{\infty}\frac{(ax)^n}{n!}\sum_{n=0}^{\infty}\frac{n!}{(ax)^n}\] So those two power series multiplied seem to be equal to that original power series.

What do?!

Answer 5

Whoops, there should be a x^(-2) in the second integral after doing by parts.

Answer 6

I dunno, I guess I wasted some peoples time here because I don't think that sum converges. I think I either did some algebra wrong or there's something fishy here hmm... Also, check this out if you're curious for an explanation of what I'm talking about.

http://mrchasemath.wordpress.com/2013/01/09/integration-by-parts-and-infinite-series/

Answer 7

what class is this?

Answer 8

calc III