I'm trying to evaluate the infinite series of e^Nx from N=0. As this expands to:
1 + e^x + e^2x + ... e^Nx
Can I treat it as a geometric series, where r = e^x, collapsing it to
1/(1 - e^x)?
$\displaystyle \displaystyle 1 + e^x + e^{2x} + ... e^{Nx} $
$\displaystyle \displaystyle \left(1 + e^x + e^{2x} + ... e^{Nx}\right) \left(\frac{(e^x - 1 )}{e^x -1 }\right) $
$\displaystyle \displaystyle \frac{e^{Nx+1} -1 }{e^x -1 }$
I do not understand that question.
You have two different ideas going on: $\displaystyle e^x~\&~N$.
If $\displaystyle x<0$, that is $\displaystyle 0<e^x<1$, then the series does converge.
Its sum is $\displaystyle \dfrac{1}{1-e^x}$.
As I said in reply #2, if $\displaystyle e^x\ge 1$ the series diverges.
So the answer depends upon the value of $\displaystyle x$.
Whereas, $\displaystyle N$ is simply an index for the sum.