# Thread: Infinite series of an exponent question.

1. ## Infinite series of an exponent question.

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)?

2. Originally Posted by DenJansen
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
If $e^x\ge 1$, As an infinite series that series diverges.

3. $\displaystyle 1 + e^x + e^{2x} + ... e^{Nx}$

$\displaystyle \left(1 + e^x + e^{2x} + ... e^{Nx}\right) \left(\frac{(e^x - 1 )}{e^x -1 }\right)$

$\displaystyle \frac{e^{Nx+1} -1 }{e^x -1 }$

4. Ah, I see. Thank you both!

5. Wait, I just noticed something. As this is an infinite series from N=0 to N->∞, wouldn't the equation above simply = ∞? Is there a way to solve this where N is not in the final equation?

6. Originally Posted by DenJansen
Wait, I just noticed something. As this is an infinite series from N=0 to N->∞, wouldn't the equation above simply = ∞? Is there a way to solve this where N is not in the final equation?
I do not understand that question.
You have two different ideas going on: $e^x~\&~N$.
If $x<0$, that is $0, then the series does converge.
Its sum is $\dfrac{1}{1-e^x}$.

As I said in reply #2, if $e^x\ge 1$ the series diverges.

So the answer depends upon the value of $x$.
Whereas, $N$ is simply an index for the sum.