The discussion revolves around understanding the manipulation of an infinite series, specifically transitioning from the summation of an exponential function to a closed form expression. The subject area is calculus, focusing on geometric series and their convergence properties.
Some participants have provided insights into the properties of geometric series, suggesting that the limit depends on the value of the variable involved. There is an exploration of the conditions under which the series converges, but no consensus has been reached regarding the specific steps of the derivation.
Participants note that questions about infinite series are typically covered in calculus courses, indicating a potential gap in the original poster's background knowledge. There is also mention of the need for clarity in notation and expression to avoid confusion in mathematical implications.
This is a geometric series.Alex_Neof said:Homework Statement
I'm reading a derivation and there is a step where the writer goes from:
## \sum_{n=0}^\infty e^{-n\beta E_0}##
to:
## \frac {1} {(1-e^{-\beta E_0})}.##
I can't see how they did this.Homework Equations
[/B]
I think it just involves equation manipulation.
The Attempt at a Solution
Can someone give me a clue, so I can attempt this problem?
Kind regards.
Although it looks very fancy, the last line should be ##= \frac {1} {1-e^{-\beta E_0}}##. The implication arrow (##\Rightarrow##) is used to show that one statement implies the following statement. What you have instead are equal expressions.Alex_Neof said:Cheers guys. I found online that "the limit of a geometric series is fully understood and depends only on the position of the number x on the real line":
So for my case, if ##|x|\lt1,##
then ##\sum_{n=0}^\infty x^{n}= \frac{1} {1-x}. ##
So,
##\sum_{n=0}^\infty e^{-n\beta E_0} = \sum_{n=0}^\infty (e^{-\beta E_0})^n ##
##\Rightarrow \frac {1} {1-e^{-\beta E_0}}##