Get a Clue: Understanding the Limit of a Series

[Alex_Neof]

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.

 

[Mark44]

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.

This is a geometric series.

Questions about infinite series are normally covered in calculus courses, so I moved this thread from the Precalc section.

 

[fourier jr]

think $\frac{a}{1-r}$

 

[Alex_Neof]

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,

$\sum_{n=0}^\infty e^{-n\beta E_0} = \sum_{n=0}^\infty (e^{-\beta E_0})^n$

$\Rightarrow \frac {1} {1-e^{-\beta E_0}}$

 

[Mark44]

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,

$\sum_{n=0}^\infty e^{-n\beta E_0} = \sum_{n=0}^\infty (e^{-\beta E_0})^n$

$\Rightarrow \frac {1} {1-e^{-\beta E_0}}$

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.