Everage energy of a gas of quantom SHO

[Jason Gomez]

Homework Statement

[/U]
The average energy of a gas of quantum SHO is

Eav= \sum_{n=0}^{\infty}n\hbar\omega e^(-n\hbar\omega/kT)\div \sum_{n=0}^{\infty}e^(-n\hbar\omega/kT)

can be solved to be

Eav=\hbar\omega\div \left \{ e^\left ( \hbar\omega/kt \right ) \right \}-1

make use of the following two sums, true when [x]<1:

\sum_{n=o}^{\infty }x^n=1/(1-x)

\sum_{n=0}^{\infty}nx^n=x/(1-x)^2

Homework Equations

I tried dividing the two equations, and I think that is the right course of action, but I am not sure what to do with the summations, what I finally get looks like this:

\left \{\sum_{n=0}^{\infty} x^n\div \sum_{n=0}^{\infty}nx^n\right \}= 1/(x-x^2)

I am not sure, but can I just say:

n=1/\left ( x-x^2 \right )
b]3. The Attempt at a Solution [/b]

 

[Dickfore]

\left \{\sum_{n=0}^{\infty} x^n\div \sum_{n=0}^{\infty}nx^n\right \}= 1/(x-x^2)

This is wrong. Use the provided formulas to plug in both of the fractions and do step by step simplification to find the correct expression. then, you will need to identify what x is in terms of the parameters given in the problem.

 

[Jason Gomez]

Thank you, I assume you mean plug in the summation equations that I am to make use of into the Eav equations provided, but I am not sure how. I think I may be making this problem harder than what it is, but I just can't rap my brain around it

 

[Dickfore]

You have a sum in the numerator and denominator of the expression. Both of them are evaluated for you in the statement of the problem.

Note:

The "\div" sign means division which is the same as a fraction line.