- #1
TheDestroyer
- 402
- 1
How to take this sum?
I'm getting this sum in the statistical mechanics for the rotating energy, anyone can help with?
[tex]
Z_r = \sum\limits_{\ell = 0}^\infty {\left( {2\ell + 1} \right)e^{{\textstyle{{T_r } \over T}}\ell \left( {\ell + 1} \right)} }
[/tex]
If tex doesn't work here is it, Where the sum is from 0 to infinity
Z=Sum[(2L+1)exp(-x L(L+1))]
Seems simple lol !
Thanks
I'm getting this sum in the statistical mechanics for the rotating energy, anyone can help with?
[tex]
Z_r = \sum\limits_{\ell = 0}^\infty {\left( {2\ell + 1} \right)e^{{\textstyle{{T_r } \over T}}\ell \left( {\ell + 1} \right)} }
[/tex]
If tex doesn't work here is it, Where the sum is from 0 to infinity
Z=Sum[(2L+1)exp(-x L(L+1))]
Seems simple lol !
Thanks