- #1

- 1,598

- 605

- TL;DR Summary
- About proving a summation rule that appears in theory of spin chains.

I ran into this kind of expression for a sum that appears in the theory of 1-dimensional Ising spin chains

##\displaystyle\sum\limits_{m=0}^{N-1}\frac{2(N-1)!}{(N-m-1)!m!}e^{-J(2m-N+1)/kT} = \frac{2e^{2J/kT-J(1-N)/kT}\left(e^{-2J/kT}(1+e^{2J/kT})\right)^N}{1+e^{2J/kT}}##

where the ##k## is the Boltzmann constant, ##J## is strength of magnetic interaction and ##T## is the absolute temperature.

It's quite simple to check that this holds for ##N=1##, ##N=2## and ##N=3##. However, it seems to be a bit more difficult to actually prove by induction than some other sums like

##\displaystyle\sum\limits_{m=1}^{N}m^2 = \frac{N(1+N)(1+2N)}{6}##,

where only the last term of the sum changes when ##N\mapsto N+1##.

If anyone wants to try to prove this as an exercise, be my guest and post it here... It's probably just about applying the binomial theorem.

##\displaystyle\sum\limits_{m=0}^{N-1}\frac{2(N-1)!}{(N-m-1)!m!}e^{-J(2m-N+1)/kT} = \frac{2e^{2J/kT-J(1-N)/kT}\left(e^{-2J/kT}(1+e^{2J/kT})\right)^N}{1+e^{2J/kT}}##

where the ##k## is the Boltzmann constant, ##J## is strength of magnetic interaction and ##T## is the absolute temperature.

It's quite simple to check that this holds for ##N=1##, ##N=2## and ##N=3##. However, it seems to be a bit more difficult to actually prove by induction than some other sums like

##\displaystyle\sum\limits_{m=1}^{N}m^2 = \frac{N(1+N)(1+2N)}{6}##,

where only the last term of the sum changes when ##N\mapsto N+1##.

If anyone wants to try to prove this as an exercise, be my guest and post it here... It's probably just about applying the binomial theorem.