Taylor expansion of an Ising-like Hamiltonian

[Silicon-Based]

Homework Statement

Compute $Z$ and $\log{Z}$ of a set of $N$ two-level systems with total energy specified by the given Hamiltonian by Taylor expanding in powers of $B$ to first and second order.

Relevant Equations

$$H(\{n_i\}) = A\sum_{i=1}^{N}n_i + B\sum_{i=1}^{N-1}n_i n_{i+1}$$
$n_i = 0, 1$ for $i = 1, ..., N$

For the case when $B=0$ I get: $$Z = \sum_{n_i = 0,1} e^{-\beta H(\{n_i\})} = \sum_{n_i = 0,1} e^{-\beta A \sum_i^N n_i} =\prod_i^N \sum_{n_i = 0,1} e^{-\beta A n_i} = [1+e^{-\beta A}]^N$$
For non-zero $B$ to first order the best I can get is:
$$Z = \sum_{n_i = 0,1} e^{-\beta(A\sum_{i=1}^{N}n_i + B\sum_{i=1}^{N-1}n_i n_{i+1})} \approx \sum_{n_i = 0,1} e^{-\beta A\sum_{i=1}^{N}n_i} \left[1-\beta B \sum_{i=1}^{N-1}n_i n_{i+1} \right]$$ $$=
[1+e^{-\beta A}]^N - \beta B \sum_{n_i = 0,1} \sum_{i=1}^{N-1}n_i n_{i+1} e^{-\beta A\sum_{i=1}^{N}n_i}
$$ At this point I'm not sure how to evaluate the sums. Obiously, the only case when the sums yield a non-zero contribution is when $n_i = n_{i+1} = 1$, but I don't know what $e^{-\beta A\sum_{i=1}^{N}n_i}$ evaluates to in that case. Should this just be $-\beta B (N-1) e^{-\beta A N}$? That doesn't seem like the right answer since I can't really evaluate $\log{Z}$ then.

 

[king vitamin]

It seems like it would be easier to turn the sums in the arguments of the exponential into products of exponentials like you did in the $B=$ case. So you get
$$
Z = \prod_i \sum_{n_i = 0,1} \exp\left( - \beta A n_i - \beta B n_i n_{i+1} \right) \approx \prod_i \sum_{n_i = 0,1} \exp\left( - \beta A n_i \right) \left[ 1 - \beta B n_i n_{i+1} \right].
$$
Now the sum over $n_i = 0,1$ is straightforward, and the same idea applies to second-order in $B$.