# Summation over multiple arguments

This might seem stupid, but I'm really stuck. I don't understand how to calculate the following explicitly:

$$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}$$

(it's the Ising model for 3 lattice sites).

I don't understand how I can calculate this in a brute force way, since the sums for $$s_1$$ and $$s_3$$ only apply to one part of the equation to be summed over.

(I know you can simplify it and end up with a much nicer expression in terms of cosh)

Help! Thanks.

\begin{align*}\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =\sum_{s_1=\pm1} \sum_{s_2=\pm1}e^{-{s_1s_2}}\left[ e^{{s_2}}+e^{-{s_2}} \right]\\ &=\sum_{s_1=\pm1}\left( e^{{s_1}}\left[ e^{{-1}}+e^{1} \right]+e^{-{s_1}}\left[ e^{1}+e^{-{1}} \right]\right)\\ &=\sum_{s_1=\pm1}\left( [e^{{s_1}}+e^{-{s_1}}]\left[ e^{{-1}}+e^{1} \right]\right)\\ &=[e^{{-1}}+e^{1}]\left[ e^{{-1}}+e^{1} \right]+[e^{{1}}+e^{-1}]\left[ e^{{-1}}+e^{1} \right]\\ &=2[e^{{-1}}+e^{1}]^2\\ &=4(\cosh(2)+1) \end{align*}
Starting with the inside sum of $$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}},$$ we can factor out the term that doesn't depend on $$s_3$$ to get $$\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}= \sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}\sum_{s_3=\pm1} e^{-{s_2s_3}}.$$ Now, let's look at the exponent of the internal sum, since $$s_2$$ is either $$+1$$ or $$-1$$, then $$-s_2$$ is either $$+1$$ or $$-1$$. Multiplying by $$s_3$$ which is either $$+1$$ or $$-1$$ results in one of each of $$+1$$ and $$-1$$. Therefore, the sum simplifies to $$\sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}\sum_{s_3=\pm1} e^{-{s_2s_3}}= \sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}(e+e^{-1})=(e+e^{-1})\sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}.$$ By applying the same argument as above, we see that the exponent of this $$e$$ is one of each of $$+1$$ and $$-1$$, so we get $$(e+e^{-1})\sum_{s_1=\pm1} \sum_{s_2=\pm1} e^{-{s_1s_2}}=(e+e^{-1})\sum_{s_1=\pm1} (e+e^{-1})=(e+e^{-1})^2\sum_{s_1=\pm1}1.$$ Since there are only two values for $$s_1$$, we get that this simplifies to $$2(e+e^{-1})^2$$, which can then be simplified in terms of $$\cosh$$.