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.
 A: Proceeding from right to left:
$$\begin{align*}\sum_{s_1=\pm1} \sum_{s_2=\pm1} \sum_{s_3=\pm1} e^{-{s_1s_2}}e^{-{s_2s_3}}& =\sum_{s_1=\pm1} \sum_{s_2=\pm1}e^{-{s_1s_2}}\left[ e^{{s_2}}+e^{-{s_2}} \right]\\
&=\sum_{s_1=\pm1}\left( e^{{s_1}}\left[ e^{{-1}}+e^{1} \right]+e^{-{s_1}}\left[ e^{1}+e^{-{1}} \right]\right)\\
&=\sum_{s_1=\pm1}\left( [e^{{s_1}}+e^{-{s_1}}]\left[ e^{{-1}}+e^{1} \right]\right)\\
&=[e^{{-1}}+e^{1}]\left[ e^{{-1}}+e^{1} \right]+[e^{{1}}+e^{-1}]\left[ e^{{-1}}+e^{1} \right]\\
&=2[e^{{-1}}+e^{1}]^2\\
&=4(\cosh(2)+1)
\end{align*}$$
A: A slightly alternate approach than in the other answer (dividing up the sum differently):
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 $\cosh$.
