What is the meaning of this Sigma notation? This is from Probability and Statistics Fourth Edition by Mark J. Schervish and Mark J. Schervish pg.48.
$Pr(\bigcup\limits_{i=1}^{n}A_i) = \sum\limits_{i=1}^{n} Pr(A_i) - \sum\limits_{\bbox[yellow]
{i<j}} Pr(A_i \cap A_j) + \sum\limits_{\bbox[yellow]
{i<j<k}} Pr(A_i \cap A_j \cap A_k) - \sum\limits_{\bbox[yellow]
{i<j<k<l}} Pr(A_i \cap A_j \cap A_k \cap A_l) + (-1)^{n+1}Pr(A_1 \cap A_2 \cap \ldots \cap A_n$  

I don't understand the highlighted parts.
For instance:  
$\sum\limits_{i<j<k} Pr(A_i \cap A_j \cap A_k)$
Does $k$ go from $3$ to $n$? and $j$ from $2$ to $n-1$? and $i$ from $1$ to $n-2$?
If I understand correctly, this is taking all $A$s three at a time, finding the intersection of each three, and adding up all the probabilities of those intersections, right?
Is this the usual notation? Is there a better way to write this?
 A: The notation $\sum_{i<j}Pr(A_i\cap A_j)$ means $\sum_{(i,j)\in\{(i,j)\in I^2\mid i<j\}}Pr(A_i\cap A_j)$. That is, the sum over all pairs $(i,j)\in I\times I$ which satisfy $i < j$. Here $I=\{1,\dots,n\}$ which isn't implied by the notation but is inferred from context in your example from the first sum, $\sum_{i=1}^n Pr(A_i)$ which, for consistency, can be written $\sum_{i\in I}Pr(A_i)$. (For that matter, the union can be written $\bigcup_{i\in I}A_i$.)
The further sums are over all triples in $I^3$ satisfying the conditions, then all quadruples, and so forth.
Since it's somewhat redundant to use set builder notation just to immediately test for membership, I've seen notation like $\sum_{i,j:i<j}Pr(A_i\cap A_j)$. And, of course, what you started with compresses this further.
If you wanted to take a different perspective on it, you could view $\sum$ as taking a set of terms and write $\sum\{Pr(A_i\cap A_j)\mid i\in I,j\in I, i < j\}$.
A: It means summing over all the intersections, but without duplicates.  So if $n=3$ then $\sum\limits_{
{i<j}} \Pr(A_i \cap A_j)$ means $\Pr(A_1 \cap A_2)+\Pr(A_1 \cap A_3)+\Pr(A_2 \cap A_3)$
You could rewrite  $\sum\limits_{
{i<j<k}} Pr(A_i \cap A_j \cap A_k)$ as $\sum\limits_{
{i=1}}^{n-2}\sum\limits_{
{j=i+1}}^{n-1}\sum\limits_{
{k=j+1}}^{n} Pr(A_i \cap A_j \cap A_k)$ or as $\sum\limits_{
{k=3}}^{n}\sum\limits_{
{j=2}}^{k-1}\sum\limits_{
{i=1}}^{j-1} Pr(A_i \cap A_j \cap A_k)$ but some might say this was less immediately clear
A: Interpreting $\sum_{i<j} P(A_i\cap A_j)$
The indices goes from $i$ to $n$ and $j$ to $n$, but only where $i<j$.
The indices can be represented in a table ($n=4$):
$$\begin{array}{|c|c|}\hline i/j&1&2&3&4  \\ \hline 1& & x& x&x  \\ \hline 2& & &x &x  \\ \hline  3& & & & x \\ \hline 4& & & &  \hline \end{array}$$
So the number of summands is $6$
For an arbitrary $n$ the closed formula for the number of summands is 
$$s(n)=\frac{n}2\cdot (n-1)$$
You just substract n cells (blank diagonal) from the total number of the cells: $n\cdot n-n$. Then you divide the result by 2. You get the number of cells above the diagonal.
In case of $n=4$ we get $s(n)=\frac{4}2\cdot (3)=2\cdot 3=6$
Consequently $\sum_{i<j}^4 P(A_i\cap A_j)$
$$=P(A_1\cap A_2)+P(A_1\cap A_3)+P(A_1\cap A_4)+P(A_2\cap A_3)+P(A_2\cap A_4)+P(A_3\cap A_4)$$
