Clarification of a symbol How can I expand $\sum\limits_{1 \le i \le j \le n}^{} \sqrt {a_i a_{j+1}} - \sqrt {a_j a_{i+1}} -a_i a_j$ for some values of $n$, for example $n=4$ or $n=5$? I can't understand thw symbol $\sum\limits_{1 \le i \le j \le n}^{}$. That means $\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n} \sqrt {a_i a_{j+1}} - \sqrt {a_j a_{i+1}} -a_i a_j?$
 A: Let $b_{ij} = \sqrt{a_i a_{j+1}} - \sqrt{a_j a_{i+1}} -a_i a_j$. By $1 \le i \le j \le n$ it is meant that $1 \le i \le n$ and for fixed $i$ the $j$ ranges on $[i, n]$. You can write then 
$$
\sum_{1 \le i \le j \le n} b_{ij}
= \sum_{i = 1}^{n} \sum_{j=i}^n b_{ij}.
$$
Another possibility is to write first a sum in $j$ and then a sum in $i$. In this case you can write
$$
\sum_{1 \le i \le j \le n} b_{ij}
= \sum_{j = 1}^{n} \sum_{i=1}^{j} b_{ij}.
$$
The above applies to any finite double sum on this set of indices. In your specific problem, you have that what is being summed is symmetric in $i, j$, meaning that $b_{ij} = b_{ji}$. A consequence of this symmetry that might be useful is that
$$
\sum_{1 \le i \le j \le n} b_{ij}
= \sum_{1 \le i \le j \le n} b_{ji}
= \sum_{1 \le j \le i \le n} b_{ij}
$$
Thus, you have that
$$
\sum_{1 \le i \le j \le n} b_{ij} + \sum_{1 \le j \le i \le n} b_{ij}
= \sum_{i=1}^n \sum_{j=1}^n b_{ij} + \sum_{i=1}^n b_{ii}^2
$$
You should try to explore these symmetries to achieve whatever you want to show. For instance, the two square roots that are being summed seem to cancel each other.
