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In the solutions she provided following our first midterm, my real analysis professor used this this summation: $$\sum_{1\leq k<j \leq n,} a_kb_ka_jb_j$$ I have never seen this type of terminology before in my course or elsewhere, so I was curious what it means. Usually, we have a bound on top of the summation as well, so I wonder why she left it blank. Is this a common notation?

To my understanding, based on the problem, it should mean something similar to: $$\sum_{k=1}^{n}\left(a_kb_k\sum_{j=1,j\neq k}^{n} a_jb_j\right)$$ However, her notation for it seems unituitve. Could someone help me see how the two are equivalent?

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    $\begingroup$ It's the sum over all $k$ and $j$ such that $1\le k<j\le n$. $k$ doesn't go from $1$ to $n$, only from $1$ to $j-1$. And $j$ only goes from $k+1$ to $n$. $\endgroup$ Commented Apr 21 at 3:29
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    $\begingroup$ It's the sum over all pairs $(k,j)$ satisfying the condition $1\leq k<j\leq n$. It can also be written as $\sum_{j=2}^n\sum_{k=1}^{j-1}$. $\endgroup$
    – Gonçalo
    Commented Apr 21 at 3:31
  • $\begingroup$ The key bit of notation in your question is hidden behind a link to an image. Please do not use images to convey vital information which can be conveyed via text. Reasons to use text when possible are outlined on Meta. I will also note that I have answered a similar question math.stackexchange.com/a/3540753 . The notation used here is similar, but dropping the set-builder notation. I suspect that there are other similar questions on the site. $\endgroup$
    – Xander Henderson
    Commented Apr 21 at 3:36
  • $\begingroup$ E.g. this one is similar: math.stackexchange.com/q/2857544 . $\endgroup$
    – Xander Henderson
    Commented Apr 21 at 3:37
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    $\begingroup$ I’m voting to close this question because OP has nothing to say. $\endgroup$ Commented Apr 24 at 6:52

2 Answers 2

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It is the sum over all $k$ and $j$ satisfying $1\leq k<j\le n$, which can be written as a double sum $$\sum_{j=1}^n\sum_{k=1}^{j-1}$$ or $$\sum_{k=1}^n\sum_{j=k+1}^n$$ Logic is simple: to find all the pairs $(k,j)$ satisfying the relation, you can say that for a fixed $j$, all $k$ from $1$ to $j-1$ are allowed, so you can write the sum in the first form. Similar reasoning for the other.

This kind of notation is very common, as writing double (and multiple sums) becomes tedious. And using the multiple sum notation just makes it harder to see what is going on, for example $$\sum_{\substack{1\leq i_1\leq i_2\leq\cdots\leq i_k\leq n\\i_1+i_2+\cdots+i_k=n}}$$ There are $k$ sums involved here. Writing in multiple sums would be a mess. But the above makes it clear what we are summing over.

Hope this helps. :)

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  • $\begingroup$ Beware: in order for $i*k<j$ to be satisfied, you should use $\sum_{j=2}^n\sum_{k=1}^{j-1}$ (specifically $j=2$) $\endgroup$ Commented Apr 25 at 9:30
  • $\begingroup$ @SergeBallesta: $\sum_{k=m}^n$ is an empty sum (equal to zero) when $m>n$. $\endgroup$ Commented Apr 25 at 9:54
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In many finite sums, the terms of the sum are indexed by elements of some finite set that set is known as the index set.

The most common type of indexing has a form tha you know, namely $$\sum_{i=1}^n a_i $$ and in this case the index set is the following subset of the natural numbers $\mathbb N$: $$\{i \in \mathbb N \mid 1 \le i \le n\} $$ You can think of this as the integer dots along the subinterval $[1,n] \subset \mathbb R$.

A more general and still common form is $$\sum_{i=a}^b a_i $$ and this index set consists of the integer dots along the interval $[a,b] \subset \mathbb N$, which we can also write as $$\{i \in \mathbb N \mid a \le i \le b\} $$

For a double sum like $$\sum_{i=a}^b \sum_{j=c}^d a_{ij} $$ the index set becomes a subset of $\mathbb N^2$, in this case the subset $$\{(i,j) \in \mathbb N^2 \mid a \le i \le b \quad\text{and}\quad c \le j \le d\} $$ If you were to draw this index set in the coordinate plane $\mathbb R^2$, it would appear as the integer coordinate dots that are contained in the rectangle $[a,b] \times [c,d]$.

But sometimes you need a different kind of index set $\mathbb N^2$, one which does not necessarily have the form of integer coordinate dots contained in some rectangle. There's nothing to keep you from using such an index set except your ability to describe the set. For the sum you are asking about, namely
$$\sum_{1 \le k < j \le n} a_k b_k a_j b_j $$ the index set is $$\{(j,k) \in \mathbb N^2 \mid 1 \le k < j \le n\} $$ You can think of this as the set of integer coordinate dots in the square $[1,n] \times [1,n] \subset \mathbb R^2$ that lie below the graph of the equation $y=x$.

Perhaps the most general format of a finite sum is this: $$\sum_{P(s)} a_s $$ In this format, $s$ varies over some un-named set that I'll temporarily denote as $S$, and $P(s)$ is a true-false predicate for the set $S$, so the index set of the sum is expressed in set builder notation as $$\{s \in S \mid P(s)\} $$ For instance, the unnamed set could be $\mathbb N^2$, and as $s = (j,k) \in \mathbb N^2$ varies the predicate could be $$P(j,k) := 1 \le k < j \le n $$ and so the index set is the same one as for your sum.

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