What Does a Double Sigma Notation Mean?

A double sigma is shorthand for adding terms indexed by two variables. It does not have one special formula that applies to every problem: the bounds and the summand determine what is being added.

For a finite sum, the basic rule is simple: calculate the inner sum for each value of the outer index, then sum those results. This article shows how to read the notation, when you can change the order, and how to recognize useful special cases.

How to read a double sum

A typical expression is

$$\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij}.$$

The inner sum, over $j$, is evaluated first. For each fixed $i$, let $j$ run from $1$ to $n$. Then add the results as $i$ runs from $1$ to $m$:

$$\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij} =\sum_{i=1}^{m}\left(a_{i1}+a_{i2}+\cdots+a_{in}\right).$$

Equivalently, this adds every entry $a_{ij}$ in an $m$-by-$n$ rectangular array. There are $mn$ terms if every pair of indices is included.

The inner sum is not necessarily written on the right. Read the bounds attached to each sigma: the index that appears in the inner sigma is the one summed inside the outer sum.

Worked example

Evaluate

$$\sum_{i=1}^{3}\sum_{j=1}^{2}(i+2j).$$

For each fixed $i$, substitute $j=1$ and $j=2$ into the summand:

$$\begin{aligned} \sum_{j=1}^{2}(i+2j) &=(i+2)+(i+4)\\ &=2i+6. \end{aligned}$$

Now sum that result over $i=1,2,3$:

$$\sum_{i=1}^{3}(2i+6) =(2+6)+(4+6)+(6+6)=30.$$

So the double sum equals $30$. You can check by listing all six terms: $3+5+4+6+5+7=30$.

When can you switch the order?

For finite sums over independent bounds, changing the order does not change the result:

$$\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij} =\sum_{j=1}^{n}\sum_{i=1}^{m}a_{ij}.$$

Both expressions include exactly the same terms, once each. In the example above, switching the sums means summing first over $i$ for each fixed $j$; the answer remains $30$.

The bounds may depend on an index, however. For example,

$$\sum_{i=1}^{3}\sum_{j=1}^{i}1$$

adds one term when $i=1$, two when $i=2$, and three when $i=3$. Its value is $1+2+3=6$. The index pairs form a triangular region, not a rectangle. If you reverse the order, you must rewrite the bounds to describe that same region; you cannot simply swap the two sigma symbols and keep the old limits.

For infinite sums, changing the order needs additional conditions. It is valid for nonnegative terms, and it is valid for absolutely convergent sums. A conditionally convergent double series can behave differently when its terms are rearranged.

A useful product formula

If the bounds are independent and the summand separates into a function of $i$ times a function of $j$, then the double sum factors:

$$\sum_{i=1}^{m}\sum_{j=1}^{n} f(i)g(j) =\left(\sum_{i=1}^{m}f(i)\right) \left(\sum_{j=1}^{n}g(j)\right).$$

For example,

$$\sum_{i=1}^{3}\sum_{j=1}^{2}ij =\left(1+2+3\right)\left(1+2\right) =6\cdot 3=18.$$

This shortcut works because every product $f(i)g(j)$ appears once for each allowed pair $(i,j)$. It does not apply directly when the bounds couple the indices, as in a triangular sum.

Common mistakes

• Treating both sums as one undifferentiated operation. The inner index runs through its full range for each fixed value of the outer index.
• Assuming every double sum has rectangular bounds. A limit such as $j=1$ to $i$ makes the range depend on the outer index.
• Switching dependent bounds without changing them. To reverse a sum, preserve the same set of index pairs.
• Using the product formula for a nonseparable term. The shortcut requires the summand to be a product of an $i$-only factor and a $j$-only factor, with independent bounds.
• Reordering an infinite sum without checking convergence. The finite-sum rule alone does not justify rearranging every infinite series.