Series with lower limit greater than upper limit

Question

I was under the impression that a summation of the form $\sum_{i=j}^{k}a_i$ where $j>k$ is regarded as an empty sum and so equal to $0$. My TI-89 Titanium calculator seems to disagree. It gives$$\sum_{i=2}^{1}i=0$$as expected, but also gives$$\sum_{i=3}^{1}i=-2$$and$$\sum_{i=4}^{1}i=-5\mbox{.}$$Is this just because the calculator implementation of $\sum$ is not meant to be used with backwards limits, or is there some other rule for backwards limits that doesn't treat them as empty sums?

Answer 1

I use the convention of empty sum as well.

I debug a programming problem for a friend years ago and in terms of coding, I am aware of people who do the following:

$$\sum_{i=p}^q a_i = \sum_{i=0}^qa_i - \sum_{i=0}^{p-1}a_i$$

$$\sum_{i=2}^1 i = \sum_{i=0}^{1}i - \sum_{i=0}^{2-1}i=0$$

$$\sum_{i=3}^1 i = \sum_{i=0}^{1}i - \sum_{i=0}^{3-1}i=-2$$

$$\sum_{i=4}^1 i = \sum_{i=0}^{1}i - \sum_{i=0}^{4-1}i=-5$$

That is using this convention, if $p > q$,

$$\sum_{i=p}^q i = \sum_{i=0}^qi - \sum_{i=0}^{p-1}i=-\sum_{i=q+1}^{p-1}i=-\frac{(p-q-1)(p+q)}2$$

They use cumulative sum to compute the summation, and perhaps they did not check if $q \ge p$. I am uncertain if this is an accepted convention or remains a bug though.