# Is ∑ with negative value solvable?

Is it possible to have a negative value in sigma?

e.g.
$y = \Sigma_{k=0}^{k=-2} k \times 10$

Will this give the result $(0 \times 10) + (-1 \times 10) + (-2 \times 10) = -30$?

Or will it be $\infty$ because $k$ will be increased with $1$ until it equals $-2$ (which is never).

Or something else?

• I would interpret the sum as being over the unordered set $\{-2,-1,0\}$. – Umberto P. Oct 9 '13 at 13:47

There are several things to remark here:

First, $$\sum_{k=0}^{k=-2}\times 10$$ is not actually the correct notation.

You seem to mean $$\sum_{k=0}^{k=-2}k\times 10$$ which is correct notation (though usually the $k=$ part is not included in the top), but would be $0$. The reason is that the notation $$\sum_{k=i}^{k=j}k$$ means "take the sum of $k$ for each $k$ which is $\geq i$ and $\leq j$". In the case at hand, there are no such $k$, and by convention, this means the sum is $0$.

To get the desired result, you can do $$\sum_{k=-2}^{k=0}k\times 10$$ or, with the more common notation, $$\sum_{k=-2}^0k\times 10$$

Depending on the application, you could probably make a convincing case for any of these. I don't think there is any standard rule for what this will actually result in.

From my point of view, what you've written there is an empty sum. $\sum^{-2}_{k=0} k$ is the sum of all integers $k$ such that $0 \leqslant k \leqslant -2$, which is obviously no values of $k$, so the sum is $0$. If, however, you were working in an application where you might want to use a different convention, you're pretty much free to do so, as long as you state clearly what you're doing and why.

No convention is more valid than any other, as long as it is justified.

Yes we can have a negative value in sigma.
You just need to write it like this:
$$y = \Sigma_{k=-2}^{k=0} k\times 10$$