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?
 A: 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, $$\sum_{k=-2}^0k\times 10$$
A: The concept of Sum has three basic definitions.

*

*Sum over  (part of) a sequence
Given a unilateral sequence
$$
x_{\,0} ,\,x_{\,1} ,\, \cdots ,\;x_{\,n} 
$$
we define a sum over a portion of it as
$$
\sum\limits_{k = \,a}^b {x_{\,k} } 
$$
where it is understood that either $a$ and $b$ are integers and that $a \le b$.
Under this acception your sum does not have meaning.
Another way of writing the sum is by imposing restriction to the index
$$
\sum\limits_{a\, \le \,k\, \le \,b} {x_{\,k} } 
$$
and if the condition is violated the sum is null.
But the sequence could be bi-lateral
$$
 \cdots ,\;x_{\, - n} ,\; \cdots x_{\, - 1} ,x_{\,0} ,\,x_{\,1} ,\, \cdots ,\;x_{\,n} ,\; \cdots 
$$
In this case you may want to write
$$
\sum\limits_{k = \,0}^b {x_{\,k} } 
$$
leaving $b$ free to address any index in the sequence, and thus understanding
$$
\sum\limits_{k = \,0}^b {x_{\,k} }  = \sum\limits_{k = \,b}^0 {x_{\,k} } 
$$
but you shall clearly state this acception.


*Sum over a set
$$
\sum\limits_{x\, \in \,A} x 
$$
and the meaning is clear.


*Indefinite sum (Antidelta)
Finally there is the concept of Indefinite Sum.
For a function $F(z)$, over the complex field in general, we define the (forward) Finite Difference as
$$
\Delta \,F(z) = F(z + 1) - F(z)
$$
If we have that
$$
\Delta  \,F(z) = F(z + 1) - F(z) = f(z)
$$
then we write
$$
F(z) = \Delta ^{\, - \,1} \,F(z) = \sum\nolimits_{\;z\;} {f(z)} + c
$$
and in particular we have
$$
F(b) - F(a) = \sum\nolimits_{\;z = \,a\,}^b {f(z)}  =  - \sum\nolimits_{\;z = \,b\,}^a {f(z)} \quad \left| {\,a,b \in \mathbb C} \right.
$$
For example
$$
\eqalign{
  & F(z) = \left( \matrix{  z \cr  2 \cr}  \right)\quad  \Leftrightarrow \quad
 \left( \matrix{  z + 1 \cr   2 \cr}  \right) - \left( \matrix{  z \cr   2 \cr}  \right)
 = \left( \matrix{  z \cr   1 \cr}  \right) = z\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \sum\nolimits_{\;x = \,a\,}^{\;b} z
  = \left( \matrix{  b \cr   2 \cr}  \right) - \left( \matrix{  a \cr   2 \cr}  \right)
 \quad \left| {\;a,b \in \mathbb C} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \sum\nolimits_{\;k = \,0\,}^{\;n} k \quad \left| {\;0 \le n \in \mathbb Z} \right.\quad
  = \sum\limits_{0\, \le \,k\, \le \,n - 1} k  = \left( \matrix{  n \cr   2 \cr}  \right)
 = {{n\left( {n - 1} \right)} \over 2} \cr} 
$$
A: 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.
A: Yes we can have a negative value in sigma.
You just need to write it like this:
$$y = \Sigma_{k=-2}^{k=0} k\times 10$$
A: Sigma notation is shorthand for iterated addition. The expression following the Sigma is the addend, and the operator for each iteration is addition.
As addition is commutative, it matters not the positive or negative direction of the iteration.
One can go from some initial index to a negative number. What may be forgotten in the notation is that there is an assumed increment. One may consider this assumption as the same as the assumed $+$ positive sign before a positive number and similarly to the trailing zeroes in decimal fraction of a Real number or Real part of a Complex number, i.e, $3.1$ is the same as $3.10,$ $3.100,$ etc..
The fact that the sign of a positive number is not always written out does not mean it does not exist. The same is true of the increment operator sometimes noted in Sigma notation.
In the convention of Sigma notation, the increment operation is defaulted as adding $+1$ to the index. When the decrement operator is used in the notation, it should not be seen as subtraction but as an increment of a negative amount.
If the upper limit of the index is lower than the initial index, one may logically assume that the operator is a negative increment. Some notationists will assist in the clarification by including the increment operator in the notation and, if not unary $(+1|-1),$ the step amount by which the index is to be positively or negatively incremented in the iteration.
