I have a sum $\frac{1}{2+a}$ which is only valid for $a \in (-\infty,-3) \cup (-1,\infty)$.
To express the some correctly, should I show the sum for the entire interval?
I guess I can say something like $$ \frac{1}{2+a} \in [-\frac{1}{2},0) \, \cup \, \cdots $$
The left side of the interval is $(-\infty,-3)$. If I put $-3$ into the sum, I get $-\frac{1}{2}$ and if I put $-\infty$ into the sum, it goes to $0$, so the sum must be between $-\frac{1}{2}$ and $0$, but none of the limits included.
But what about the right side of the interval $(-1,\infty)$? If I substitute a=-1, I get 1, so is the sum between 0 and 1, none of them included?
I think $\dfrac{1}{2+(-3)}=-1$ while $\dfrac{1}{2+(-1)}=1$
so your function will take values in $(-1,0)\cup(0,1)$ if $a \in (-\infty,-3)\cup(-1,\infty)$.