Formula for volume of $n$-ball for negative $n$ Does the expression $$\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n,$$ which gives the volume of an $n$-dimensional ball of radius $R$ when $n$ is a nonnegative integer, have any known significance when $n$ is an odd negative integer?
 A: I will answer regarding the dimension $-1$.
An example of such space is a set of periodic lattices on a real line.
Indeed, you can see that the Hausdorff dimension of a periodic lattice is $-1$: If we scale down the lattice twice, it would be able to include two original lattices. Since scaling the fractal down twice makes it twice as big, its dimension is $\frac{\ln 2}{\ln (1/2)}=-1$.
The  formula for the volume of a ball gives $\frac1{\pi R}$ for $n=-1$. This means the unit ball ($R=1$) includes only one lattice: the $\pi$-periodic one.
If we reduce the radius (the step of the lattice divided by $\pi$), its volume increases: a lattice with step $1/2$ can be represented as two lattices of step $1$. Thus the volume of a lattice of step $1/2$ consists of two "points" ($1$-periodic lattices), and so has volume twice the volume of $1$-periodic lattice.
Alternatively, we can consider it being 1 point with "weight" 2 (the weight being proportional to the density of a lattice, so that our space has fuzzy membership function).
Similarly, if we increase the lattice step ("radius" of the $-1$-sphere), we can consider it a lattice with weight below $1$, the same as its volume.
So, the volume of the $-1$-ball is the density of the lattice.
Interesting observation: in positive-dimentional space the ball becomes a point with reduction of its radius to zero. In zero-dimensional Euclidean space the ball is always a point, disregarding the "radius" and in this $-1$-dimensional sphere the ball becomes a zero-volume point when we increase the radius infinitely.
You can find some further ideas here.
A: I want to make another answer to this question, although it is is not really an answer, and I do not know whether it is relevant or not, but it may be useful.
Let's take 
$$V(n,R)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n$$
Also, using the reflection formula for the Zeta function we can see:
$$n\zeta(1-n)\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}=(1-n)\zeta(n)\frac{\pi^{\frac{1-n}{2}}}{\Gamma(\frac{1-n}{2}+1)}$$
Interestingly enough, in the algebra of divergent integrals described here, there is a rule $\operatorname{reg}\omega_+^n=-n\zeta(1-n)$, (where $\omega_+=\int_{-1/2}^\infty dx$ is an infinite divergent integral and $\operatorname{reg}$ denotes regularization), the overall formula becomes
$$\operatorname{reg} V(n, \omega_+)=\operatorname{reg} V(1-n, \omega_+)$$
This is a very nice-looking relation but its deep meaning is unclear. Particularly because it is not evident what meaning should have the balls with infinite radius, and especially the balls of infinite radius in negative dimension :-). Still formally this relation looks beautiful.
See also this question: What is the connection between the Riemann Xi-function and n-sphere?
