Volumes of n-balls: what is so special about n=5? I am reposting this question from math.stackexchange where it has not yet generated an answer I had been looking for. 


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*The volume of an $n$-dimensional ball of radius $R$ is given by the classical formula
$$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$ 
It is not difficult to see that the "dimensionless" ratio $V_n(R)/R^n$ attains its maximal value when $n=5$.

*The "dimensionless" ratio $S_n(R)/R^n$ where $S_n(R)$ is the $n$-dimensional volume of an $n$-sphere attains its maximum when $n=7$.
Question. Is there a purely geometric explanation of why the maximal values in each case are attained at these particular values of the dimension?
[EDIT. Thanks to all for the answers and comments.]
 A: There are many "dimensionless" ratios: choosing $R$ as the linear measurement is arbitrary.
For instance, the ratio of the volume of the sphere to the volume of the circumscribed cube
has a maximum at $n=1$.  The ratio to the volume of the inscribed cube never attains a maximum.  There are intermediate geometrically-related "midscribed" cubes, where all faces of some dimension are tangent to the unit sphere.  Here is the graph for the ratio of
volumes when the codimension 2 faces of a cube are tangent. It attains the maximum for $n = 12$ (just barely more than for $n=11$).  There are many other reasonable dimensionless comparisons, for instance comparing to a simplex, etc. etc.  Since the Gamma function
grows super-exponentially, these simple geometric variations tend to shift the maximum --- there's nothing special about 5 or 7. 
       (source: Wayback Machine)
A: Brian Hayes has very nice article (Wayback Machine) about the volume of the n-ball in the current issue of American Scientist  (Nov 2011). In particular, he discusses the surprising fact that the maximum volume occurs at $n=5$.
A: A very celebrated result on a related subject is the following, which was a very major advance in the Busemann-Petty problem (don't worry, the math review has all you need to know). EDIT by popular demand, the review can be seen here (Wayback Machine).
@incollection {MR950983,
    AUTHOR = {Ball, Keith},
     TITLE = {Some remarks on the geometry of convex sets},
 BOOKTITLE = {Geometric aspects of functional analysis (1986/87)},
    SERIES = {Lecture Notes in Math.},
    VOLUME = {1317},
     PAGES = {224--231},
 PUBLISHER = {Springer},
   ADDRESS = {Berlin},
      YEAR = {1988},
   MRCLASS = {52A40},
  MRNUMBER = {950983 (89h:52009)},
MRREVIEWER = {G. D. Chakerian},
       DOI = {10.1007/BFb0081743},
       URL = {http://dx.doi.org/10.1007/BFb0081743},
}

A: In my opinion, nothing is special about $n = 5$.
The "dimensionless ratio" $V_n(R) / R^n$ is the ratio of the volume of the $n$-ball of radius $1$ to the volume of the $n$-cube of side-length $1$.  So this is maximized at $n =5$, but bluntly, so what?
More interesting geometrically might be the equally dimensionless ratio $V_n(R) / (2R)^n$, which is the ratio of the volume of the $n$-ball to the volume of the smallest $n$-cube containing it.  This is monotonic decreasing (for $n \geq 1$), showing that balls decrease in volume relative to their smallest cubical container, as the dimension increases.  This has more geometric content, since there is a simple geometric relationship between the sphere and cube here.  
One could consider many similar problems, involving inscribing a cube inside a sphere (instead of the other way around), or using an orthoplex or polycylinder or other figure instead of a cube.  All of these have some geometric content, and are expressed as a sequence of dimensionless ratios.
