Richard Guy wrote:

<< I don't see why one shouldn't use very elementary summations

I mentioned that only to make the problem more of a challenge.

<<
The scalar product of (the numerators of) the
two vectors is

     1*(n-1) + 2*(n-2) + ... + (n-1)*1

which is the (n-1)th tetrahedral number (see Fig 2.34 on p.46 of the Book of Numbers, where one sums a diagonal of the multiplication table):  (n+1)*n*(n-1)/6 
so the scalar product is (n+1)*(n-1)/6n.

The lengths are just the sum of the first (n-1) squares, divided by  n^2:  n*(n-1)*(2n-1)/6n^2.  So cosine of angle twixt the two is

                      (n+1)*(n-1)/6n / (n-1)*(n-1)*(2n-1)*(2n-1)/36n^2

= 6n*(n+1)/(n-1)*( 2n-1)*(2n-1)  which tends to 0.
>>

Since the (equal) lengths of the integer vectors have been multiplied the to get the sum of the first n-1 squares n*(n-1)*(2n-1)/6, there's no need to square this again in the denominator.  And since (1,2,...,n-1) and (n-1,...,2,1) are just scaled versions of the vectors I asked about, they will have the same angle and we may as well just consider these.  According to the above dot product and length calculation, the cosine at the nth stage is

                            ((n+1)*n*(n-1)/6) / (n*(n-1)*(2n-1)/6),

and so the limit as n --> oo is 1/2, and the angle is pi/3.

Dan Asimov