ok, nobody is thinking about this, so here is the answer (I had to work it out myself, as all I had initially was an excel sheet showing that as N got large the ratio became ever closer to pi)

If we join the points of a polygon with N sides (length 1) to the centre, the angles at the centre are all 2*pi/N

a = 2*pi/N

The cosine rule is:

A^2 = B^2 + C^2 - 2BC.cos(a)

we can use this to derive the length from the centre of the polygon to its corners (r):

r = 1/sqrt(2(1- cos(a))

As the perimeter of the star is 2N sides, each of this length:

perimeter_of_star = 2N/sqrt(2(1- cos(a))

the perimeter of the polygon itsefl is N, so, if we call our ratio x, which is:

polygon_perimeter^2/star_perimeter

then we can construct the equation:

x = N.sqrt(2(1-cos(a))/2

we want to work out what the value of x is, as N tends towards infinity.

we can simplify our subsequent work, if we observe that 1-cos(a) is a small positive number, for very large N, so we can square both sides of this equation, and later take the positive square root of them, and this will still be unambiguously the value of x

This gives us the equation:

x^2 = N^2(1-cos(a))/2

the problem here is that N => infinity, (1-cos(a)) => 0, so the value of x^2 as N=>infinity is not obvious.

We can use l'Hopitals rule to proceed, basically this states that if we have a fraction, and the top and bottom of the fraction both tend to zero (or both tend to +- infinity) then we can take the differential of the top and the bottom, and the resultant new fraction will tend to the same thing as the original fraction.

If we recast our expression so that it is a fraction:

x^2 = (1-cos(a)) / (2/N^2)

we can see that, as required, both the top and the bottom tend to O as N=> infinity, so we can differentiate them:

d/dn { 2/N^2 } = -4/N^3

d/dN { 1 - cos(a) } = -a.sin(a)/N

so:

lim {N=>inf} x^2 = lim {N=>inf} N.pi.sin(a)/2

using lim, to stand for lim {N=>inf}

lim x^2 = (pi/2) lim sin(a)/(1/n)

once again, the top and the bottom of the fraction tend to 0 as N=> inf, so we can again apply L'hopital:

d/dn {sin(a)) = -2.pi.cos(a)/N^2

d/dn {1/n} = -1/N^2

so:

lim x^2 = (pi/2) lim 2.pi.cos(a)

as, N => inf, clearly cos(a) => 1, so this evaluates to:

lim x^2 = pi^2

and taking square roots:

lim x = lim {N=>inf} N^2(1-cos(a))/2 = pi

so our ratio tends to pi, as the number of the sides of the polygon tend to infinity.