It's obvious that if we think not of straight lines but on a curve, we get a very dense spiral and the distance becomes infinite, as the circles of the spiral can be infinitely close one to another.

But when trying to adapt to that spiral a finite number of parts N an idea comes. And is this:

Consider a circle with origin in B and being A a point of it. Now if the number of segments is N consider dividing the distance between A and B in N equal parts. Now trace the circle with centre in B and radio the distance from B to each part of the division. We get N concentric circles with origin in B. 

Now we trace from A a tangential segment to the immediate circle, and from that point a segment tangencial to the next lower circle, and so to arrive to B. Each segment departs from a circle and is tangent to the immediately inferior concentric circle. When N is a very big number I suppose that this is going to fit very well into the spiral. 

And is mathematically simple because the length of each segment is the producto of the radio of the circle for the sin of the angle between both radios

For segment in position  i: L(i)= r(i) sen a(i)

But r(i)=(n-i)/n

and cos a(i)=(n-i-1)/n-i, 

so we have r (i) and sin a(i) as function of n.

The lenght of each segment is then:

And the formula for the distance is: 

From here is clear that this is not the solution as for N=2 we can have a distance of 1,4142 (bigger than mine). 

We can adjust the radios in a non linear but quadratic way. I mean making the concentric circles have each an area in arithmetic proportion. Then the radio of the circle i should be

When we do this and use the formulas given before everything simplifies and we get

d=sigma 1/sqrN  and then ae there are N terms

which is a better result.