You can create an infinite overhang.

Let us reverse the problem: how far can brick 1 be from brick 0?

Let us assume that the brick is of length 1.

To determine the place of the center of mass a(n): a(1)=1/2 a(n)=1/n[(n-1)*a(n-1)+[[a(n-1)+1/2?=a(n-1)+1/(2n) Thus

n 1 n 1

a(n)=Sum -- = 1/2 Sum - = 1/2 H(n)

m=1 2m m=1 m

Needless to say the limit for n->oo of half the Harmonic series is oo.

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