<pre>
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)<code>1/n~[[(n-1)*a(n-1) + a(n-1) + 1/2]]</code>a(n-1)+1/(2n)
Thus
<pre>
       n   1        n  1
 a(n)<code>Sum -- </code> 1/2 Sum - = 1/2 H(n)
      m<code>1 2m       m</code>1 m
</pre>
Needless to say the limit for n->oo of half the Harmonic series is oo.
</pre>
