Infinite Jenga Half Bridge

How much can you make the top Jenga brick overhang the base by stacking them together?

Surprisingly, you can go as far as you want.

Suppose your bricks are length $l$ and you have one brick (not including the base).  Obviously you can overhang by $\frac{l}{2}$ without the centre of mass being unsupported.  What if you have two?  Now you have two conditions

1. The centre of mass of the top brick is supported
2. The centre of mass of the top 2 bricks are supported

A quick calculation gives us that if the top brick is displaced (relative to the one below) by $\frac{l}{2}$ then the one below could be displaced by at most $\frac{l}{4}$ (relative to the one below it).

Now suppose you have $n$ bricks (not including the base), then  you have $n$ conditions.  Let's guess the answer based on the result for $n=2$ and let's set $d_k = \frac{l}{2}\frac{1}{k}$ where $d_k$ is the displacement relative to the brick below and $k$ is the brick number starting at the top.  Then the centre of mass of the $n$ bricks is
$$
\begin{align}
\frac{1}{n} \sum_{k=1}^{n} k d_k &= \frac{1}{n} \sum_{k=1}^{n} k \frac{l}{2}\frac{1}{k} \\
&= \frac{1}{n} \sum_{k=1}^{n} \frac{l}{2} \\
&= \frac{l}{2}
\end{align}
$$
which shows that the centre of mass of the $n$ bricks is in fact supported. But since the $n$ could be replaced by any value from $1..n$ in the above and still come up with the same value it shows that the top $m$ bricks are supported for any $1 \le m \le n$.

So, how far does the overhang $O_n$ get?  Well the total is obviously
$$
\begin{align}
O_n &= \sum_{k=1}^{n} d_k \\
&=  \frac{l}{2} \sum_{k=1}^{n} \frac{1}{k}
\end{align}
$$
which is unbounded.  To see why note that $\sum_{k=1}^{n} \frac{1}{k}$ is an upper bound for $\int_{1}^{n+1}\frac{1}{x}dx$ which is equal to $ln(n+1)$, an unbounded function.  It's also a lower bound for $1+\int_1^n \frac{1}{x}dx$ so, approximately
$$
O_n \approx \frac{l}{2} ln(n)
$$
So, although you can get as far as you want, you'll need to use a lot of blocks.  Specifically for each $\frac{l}{2}$ further you need $\approx 2.718$ times as many blocks!