Here, on slide 27, it says that
$\pi = \{3, 25/8, 201/64, ... | 4, 7/2, 13/4, ... \}$
The largest number on the left will be $3 + 1/8 + 1/64 + \dots$ which I evaluated as \begin{align} 2 + (1 + 1/8 + 1/64 + \ldots) = 2 + 1/(1-1/8) = 3.142847142857143 \end{align}
Which is larger than $\pi$?
And the smallest number on the right I evaluated as \begin{align} 4 - 1/2 - 1/4 - ... &= 4 + 1 - 1 - 1/2 - 1/4 - .... \\ &= 5 - (1 + 1/2 + 1/4 + ....) \\ &= 5 - 2 \\ &= 3 \\ \end{align}
Since the smallest number on the right is larger than the largest number on the right, shouldn't this not equate to a number? How does it equate to $\pi$?
Giving some extra details to supplement Dustan Levenstein's answer, the left hand side contains an infinite sequence of values $a_0, a_1, a_2, \dots$ with $a_0 = 3, a_1 = 25/8, a_2 = 201/64$. How should the sequence be defined in general? One answer is that $a_n = c_n / 2^n$, where $c_n$ is the largest integer that is less than or equal to $2^n \pi$. This insures that $a_n < \pi$ for all $n$ and $\lim\limits_{n \rightarrow \infty} a_n = \pi$. (This does not quite give the original sequence because it includes some repetition. In fact, we have $a_0 = a_1 = a_2 = 3$, $a_3 = a_4 = a_5 = 25/8$, $a_6 = 201/64$. Still it gives the same set of values for the left-hand side.)
The right-hand side is similarly all of the values from a sequence $b_n$ with $b_n = d_n / 2^n$, where $d_n$ is the smallest integer that is greater than or equal to $2^n \pi$. Then $b_n > \pi$ for all $n$ and $\lim\limits_{n \rightarrow \infty} b_n = \pi$.
Note that, as already pointed out, these sequences are not geometric, so you cannot sum a geometric series to find the value of either side. Also, we use dyadic rational numbers to approximate $\pi$ because dyadic rational numbers correspond to the games that have numbers as values and have finite birthdays. Thus $\pi$ is the value of a game "born on day $\omega$''.