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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 $\pi$?

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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 $\omega$''.

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  • $\begingroup$ I can't vote up because I lack the reputation, but thanks for this. I'm new to the field and this detailed answer helped a lot. $\endgroup$ – Ingrid Morstrad Apr 9 '14 at 18:00
  • $\begingroup$ If possible, could you explain this: This insures that an<π for all n and limn→∞ an=π $\endgroup$ – Ingrid Morstrad Apr 9 '14 at 18:15
  • $\begingroup$ Well $a_n < \pi$ because, by construction, $c_n < 2^n \pi$, so $a_n = c_n / 2^n < \pi$. But also, by the choice of $c_n$, we have that $c_n + 1 > 2^n \pi$ (otherwise $c_n$ would not be the largest integer less than $2^n \pi$!), so $a_n > \pi - 1/2^n$. Since $\pi - 1/2^n < a_n < \pi$, by the squeeze theorem, $\lim\limits_{n \rightarrow \infty} a_n = \pi$. $\endgroup$ – Michael Joyce Apr 9 '14 at 18:29
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You seem to be inferring some unintended patterns in the truncated list as it was given. The LHS is simply supposed to be a sample of dyadic fractions which fall short of pi, chosen so that the supremum of the sample is precisely pi, and similarly the RHS a sample of dyadic fractions strictly greater than pi chosen so that the infimum is pi.

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  • $\begingroup$ I thought both the left and right games contain members of an infinite series and since 25/8 = 3 + 1/8 and 201/64 = 3 + 1/8 + 1/32, I evaluated it as described above. Is it wrong to assume they are an infinite series? If so, how do we know the next number in the series (and so the supremum). $\endgroup$ – Ingrid Morstrad Apr 9 '14 at 6:07
  • $\begingroup$ I don't know how to edit my comment. I mean 201/64 = 3 + 1/8 + 1/64 $\endgroup$ – Ingrid Morstrad Apr 9 '14 at 7:44
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    $\begingroup$ It is not wrong to assume each side can be converted into an infinite series. It IS wrong to assume that it is a geometric series with a particular ratio. By no means is $\pi$ the sum of any geometric series of rational numbers: if so, $\pi$ would be rational, which it ain't. $\endgroup$ – Lee Mosher Apr 9 '14 at 12:21

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