# The Prisoner's Dilemma

### D1. Continuing Irrationality

Contributed by Christopher Havens, PMP

One interesting thing about continued fractions is that every infinite continued fraction represents an irrational number. But in our first dilemma, we ask what happens when you have an infinite sequence of finite continued fractions with more and more terms.

Recall that the binomial coefficient $a\choose b$ has the formula $\frac{a!}{b!(a-b)!}$. Now for any integer $m\geq 0,$ define the $m+1$st term in our sequence by $C_{m+1} = {m\choose0} + \cfrac1{{m-1\choose1} + \cfrac{1}{{m-2\choose2}+\cdots+\cfrac1{m-k\choose k}}},$ where $k= \lfloor \frac{m}2 \rfloor.$ (The notation $\lfloor x \rfloor$ means “the greatest integer less than or equal to $x$,” so $k = \frac{m}2$ when $m$ is even and $k=\frac{m-1}2$ when $m$ is odd.)

For example, we have $C_6 = {5\choose0} + \cfrac1{{4\choose1}+\cfrac1{3\choose2}} .$

As $n$ becomes infinitely large, that is, as $n\to\infty$, the continued fraction $C_n$ has more and more terms. Does that mean that $C_n$ approaches an irrational number as $n\to\infty$? What is $\lim_{n\to\infty} C_n$, if it exists?

This problem originally appeared in the Prisoner’s Dilemma in the 2021 Spring issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

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Solution.

We received a submission for this dilemma from PMP member William Burns, who correctly identified that the limit of this sequence is one, not an irrational number. So although a single infinite continued fraction must represent an irrational number, a sequence of ever-longer finite continued fractions can tend to a rational number.

Here’s why $\lim_{n\to\infty} C_n = 1.$ Notice that the leading term of $C_{m+1}$ is $\binom m0,$ which is always equal to one. So for $m>1,$ $C_{m+1}$ is one plus something, and that something is always positive. And the second term of $C_{m+1}$ is $\binom{m-1}1,$ or just $m-1,$ so the quantity that we are adding to that initial one is a fraction with numerator 1 and denominator $m-1$, possibly plus something positive. In any case, the quantity added is less than $1/(m-1).$ In other words, we have shown that for $n>2,$ $1 < C_n < \frac1{n-2}.$ (we have $n-2$ in the denominator here as $n$ and $m$ differ by one because of the definition of the sequence $C.$) But as $n\to\infty,$ $1/(n-2) \to 0.$ So by the Squeeze Theorem for sequences, we must also have that $C_n \to 1.$ ■