The Prisoner's Dilemma

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 $\underset{n\to \mathrm{\infty }}{lim}{C}_n=1$. Notice that the leading term of $C_{m+1}$ is $(\genfrac{}{}{0ex}{}{m}{0})$, which is always equal to one. So for , $C_{m+1}$ is one plus something, and that something is always positive. And the second term of $C_{m+1}$ is $(\genfrac{}{}{0ex}{}{m-1}{1}),$ 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 , (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 $n\to \mathrm{\infty }$, $1/(n-2)\to 0.$ So by the Squeeze Theorem for sequences, we must also have that $C_n\to 1.$ ■