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 $(\genfrac{}{}{0ex}{}{a}{b})$ has the formula $\frac{a!}{b!(a-b)!}$. Now for any integer $m\ge 0$, define the $m+1$st term in our sequence by $${C}_{m+1}=(\genfrac{}{}{0ex}{}{m}{0})+\frac{{\textstyle 1}}{{\textstyle (\genfrac{}{}{0ex}{}{m-1}{1})+\frac{{\textstyle 1}}{{\textstyle (\genfrac{}{}{0ex}{}{m-2}{2})+\cdots +\frac{{\textstyle 1}}{{\textstyle (\genfrac{}{}{0ex}{}{m-k}{k})}}}}}},$$ 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 $${C}_{6}=(\genfrac{}{}{0ex}{}{5}{0})+\frac{{\textstyle 1}}{{\textstyle (\genfrac{}{}{0ex}{}{4}{1})+\frac{{\textstyle 1}}{{\textstyle (\genfrac{}{}{0ex}{}{3}{2})}}}}.$$

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