# A Family Of Puzzling Infinite Continued Fractions

We can easily figure out that

$\large 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}}}=\phi$

where $\phi$ is the 'golden ratio' $1.6180339...$.

(Proof: $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}}}$ is equal to $1$ plus its reciprocal, or $x=1+\frac{1}{x}$. After multiplying both sides by $x$ and moving all of the terms to the left, we obtain

$x^2-x-1=0$

which can be evaluated to give the solution $x=\frac{1\pm\sqrt5}{2}$.)

One can see that a continued fraction of this form can be generalized for any continued fraction of the form

$n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\ddots}}}}}}$

which has a solution $\frac{n\pm\sqrt{n^2+4m}}{2}$. I will omit the derivation, which is easy to derive by modifying the proof above.

$\large 1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\frac{5}{6+\frac{6}{7+\frac{7}{\ddots(n-1)+\frac{n-1}{n}\ddots}}}}}}}$

What is the limit of this continued fraction as $n\to\infty$? What is the limit as $n\to\infty$ of any continued fraction of the form

$\large m+\frac{m}{(m+1)\frac{m+1}{(m+2)+\frac{m+2}{(m+3)+\frac{m+3}{(m+4)+\frac{m+4}{(m+5)+\frac{m+6}{(m+7)+\frac{m+7}{\ddots(m+n-1)+\frac{m+n-1}{\ddots}}}}}}}}$ Note by Andrei Li
2 years, 3 months ago

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