Infinite Series And Continued Fractions

We can easily figure out that

\[1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}}}=\phi\]

where \(\phi\) is the 'golden ratio' \(1.6180339...\).

(Proof: \(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}}}\) 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

\[n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+...}}}}}}\]

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.

However, what about

\[1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\frac{5}{6+\frac{6}{7+\frac{7}{...(n-1)+\frac{n-1}{n}...}}}}}}}\]

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

\[m+\frac{m}{(m+1)\frac{m+1}{(m+2)+\frac{m+2}{(m+3)+\frac{m+3}{(m+4)+\frac{m+4}{(m+5)+\frac{m+6}{(m+7)+\frac{m+7}{...(m+n-1)+\frac{m+n-1}{m+n}...}}}}}}}\]

Note by Andrei Li
2 weeks, 1 day ago

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