A Family Of Puzzling Infinite Continued Fractions

We can easily figure out that

1+11+11+11+11+11+11+=ϕ\large 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}}}=\phi

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

(Proof: 1+11+11+11+11+11+11+1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}}} is equal to 11 plus its reciprocal, or x=1+1xx=1+\frac{1}{x}. After multiplying both sides by xx and moving all of the terms to the left, we obtain

x2x1=0x^2-x-1=0

which can be evaluated to give the solution x=1±52x=\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+mn+mn+mn+mn+mn+mn+n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\frac{m}{n+\ddots}}}}}}

which has a solution n±n2+4m2\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+12+23+34+45+56+67+7(n1)+n1n\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 nn\to\infty? What is the limit as nn\to\infty of any continued fraction of the form

m+m(m+1)m+1(m+2)+m+2(m+3)+m+3(m+4)+m+4(m+5)+m+6(m+7)+m+7(m+n1)+m+n1\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
1 year, 2 months ago

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