Benedict Irwin edited Investigation.tex  over 9 years ago

Commit id: 612773019c08dc1d98c10ec2a10ffcabe9308d1a

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\lim_{n\to\infty} \frac{1|2;n}{1|8;n} =0.\overline{6470588235294117} = \frac{11}{17} \\  \end{equation}  Which would at then imply \begin{equation}  \lim_{n\to\infty} \frac{1|d_1;n}{d_2|8;n}=\frac{(1|d_1) - 1}{(d_2|8) - d_2}  \end{equation}