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I saw this link (written in Japanese) and found an interesting problem:

Calculate $1 + \frac 2{3 + \frac 4 {5 + \cdots}}$.

The link provides the answer ($\frac 1 {\sqrt e - 1}$) and a hint that one uses Maclaurin series, but doesn't provide a detailed answer.

Can someone explain this equality? If possible, can someone also calculate

  1. $1^2 + \frac {2^2}{3^2 + \frac {4^2} {5^2 + \cdots}}$ and
  2. $1^n + \frac {2^n}{3^n + \frac {4^n} {5^n + \cdots}}$?

These two are also in the link, and answers of them are not provided and seem unsolved.

Answer of Continued fraction for $\frac{1}{e-2}$ might help.

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    $\begingroup$ See this link $\endgroup$ – Yuriy S Jun 24 '16 at 8:13
  • $\begingroup$ The first question was easy: just substitute -1/2 for z in the link. Thank you. $\endgroup$ – H Koba Jun 24 '16 at 8:39

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