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A couple of years ago I found the following continued fraction for $\frac1{e-2}$:

$\frac{1}{e-2} = 1+\cfrac1{2 + \cfrac2{3 + \cfrac3{4 + \cfrac4{5 + \cfrac5{6 + \cfrac6{7 + \cfrac7{\cdots}}}}}}}$

from fooling around with the well-known continued fraction for $\phi$. Can anyone here help me figure out why this equality holds?

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    @J. Mangaldan: One man's CF is another man's simple CF :-) I guess that clears it up :-)2010-08-26

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Euler proved in "De Transformatione Serium in Fractiones Continuas" Reference: The Euler Archive, Index number E593 (On the Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that

$s=\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$

Here his an explanation on how he proceeded. He stated that if

$\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=s,$

then

$a+\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=\dfrac{s}{1-s}.$

Since, in this case, we have $a=1,b=2,c=3,\ldots $ it follows

$1+\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-2}.$

Edit: Euler proves first how to transform an alternating series of a particular type into a continued fraction and then uses the expansion

$e^{-1}=1-\dfrac{1}{1}+\dfrac{1}{1\cdot 2}-\dfrac{1}{1\cdot 2\cdot 3}+\ldots .$


REFERENCES

The Euler Archive, Index number E593, http://www.math.dartmouth.edu/~euler/

Translation of Leonhard Euler's paper by Daniel W. File, The Ohio State University.

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    Dear Américo, once again, thanks a lot for this $p$ost. I printed the original article you linked to, took my dusty Sto$w$asser (standard Latin-German dictionar$y$) out of my shelf and had some of the more joyful hours in a long time while reading (deci$p$hering, rather -- my Latin has become more than rusty over the years). What a great and inspiring contribution to this site!2011-08-06
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Another possibility: remember that the numerators and denominators of successive convergents of a continued fraction can be computed using a three term recurrence.

For a continued fraction

$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\dots}}$

with nth convergent $\frac{C_n}{D_n}$, the recurrence

$\begin{bmatrix}C_n\\\\D_n\end{bmatrix}=b_n\begin{bmatrix}C_{n-1}\\\\D_{n-1}\end{bmatrix}+a_n\begin{bmatrix}C_{n-2}\\\\D_{n-2}\end{bmatrix}$

with starting values

$\begin{bmatrix}C_{-1}\\\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\\\0\end{bmatrix}$, $\begin{bmatrix}C_{0}\\\\D_{0}\end{bmatrix}=\begin{bmatrix}b_0\\\\1\end{bmatrix}$

holds.

With $b_j=j+1$ and $a_j=j$, you now try to find a solution for those two difference equations.

Skipping details, the solution of those two recursions are

$C_n=\frac{(n+3)!}{n+2}\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}$

and

$D_n=\frac{(n+3)!}{n+2}\left(1-2\sum_{j=0}^{n+3}\frac{(-1)^j}{j!}\right)$

are solutions to the two difference equations.

Divide $C_n$ by $D_n$ and take the limit as $n\to\infty$; you should get the expected result.

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    @J. Mangaldan: +1 for using the continued fraction recurrence and linking to paper that shows how to find $A_n,B_n$ from $a_j,b_j$.2010-08-30