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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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What do you mean by "actually works"? –  Asaf Karagila Aug 26 '10 at 15:09
    
I meant "why would this continued fraction converge to said value"? –  Yonatan N Aug 26 '10 at 15:19
    
It's easy to get from this to the continued fraction for ee (or back) so this question is equivalent to finding a continued fraction for ee. According to MathWorld there is a proof of this on page 348 of Wall, H. S. Analytic Theory of Continued Fractions. –  anon Aug 26 '10 at 15:27
    
If you think about it, every time you add a slightly smaller number, and it's quite simply to verify that this series converges. The rather amazing theorem about continued fractions is that they give best approximations. –  Asaf Karagila Aug 26 '10 at 15:36
    
muad: I've a copy of the book, and there is a CF expansion of the exponential function, but not of the constant Yonatan is interested in on that page. Are you sure of the page number? –  J. M. Aug 26 '10 at 15:41
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2 Answers

up vote 20 down vote accepted

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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Tavares: Awesome! –  anonymous Aug 27 '10 at 1:56
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@Chandru1: Thanks! And thanks to Euler. –  Américo Tavares Aug 27 '10 at 21:57
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This is stunning and beautiful, thank you for this answer! (and the explanation of the coincidence) –  t.b. Aug 5 '11 at 23:35
    
@Theo: Many thanks! –  Américo Tavares Aug 6 '11 at 9:22
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Dear Américo, once again, thanks a lot for this post. I printed the original article you linked to, took my dusty Stowasser (standard Latin-German dictionary) out of my shelf and had some of the more joyful hours in a long time while reading (deciphering, rather -- my Latin has become more than rusty over the years). What a great and inspiring contribution to this site! –  t.b. Aug 6 '11 at 17:54
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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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The solutions to both difference equations are related to the so-called "subfactorial": mathworld.wolfram.com/Subfactorial.html –  J. M. Aug 26 '10 at 16:36
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If you wonder where these formulas come from, you might want to read Henry Cohn's exposition: arxiv.org/abs/math.NT/0601660 –  David Speyer Aug 27 '10 at 15:24
    
David: Nice paper, though that is for the SCF expansion of $e$. Still, it is useful to note that the numerator and denominator polynomials of the Padé approximant for $\exp(z)$ are expressible as incomplete gamma functions, and the subfactorial used in the solution above is related to these incomplete gammas. –  J. M. Aug 27 '10 at 15:42
    
The continued fraction for $e$ starts off $2+1/(a+1/(b+\cdots))$. So $1/(e-2) = a+1/(b+\cdots)$. If you understand one, you understand the other. –  David Speyer Aug 27 '10 at 17:39
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David, you're having the same sort of confusion Moron had in the comments (see the extended discussion above); the one in the paper is the simple continued fraction, with $a_j=1$. The one considered here, on the other hand, has $a_j=j$. You're telling me there's an equivalence transformation relating the two? I think Moron and me would like to hear it. (Though again, there is the nice connection that partial sums of the exponential series are expressible as incomplete gamma functions. This manifests itself in both CFs.) –  J. M. Aug 27 '10 at 23:09
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