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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?

Américo Tavares
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Yonatan N
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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. –  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. ain't a mathematician Aug 26 '10 at 15:41
  • I have a translation of Leonhard Euler's paper by Daniel W. File, The Ohio State University. See my answer. – Américo Tavares Aug 26 '10 at 16:02
  • Techinically speaking this isn't a continued fraction. – Aryabhata Aug 26 '10 at 16:15
  • What I meant was: the theorems about CF being 'best' approximations etc need the numberators to be 1, don't they? – Aryabhata Aug 26 '10 at 16:24
  • Moron: no, not necessarily. However all continued fractions can by "equivalence transformations" be transformed such that all the numerator terms are unity. – J. M. ain't a mathematician Aug 26 '10 at 16:27
  • In fact, one can have arbitrary functions in the numerators and denominators of a continued fraction. So yes, what Yonatan has is a perfectly valid continued fraction. – J. M. ain't a mathematician Aug 26 '10 at 16:29
  • @J. Mangaldan: My comment was mainly in response to Asaf's statment that "the rather amazing theorem about CF is that they give the best approximations". I am doubtful that putting in any numbers in the numerators will work for that statement to be true. i.e. it is not necessary that what we see in the question will give us only the convergents of 1/(e-2). – Aryabhata Aug 26 '10 at 16:34
  • Moron: Well, admittedly, the theorems on proving the convergence of a continued fraction can be subtle (one frequently used convergence criterion rests on showing that a related series is divergent!); as I said, though, there's always a way to transform continued fractions so that they have numerator terms or denominator terms of unity. See Lorentzen and Waadeland's "Continued Fractions with Applications" for the juicy details. – J. M. ain't a mathematician Aug 26 '10 at 16:52
  • The most straightforward way to show convergence, of course, would proceed like in my answer: determine expressions for the numerators and denominators of the nth convergent, and check that the limit of the quotient as n approaches infinity does in fact exist. – J. M. ain't a mathematician Aug 26 '10 at 16:58
  • @J.Mangaldan: Do you have any online reference? The only equivalence transform I have seen is in a general setting, where you are allowed to have any arbitrary real/complex numbers. Trying to make the numerators 1, might involve making some of the other numbers non integral (i.e. multiply by 1/a etc). Btw, by convergents I meant: the fractions you get by using the unique CF of the number. I think there is a disconnect there. – Aryabhata Aug 26 '10 at 17:01
  • Ah, you meant the simple continued fraction expansion of a real number. Then, yes, the "best approximant" label does not always apply. Sorry for the confusion, my dealings with CFs have been mostly with complex numerators and denominators. – J. M. ain't a mathematician Aug 26 '10 at 17:11
  • Also, I was thinking of the general meaning of the term "convergent" in the theory of continued fractions, which is the concept applied in my answer. – J. M. ain't a mathematician Aug 26 '10 at 17:16
  • @J. Mangaldan: One man's CF is another man's simple CF :-) I guess that clears it up :-) – Aryabhata Aug 26 '10 at 17:17

2 Answers2

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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 is an explanation of how he proceeded.

New Edit in response to @A-Level Student's comment. I transcribed the following assertion from the available translation to English of Euler's article. Now I checked the original paper and corrected equation (1b).

He stated that "we are able to demonstrate without much difficulty, that if

$$\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=s,\tag{1a}$$

then

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

Since, in this case, we have $s=1/(1-e)$, $a=1,b=2,c=3,\ldots $ it follows

$$1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cdots }}}=\dfrac{1}{e-2}.$$

Edit: Euler proves first how to form a continued fraction from an alternating series of a particular type [Theorem VI, §40] 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.

Américo Tavares
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    This is stunning and beautiful, thank you for this answer! (and the explanation of the coincidence) – t.b. Aug 05 '11 at 23:35
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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 06 '11 at 17:54
  • Hi, please could you explain how you get $$a+\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=\dfrac{s}{1-s}?$$ Thank you. – A-Level Student Jun 17 '21 at 11:08
  • @A-LevelStudent Thank you very much! This equation was wrong. Please see the edit. – Américo Tavares Jun 19 '21 at 22:33
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    @AméricoTavares You're welcome! Thanks for editing. I find your answer fascinating. – A-Level Student Jun 20 '21 at 09:03
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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.

J. M. ain't a mathematician
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  • The solutions to both difference equations are related to the so-called "subfactorial": http://mathworld.wolfram.com/Subfactorial.html – J. M. ain't a mathematician 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: http://arxiv.org/abs/math.NT/0601660 – David E 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. ain't a mathematician 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 E 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. ain't a mathematician Aug 27 '10 at 23:09
  • @J. Mangaldan: +1 for using the continued fraction recurrence and linking to paper that shows how to find $A_n,B_n$ from $a_j,b_j$. – Américo Tavares Aug 30 '10 at 12:01