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Lately, I learned about the following continued fraction for the exponential function:

$$\exp(x)=1+\cfrac{x}{1-\cfrac{x/2}{1+x/2-\cfrac{x/3}{1+x/3-\cfrac{x/4}{1+x/4-\dots}}}}$$

I thought it was something new, but evaluating the successive convergents of this continued fraction was a disappointment, as they are nothing more than the partial sums of the usual series $\exp(x)=\sum_{j=0}^{\infty}\frac{x^j}{j!}$.

So, there must be some way to obtain the continued fraction from the series. How might this be done?

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Can you explain how you get the usual partial sums? Truncating after two steps, I get $(1+x/2)/(1-x/2)$, which isn't even a polynomial. –  Hans Lundmark Apr 1 '11 at 7:46
    
@Hans: the truncation is supposed to be $$1+\cfrac{x}{1-\cfrac{x/2}{1+x/2}}$$ in that case. You forgot the partial denominator. –  gorilla Apr 1 '11 at 7:57
    
God, I'm stupid sometimes! Thanks... –  Hans Lundmark Apr 1 '11 at 8:02
    
(I've made that mistake too, which is why I could answer your question. :D ) –  gorilla Apr 1 '11 at 8:04

1 Answer 1

This transformation of a series into its equivalent continued fraction, with the series partial sums being equal to the continued fraction convergents, is due to Euler. The series $$\sum_{n\geq 0}c_{n}=c_0+c_1+\dots+c_n+\dots$$ is transformed into the continued fraction

$$b_0+\mathbf{K}\left( a_{n}|b_{n}\right) =b_0+\dfrac{a_{1|}}{|b_{1}}+\dfrac{a_{2}|}{% |b_{2}}+\cdots +\dfrac{a_{n}|}{|b_{n}}+\cdots ,$$

whose elements $a_{n}$, $b_{n}$ are expressed in terms of $c_n$ as follows: $b_0=c_0$, $a_1=c_1$, $b_1=1$ and $$a_{n}=-\dfrac{c_{n}}{c_{n-1}},\qquad b_{n}=1+\dfrac{c_{n}}{c_{n-1}}\qquad n\ge 2.$$

For the power series $e^x=\sum_{n\geq 0}\dfrac{1}{n!}x^{n}$, we have $c_{n}=\dfrac{1}{n!}x^{n}$, and get $$a_{n}=-\dfrac{c_{n}}{c_{n-1}}x=-\dfrac{1}{n}x,\qquad b_{n}=1+\dfrac{c_{n}}{c_{n-1}}=1+\dfrac{1}{n}x\qquad n\ge 2.$$

Thus

$$\begin{eqnarray*} e^x &=&\sum_{n\geq 0}\frac{1}{n!}x^{n}=1+\sum_{n\geq 1}\frac{1}{n!}x^{n} \\ &=&1+\frac{x|}{|1}-\frac{\frac{1}{2}x|}{|1+\frac{1}{2}x}-\cdots -\frac{\frac{% 1}{n}x|}{|1+\frac{1}{n}x}-\cdots, \end{eqnarray*}$$

which is equivalent to

$$1+\frac{x|}{|1}-\frac{x|}{|2+x}-\frac{2x|}{|3+x}-\cdots -\frac{nx|}{|n+1+x}+\cdots.$$

This is explained in p.17 of Die Lehre von den Kettenbrüchen Band II by Oskar Perron and proved in Theorem 4.2 of Orthogonal Polynomials and Continued Fractions From Euler´s Point of View by Sergey Khrushchev. It is derived from a theorem that establishes the equivalence between a sequence and a continued fraction.

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