## A fracção contínua de 1 / (e – 2)

Nesta questão  de Agosto de 2010, no MSE, Yonatan N perguntou porque é que uma fracção contínua de $\dfrac{1}{e-2}$, que indicou, era válida.

A resposta remonta a Euler. Eis a minha resposta, no original:

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

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

Anúncios

## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Fracções Contínuas, Matemática, Mathematics Stack Exchange com as etiquetas , . ligação permanente.