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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.2011-04-01
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    @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.2011-04-01
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    God, I'm stupid sometimes! Thanks...2011-04-01
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    (I've made that mistake too, which is why I could answer your question. :D )2011-04-01

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