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From Euler's continued fraction formula, we have $$x = \cfrac{1}{1 - \cfrac{r_1}{1 + r_1 - \cfrac{r_2}{1 + r_2 - \cfrac{r_3}{1 + r_3 - \ddots}}}}\,$$ and $$x = 1 + \sum_{i=1}^\infty r_1r_2\cdots r_i = 1 + \sum_{i=1}^\infty \left( \prod_{j=1}^i r_j \right)\,$$ For example: $$e^z = 1 + \sum_{n=1}^\infty \frac{z^n}{n!} = 1 + \sum_{n=1}^\infty \left(\prod_{j=1}^n \frac{z}{j}\right)=$$$$=\cfrac{1}{1 - \cfrac{z}{1 + z - \cfrac{\frac{1}{2}z}{1 + \frac{1}{2}z - \cfrac{\frac{1}{3}z} {1 + \frac{1}{3}z - \cfrac{\frac{1}{4}z}{1 + \frac{1}{4}z - \ddots}}}}}.\,$$ (by Wikipedia).

Well, my question is the following: are there any version of Euler's continued fraction for double series?

Thanks.

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  • $\begingroup$ Double series is still just a series (you only need to expand them to see this). So, you can still use Euler's continued fraction for double series as well $\endgroup$ – Yuriy S Mar 22 '16 at 13:57

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