I've found this formula, but I don't know how prove it? What's your idea for proof? $$1+{1\over1}\left(1+{1\over2}\left(1+{1\over3}\left(1+{1\over4}\left(1+{1\over5}\left( ... \right)\right)\right)\right)\right)=e.$$
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6Expand the first few terms. It will look awfully familiar. – André Nicolas Oct 04 '13 at 06:17
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3You know $e=1+1+\frac12+\frac1{3!}+\frac1{4!}+\frac1{5!}+\cdots$? – Jonas Meyer Oct 04 '13 at 06:17
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Yes, I know.... – Oct 04 '13 at 06:18
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This is called "nested form polynomial" (although it's not exactly a polynomial in this case). – DanielV Oct 04 '13 at 06:34
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The expression in the OP is an added fraction or continued numerator, which here can be treated with a base which incresing denominator as one goes on. It's the sort of fraction that evolved into decimals.
You get a similar idea when you divide a length in yards into a feet, and then into b inches, then into c sixteenths, as $y \frac a3 \frac b{12} \frac {c}{16}$ yards.
A number in a reglare base B, like 10, might be written as $1 \frac aB \frac bB \frac cB \cdots $, which is a sum of fractions, the denominator of the last being the product of it and all of the ones to the right. So in the base expression, the $c$ is divided by $B^3$.
The expression for $e$ can be written like this:
$$ e = 1 \frac 11 \frac 12 \frac 13 \frac 14 \frac 15 \frac 16 \dots $$
In the expression for $e$, we see that it is a sum $\sum \frac 1{n!} $, which means that this expression, because that's what the running product of denominators is.
One can see from this representation, that any digit as a numerator in any place leads to a unit fraction, and thus one can write any $a/b$ in $m-1$ unit fractions, where $b\mid m!$, however, smaller solutions are allowable. (ie it's an upper limit).
wendy.krieger
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