Fractional approximation of $e$

Question

For $\pi$ there are the fractional approximations $\frac{22}{7}$ and the only slightly longer but much more accurate $\frac{355}{113}$.

I am aware of an analogous fractional approximation for $e$, $\frac{19}{7}$. However, this is only accurate to $1$ decimal place ($3$ digits), so not terribly practical. I would be interested to find some more accurate and hopefully not too much longer fractional representations of $e$. Is there a fractional representation for $e$ as compact as $\pi$'s $\frac{355}{113}$ with a similar accuracy?

Answer 1

The continued fraction for $\pi$ is$$3+\cfrac1{7+\cfrac1{15+\cfrac1{1+\cfrac1{\color{red}{292}+\cdots}}}}$$That $292$ is a huge number in this context, and it's because of it that we have the excellent approximation$$\pi\approx\frac{355}{113}=3+\cfrac1{7+\cfrac1{15+\cfrac11}}.$$Nothing similar occurs in the case of $e$, whose continued fraction is$$e=2+\cfrac1{1+\cfrac1{2+\cfrac1{1+\cfrac1{1+\cfrac1{4+\cfrac1{1+\cfrac1{1+\cfrac1{6+\cfrac1{1+\cdots}}}}}}}}}$$

Answer 2

$$e \approx \frac{2721}{1001} = 2.\overline{718281}$$

This is the smallest-denominator fraction that has an absolute error ($1.102 \times 10^{-7}$) less than $|\pi - \frac{355}{113}| \approx 2.668 \times 10^{-7}$.

Answer 3

If you like those things, then you might like this too:

$$e \approx \frac{53035 \pi }{61294} \approx 2.7182818282888(...)$$

Which yelds also to

$$\frac{e}{\pi} \approx \frac{53035}{61294}$$

and the reciprocal.

Answer 4

One approximation that links to other problems is to render the Maclaurin series for $\exp(x)$ at $x=-1$ and simply truncate it, taking the reciprocal of the result as the approximation for $e$. Thus for instance:

$\exp(-1)\approx(1/2)-(1/6)+(1/24)-(1/120)=44/120,$

$e\approx(120/44)=(30/11)=2.72727...$

If the approximation is left with the factorial in the numerator, the denominator us the corresponding number of derangements of the corresponding number of objects (here, $44$ derangements for five objects).

Successive approximations by this method bracket $e$ tightly.