# A Continued Fraction for $e$

I was a little surprised to see a continued fractions conjecture on Wikipedia, stating:

$e = { 3 + \frac{-1}{4 + \frac{-2}{5 + \frac{-3}{6 + \dots}}}}$

Obviously, my first thought was, “that doesn’t look too hard to prove”. My second thought was “I imagine it already has been”, and such is the case: an arXiv paper and a MathOverflow comment both beat me to the punch.

But just because something has been proved doesn’t mean it’s not worth proving again!

### Playing with the convergents

I am not a native continuedfractionologist. I didn’t touch them during my degree, and I’ve only recently started meddling with them. I tend to approach them with a bit of trepidation, and gingerly work out the first few *convergents* - the fractions you get from truncating the expansion as you go along.

Here, the first approximation is 3. That’s not a bad approximation to $e$, as these things go.

The second approximation is $3 + -1/4$, or $\frac{11}{4}$. 2.75: definitely in the right ballpark.

Then we have $3 + -1/(4 + -2 / 5)$, which works out to be $\frac{49}{18}$, 2.7222…. It looks plausible at the very least!

In fact, as we work it through, we notice that the $n$th convergent seems to correspond to the $n$th partial sum in $3 - \frac{1}{4} - \frac{1}{36} - \dots - \frac{1}{(n+1)!(n+1)n} - \dots$.

And that’s *also* listed in the representations of $e$ article as a valid series. If we can prove that the two are the same, we’ve proved it!

### A way in!

I know I’ve told the joke repeatedly that in maths, discoveries are typically named after the *second* person to discover it, because otherwise everything would be named after Euler.

$e$ is named after Euler ((He didn’t discover it)). Euler’s continued fraction formula is also named after him ((He *did* discover this.))

It states that, if you can write a number as a convergent sum $a_0(1 + a_1(1+ a_2(1 + \dots)))$, then its continued fraction can be written as:

$\frac{a_0}{1 + \frac{a_1}{(1+a_1) - \frac{a_2}{(1+a_2 - \frac{a_3}{(1+a_4 - \dots)}}}}$

As in the example we’re looking at, the $n$th convergent matches up with the $n$th partial sum.

To make things easier for my simple mind, I’m going to start from the equivalent statement $3-e = \frac{1}{4} + \frac{1}{36} + \frac{1}{288} + \dots + \frac{1}{(n+2)!(n+2)(n+1)}$.

Alternatively, this is $3-e = \frac{1}{4}\left(1 + \frac{1}{9}\left(1 + \frac{2}{16}\left(1 + \dots \right)\right)\right)$. The $a_i$s here are $a_0 = \frac{1}{4}$, $a_1 = \frac{1}{9}$, $a_2 = \frac{2}{16}$, and generally $a_k \frac{k}{(k+2)^2}$ for $k \ge 1$.

And we can apply Euler directly to write down a continued fraction representation:

$3-e = \frac{ 1/4 }{1 - \frac{1/9}{10/9 - \frac{2/16}{18/16 - \frac{\dots}{\dots k/(k+2)^2}{1 + k/(k+2)^2 - \frac{(k+1)/(k+3)^2}{\dots}}}}}$

### Ooft.

Now we’re going to repeatedly flush out the fractions. In general, $\frac{a}{b + \frac{c}{d + \dots}} = \frac{pa}{pb + \frac{pc}{d + \dots}}$, so if we multiply the top of a fraction, the lead term of the bottom, and the top of the next fraction by the same constant, we don’t change the value of the continued fraction. In particular, if I want to turn the 1/4 on top into 1, I need to multiply the 1 and the 1/9 on the next line by 4 as well:

$3-e = \frac{1}{4 - \frac{4/9}{10/9 - \frac{2/16}{18/16 - \dots}}}$

I can repeat the process, multiplying the next bit down by 9/2:

$3 - e = \frac{1}{4 - \frac{2}{5 - \frac{9/16}{18/16 - \frac{3/25}{\dots}}}}$

On the next line, the multiplier is 16/3:

$3 - e = \frac{1}{4 - \frac{2}{5 - \frac{3}{6 - \frac{16/25}{\dots}}}}$

That’s looking pretty good so far – and the basis of an induction proof!

### Induction (sketch)

If I have $\frac{(k+1)^2/(k+2)^2}{(1 + k/(k+2)^2) - \frac{(k+1)/(k+3)^2}{\dots}}$ at some point in the fraction structure, I can rewrite the lead term in the denominator as $((k+2)^2+ k)/(k+2)^2$. The top of that expands as $k^2 + 5k + 4$, or $(k+1)(k+4)$.

We can therefore multiply all three of the relevant terms by $(k+2)^2 / (k+1)$ to leave us with $\frac{k+1}{k+4 - \frac{(k+2)^2/(k+3)^2}{\dots}}$

Using $k=1$ as the base case (we deal with the fraction starting with 4/9 in the previous section), this will show that following the flushing out will always give us the correct form: the top will become $k+1$, and the bottom line $k+4 - \frac{(k+2)^2/(k+3)^2}{\dots}$. The integer parts match up with the continued fraction we’re trying to reach, and the next fraction to deal with is the ratio of (the correct) consecutive squares so we can apply the same trick to the next line of the fraction.

I’ve left some details for the reader, so I won’t throw out a $\blacksquare$ here, but I believe I’ve given you enough scaffolding to make it rigorous if you’d like to!

I thought that was a neat thing, and I’m always keen to learn a little more about continued fractions. What did you make of it?