Intereting Posts

Is this a wrong solution to the smallest enclosing circle problem?
$x^5 – y^2 = 4$ has no solution mod $m$
Directly indecomposable rings
Dense subset of the plane that intersects every rational line at precisely one point?
Ring theory notation question
How deal with Gothic letters, like $\mathfrak{ A,B,C,D,a,b,c,d}\dots$, when writing by hand?
Why must a field with a cyclic group of units be finite?
What's the arc length of an implicit function?
$G$ finite group, $H\leq G$ such that $C_G(x)\subseteq H\quad\forall x\in H$ such that $p\mid o(x)$
How many $4$ digit even numbers have all $4$ digits distinct?
connected manifolds are path connected
Is this metric space incomplete?
Can an odd perfect number be a nontrivial multiple of a triangular number?
Needing help picturing the group idea.
How would I determine if there is at least 1 integer that can satisfy 5 congruences all at once?

What is $1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cdots}}}$ ?

What is $1+\cfrac{2}{1+\cfrac{3}{1+\cdots}}$ ?

It does bear some resemblance to the continued fraction for $e$, which is $2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cdots}}}$.

- the ratio of jacobi theta functions and a new conjectured q-continued fraction
- How to do a very long division: continued fraction for tan
- a q-continued fraction related to the octahedral group
- For which $a$ is $n\lfloor a\rfloor+1\le \lfloor na\rfloor$ true for all sufficiently large $n$?
- Fundamental unit in the ring of integers $\mathbb Z$
- Smallest number $N$, such that $\sqrt{N}-\lfloor\sqrt{N}\rfloor$ has a given continued fraction sequence

Another thing I was wondering: can all transcendental numbers be expressed as infinite continued fractions containing only rational numbers? Of course for almost all transcendental numbers there does not exist any method to determine all the numerators and denominators.

- $\sqrt{2}\notin\mathbb{Q}$ but …
- Approximating $\arctan x$ for large $|x|$
- If $x - \lvert x \rvert + \frac{1}{x} - \lvert \frac{1}{x} \rvert = 1$ then $x$ is irrational
- Show that $\arctan(n)$ is irrational for all $n \in \mathbb{N}$
- Can every transcendental number be expressed as an infinite continued fraction?
- Minimum of $|az_x-bz_y|$
- Continued fraction for $c= \sum_{k=0}^\infty \frac 1{2^{2^k}} $ - is there a systematic expression?
- The relationship between tan(x) and square roots
- Continued fraction for $\int_{0}^{\infty}(e^{-xt}/\cosh t)\,dt$
- Proving that for each prime number $p$, the number $\sqrt{p}$ is irrational

The first one is expressible in terms of the modified Bessel function of the first kind:

$$1+\cfrac1{2+\cfrac1{3+\cfrac1{4+\cdots}}}=\frac{I_0(2)}{I_1(2)}=1.433127426722\dots$$

The second one, through an equivalence transformation, can be converted into the following form:

$$1+\cfrac1{\frac12+\cfrac1{\frac23+\cfrac1{\frac38+\cfrac1{b_4+\cdots}}}}$$

where $b_k=\dfrac{k!!}{(k+1)!!}$ and $k!!$ is a double factorial. By Van Vleck, since

$$\sum_{k=1}^\infty \frac{k!!}{(k+1)!!}$$

diverges, the second continued fraction converges. This CF can be shown to be equal to

$$\frac1{\tfrac1{\sqrt{\tfrac{e\pi}{2}}\mathrm{erfc}\left(\tfrac1{\sqrt 2}\right)}-1}=1.904271233329\dots$$

where $\mathrm{erfc}(z)$ is the complementary error function.

**Establishing the value of the “continued fraction constant” (a short sketch)**

From the modified Bessel differential equation, we can derive the difference equation

$$Z_{n+1}(x)=-\frac{2n}{x}Z_n(x)+Z_{n-1}(x)$$

where $Z_n(x)$ is any of the two solutions $I_n(x)$ or $K_n(x)$. Letting $x=2$, we obtain

$$Z_{n+1}(2)=-n\,Z_n(2)+Z_{n-1}(2)$$

We can divide both sides of the recursion relation with $Z_n(2)$ and rearrange a bit, yielding

$$\frac{Z_n(2)}{Z_{n-1}(2)}=\cfrac1{n+\cfrac{Z_{n+1}(2)}{Z_n(2)}}$$

A similar manipulation can be done in turn for $\dfrac{Z_{n+1}(2)}{Z_n(2)}$; iterating that transformation yields

$$\frac{Z_n(2)}{Z_{n-1}(2)}=\cfrac1{n+\cfrac1{n+1+\cfrac1{n+2+\cdots}}}$$

Now, we don’t know if $Z$ is $I$ or $K$; the applicable theorem at this stage is Pincherle’s theorem. This states that $Z$ is necessarily the *minimal* solution of the associated difference equation if and only if the continued fraction converges (which it does, by Śleszyński–Pringsheim). Roughly speaking, the minimal solution of a difference equation is the unique solution that “decays” as the index $n$ increases (all the other solutions, meanwhile, are termed *dominant* solutions). From the asymptotics of $I$ and $K$, we find that $I$ is the minimal solution of the difference equation ($K$ and any other linear combination of $I$ and $K$ constitute the dominant solutions). By Pincherle, then, the continued fraction has the value $\dfrac{I_n(2)}{I_{n-1}(2)}$. Taking $n=1$ and reciprocating gives the first CF in the OP.

Here’s a short *Mathematica* script for evaluating the “continued fraction constant”, which uses the Lentz-Thompson-Barnett method for the evaluation:

```
prec = 50;
y = N[1, prec]; c = y; d = 0; k = 2;
While[True,
c = k + 1/c; d = 1/(k + d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^(-prec), Break[]];
k++];
y
1.4331274267223117583171834557759918204315127679060
```

We can check the agreement with the closed form:

```
y - BesselI[0, 2]/BesselI[1, 2] // InputForm
0``49.70728038020511
```

**Alternative expressions for the second continued fraction**

Just to thoroughly beat the stuffing out of this question, I’ll talk about a few other expressions that are equivalent to the OP’s second CF.

One can build the Euler-Minding series of the continued fraction:

$$1+\sum_{k=0}^\infty \frac{(-1)^k (k+2)!}{B_k B_{k+1}}$$

where $B_k$ is the denominator of the $k$-th convergent of the continued fraction, which satisfies the difference equation $B_k=B_{k-1}+(k+1)B_{k-2}$, with initial conditions $B_{-1}=0$, $B_0=1$. OEIS has a record of this sequence, but there is no mention of a closed form.

One can also split the original continued fraction into odd and even parts, yielding the following contractions:

$$3-\cfrac{6}{8-\cfrac{20}{12-\cfrac{42}{16-\cdots}}}\qquad \text{(odd part)}$$

$$\cfrac1{1-\cfrac{2}{6-\cfrac{12}{10-\cfrac{30}{14-\cdots}}}}\qquad \text{(even part)}$$

The utility of these two contractions is that they converge twice as fast as the original continued fraction, as well as providing “brackets” for the value of the continued fraction.

**Much, much later:**

Prompted by GEdgar’s question, I have found that the second CF does have a nice closed form. Here is a derivation:

The iterated integrals of the complementary error function, $\mathrm{i}^n\mathrm{erfc}(z)$ (see e.g. Abramowitz and Stegun) satisfy the difference equation

$$\mathrm{i}^{n+1}\mathrm{erfc}(z)=-\frac{z}{n+1}\mathrm{i}^n\mathrm{erfc}(z)+\frac1{2(n+1)}\mathrm{i}^{n-1}\mathrm{erfc}(z)$$

with initial conditions $\mathrm{i}^0\mathrm{erfc}(z)=\mathrm{erfc}(z)$ and $\mathrm{i}^{-1}\mathrm{erfc}(z)=\dfrac2{\sqrt\pi}\exp(-z^2)$.

This recurrence can be rearranged:

$$\frac{\mathrm{i}^n\mathrm{erfc}(z)}{\mathrm{i}^{n-1}\mathrm{erfc}(z)}=\frac1{2z+2(n+1)\tfrac{\mathrm{i}^{n+1}\mathrm{erfc}(z)}{\mathrm{i}^n\mathrm{erfc}(z)}}$$

Iterating this transformation yields the continued fraction

$$\frac{\mathrm{i}^n\mathrm{erfc}(z)}{\mathrm{i}^{n-1}\mathrm{erfc}(z)}=\cfrac1{2z+\cfrac{2(n+1)}{2z+\cfrac{2(n+2)}{2z+\dots}}}$$

(As a note, $\mathrm{i}^n\mathrm{erfc}(z)$ can be shown to be the minimal solution of its difference equation; thus, by Pincherle, the CF given above is correct.)

In particular, the case $n=0$ gives

$$\frac{\sqrt\pi}{2}\exp(z^2)\mathrm{erfc}(z)=\cfrac1{2z+\cfrac2{2z+\cfrac4{2z+\cfrac6{2z+\dots}}}}$$

If $z=\dfrac1{\sqrt 2}$, then

$$\frac{\sqrt{e\pi}}{2}\mathrm{erfc}\left(\frac1{\sqrt 2}\right)=\cfrac1{\sqrt 2+\cfrac2{\sqrt 2+\cfrac4{\sqrt 2+\cfrac6{\sqrt 2+\dots}}}}$$

We now perform an *equivalence transformation*. Recall that a general equivalence transformation of a CF

$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$

with some sequence $\mu_k, k>0$ looks like this:

$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$

If we apply this to the CF earlier with $\mu_k=\dfrac1{\sqrt 2}$, then

$$\sqrt{\frac{e\pi}{2}}\mathrm{erfc}\left(\frac1{\sqrt 2}\right)=\cfrac1{1+\cfrac1{1+\cfrac2{1+\cfrac3{1+\dots}}}}$$

Thus,

$$\frac1{\tfrac1{\sqrt{\tfrac{e\pi}{2}}\mathrm{erfc}\left(\tfrac1{\sqrt 2}\right)}-1}=1+\cfrac2{1+\cfrac3{1+\cfrac4{1+\dots}}}$$

I don’t know if either of the continued fractions can be expressed in terms of common functions and constants. However, all real numbers can be expressed as a continued fractions containing only integers. The continued fractions terminate for rational numbers, repeat for a quadratic algebraic numbers, and neither terminate nor repeat for other reals.

**Shameless plug:** There are many references out there for continued fractions. I wrote a short paper that is kind of dry and covers only the basics (nothing close to the results that J. M. cites), but it goes over the results that I mentioned.

I know how to do these. Here is the second question.

First, a more natural one:

$$

1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}=

\frac{1}{\displaystyle e^{1/2}\sqrt{\frac{\pi}{2}}\;\mathrm{erfc}\left(\frac{1}{\sqrt{2}}\right)}

\approx 1.525135276\cdots

$$

So the original one is

$$

1+\cfrac{2}{1+\cfrac{3}{1+\ddots}} =

\frac{1}{\displaystyle \frac{1}{ e^{1/2}\sqrt{\frac{\pi}{2}}\;\mathrm{erfc}\left(\frac{1}{\sqrt{2}}\right)}-1}

\approx 1.9042712\cdots

$$

[erfc is here ]

- Rate of convergence of series of squared prime reciprocals
- Is “Partition of Unity” a property of B-spline bases
- Equivalence of weak forms of Hilbert's Nullstellensatz
- Is $\{n\>mod\> \pi: n \in \mathbb{N}\}$ dense in $$?
- Least number of weights required to weigh integer weights
- induced representation, dihedral group
- Explanation of the binomial theorem and the associated Big O notation
- Peculiar numbers
- Why $ \sum_{k=0}^{\infty} q^k $ sum is $ \frac{1}{1-q}$ when $|q| < 1$
- Using $\epsilon$-$\delta$ definition to prove that $\lim_{x\to-2}\frac{x-1}{x+1}=3$.
- Is $\sqrt{2}$ contained in $\mathbb{Q}(\zeta_n)$?
- Trigonometric/polynomial equations and the algebraic nature of trig functions
- Inverse function of $y=\frac{\ln(x+1)}{\ln x}$
- Math and mental fatigue
- Is there an empty set in the complement of an empty set?