Polynomial Continued Fractions

TL;DR

This paper explores polynomial continued fractions with higher degrees in numerator and denominator, extending classical work and analyzing limits that can be rational or irrational.

Contribution

It introduces new analyses for polynomial continued fractions with higher degrees, especially where numerator and denominator degrees are equal, expanding on Ramanujan's classical results.

Findings

Extended analysis of polynomial continued fractions with higher degrees

Identified conditions for rational and irrational limits

Generalized Ramanujan's work on continued fractions

Abstract

Continued fractions whose elements are polynomial sequences have been carefully studied mostly in the cases where the degree of the numerator polynomial is less than or equal to two and the degree of the denominator polynomial is less than or equal to one. Here we study cases of higher degree for both numerator and denominator polynomials, with particular attention given to cases in which the degrees are equal. We extend work of Ramanujan on continued fractions with rational limits and also consider cases where the limits are irrational.

Tables

Table 1. Table 1. Some infinite families of solutions to ( 2.2 ) for G n subscript 𝐺 𝑛 G_{n} of degree 3.

$G_n$	$a_n$	$b_n$
$\left(n^2-1\right)mn+1$	$2mn\left(n-1\right)-1$	$-2m\left(n^2-4n+3\right)+2$
$\left(2n^2+3n+1\right)mn+1$	$-2n^{2}m\left(m-2\right)+m^{2}n$	$-n^2\left(4m-2m^2\right)$
	$+m-1$	$-n\left(7m^2-12m\right)$
		$+6m^2-7m+2$
$\left(n^2+3n+2\right)mn+1$	$2n\left(n+1\right)m-1$	$-2n\left(n-2\right)m+2$
$\left(n^3+6n^2+11n+6\right)m$	$2m{n}^2+6mn$	$-2m{n}^2+2m+2$
$+1$	$+4m-1$
$m{n}^3+3m{n}^2$	$-m^2{n}^2-n\left(m^2+m\right)$	$m^2{n}^2-n\left(2m^2-m\right)$
$+n\left(2m-3\right)-2$	$+m-1$	$-4m+2$

Equations

$$\frac{4}{\pi }=1+\underset{n=1}{\overset{\infty }{K}}\frac{(2n-1{)}^2}{2},$$

$$e=2+\frac{1}{1+\underset{n=1}{\overset{\infty }{K}}\frac{n}{n+1}}.$$

$$\underset{n=1}{\overset{N}{K}}\frac{a_n}{b_n}:=\frac{a_1}{b_1+\frac{a_2}{b_2+\frac{a_3}{b_3+\dots +\frac{a_N}{b_N}}}}$$

$$=\frac{a_1}{b_1+}\frac{a_2}{b_2+}\frac{a_3}{b_3+}\dots \frac{a_N}{b_N}.$$

$$\{\begin{array}{ll}{b}_n\ge |a_n|& \\ b_n\ge |a_n|+1,& for{a}_{n+1}<0,\end{array}$$

$$\sum _{n\ge 0}{a}_n=a_0+\frac{a_1}{1+}\frac{-a_2}{a_1+a_2+}\frac{-a_1{a}_3}{a_2+a_3+}\frac{-a_2{a}_4}{a_3+a_4+}\frac{-a_3{a}_5}{a_4+a_5+\dots }.$$

$$C=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1{)}^2}$$

$$C=\frac{1}{1+}\frac{1^4}{8+}\frac{3^4}{16+}\frac{5^4}{24+\dots }.$$

$$\begin{gathered} \frac{a_1{b}_2}{a_2+b_1{b}_2+}\frac{-a_2{a}_3{b}_4}{a_3{b}_4+b_2(a_4+b_3{b}_4)+} \\ \frac{-a_4{a}_5{b}_2{b}_6}{a_5{b}_6+b_4(a_6+b_5{a}_6)+}\frac{-a_6{a}_7{b}_4{b}_8}{a_7{b}_8+b_6(a_8+b_7{a}_8)+\dots }. \end{gathered}$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{x+n}{x+n-1}=1.$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{n^{\alpha }+1}{n^{\alpha }}=1,\text{for}\alpha >0.$$

$$1-\frac{1}{1+\underset{n=1}{\overset{\infty }{K}}\frac{n^2}{n^2+2n}}=J_0(2),$$

$$2+\underset{n=1}{\overset{\infty }{K}}\frac{2n^2+n}{2n^2+5n+2}=\frac{1}{\sqrt{2}\csc (\sqrt{2})-1}.$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{n^{12}+2n^{11}+n^{10}+4n+5}{n^{12}+4n-4}=4.$$

$$\underset{n=2}{\overset{\infty }{K}}\frac{6n^7+6n^6+2n^5+3n+2}{6n^7-6n^6+2n^5+3n-5}=\frac{19}{7}.$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{3n^6-3n^4+n^3+6n^2+6n-1}{3n^6-9n^5+6n^4+n^3+6n^2-12n-2}=1.$$

$$G_n=a_n{G}_{n-2}+b_n{G}_{n-1}.$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{m{b}_n+m^2}{b_n}=m.$$

$$\underset{n\to \infty }{\lim }{G}_n/B_n=\underset{n\to \infty }{\lim }(-1{)}^{n+1}{m}^{n+1}/B_n=0⟹\underset{n=1}{\overset{\infty }{K}}\frac{a_n}{b_n}=-G_0/G_{-1}=m.$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{b_n+1}{b_n}=1.$$

$$\frac{x+a}{a+\underset{n=1}{\overset{\infty }{K}}\frac{(x+na{)}^2-a^2}{a}}=1.$$

$$\frac{x+a}{a+\underset{n=1}{\overset{\infty }{K}}\frac{(x+na{)}^2-a^2}{a}}=\frac{\frac{x}{a}+1}{1+\underset{n=1}{\overset{\infty }{K}}\frac{{\left(\frac{x}{a}+n\right)}^2-1}{1}}.$$

$$m=\underset{n=1}{\overset{\infty }{K}}\frac{(m+n{)}^2-1}{1}=\underset{n=1}{\overset{\infty }{K}}\frac{(m+n-1)(m+n+1)}{1}.$$

$$B_{2k+1}=(k+1)\prod _{i=2}^{2k+1}(m+i),\text{and}{B}_{2k}=(m+k+1)\prod _{i=2}^{2k}(m+i).$$

$$\underset{n=1}{\overset{\infty }{K}}\frac{mn{b}_n+m^{2}n(n+1)}{b_n}=m.$$

$$a_n{G}_{n-2}+b_n{G}_{n-1}-G_n=0$$

$$s_n{G}_{n-2}^{\prime }+t_n{G}_{n-1}^{\prime }-G_n^{\prime }=(-1{)}^{n-1}(a_n{G}_{n-2}+b_n{G}_{n-1}-G_n)=0.$$

$$\frac{x+a+1}{x+1}=\underset{n=1}{\overset{\infty }{K}}\frac{x+na}{x+(n-1)a-1}.$$

$$d=-f$$

$$e$$

$$g$$

Full text

have been carefully studied mostly in the cases where the degree of the numerator polynomial is less than or equal to two and the degree of the denominator polynomial is less than or equal to one. Here we study cases of higher degree for both numerator and denominator polynomials, with particular attention given to cases in which the degrees are equal. We extend work of Ramanujan on continued fractions with rational limits and also consider cases where the limits are irrational.

1. Introduction

A polynomial continued fraction is a continued fraction $K_{n=1}^{\infty}\displaystyle{a_{n}/b_{n}}$ where $a_{n}$ and $b_{n}$ are polynomials in $n$. Most well known continued fractions are of this type. For example the first continued fractions giving values for $\pi$ (due to Lord Brouncker, first published in  [10]) and $e$ ([3]) are of this type:

[TABLE]

[TABLE]

Here we use the standard notations

[TABLE]

[TABLE]

We write $\displaystyle{A_{N}/B_{N}}$ for the above finite continued fraction written as a rational function of the variables $a_{1},...,a_{N},b_{1},...,b_{N}$. By $K_{n=1}^{\infty}\displaystyle{a_{n}/b_{n}}$ we mean the limit of the sequence {$\displaystyle{A_{n}/B_{n}}$} as  $n$ goes to infinity, if the limit exists.

The first systematic treatment of this type of continued fraction seems to be in Perron [7] where degrees through two for $a_{n}$ and degree one for $b_{n}$  are studied. Lorentzen and Waadeland [6] also study these cases in detail and they evaluate all such continued fractions in terms of hypergeometric series. There is presently no such systematic treatment for cases of higher degree in the and examples in the literature are accordingly scarcer. Of particular interest are cases where the degree of $a_{n}$ is less than or equal to the degree of $b_{n}$. These cases are interesting from a number theoretic standpoint since the values of the continued fraction can then be approximated exceptionally well by rationals and irrationality measures may then be given. When the degrees are equal, the value may be rational or irrational and certainly the latter when the first differing coefficient is larger in $b_{n}$. (Here we count the first coefficient as the coefficient of the largest degree term.) Irrationality follows from the criterion given by Tietze, extending the famous Theorem of Legendre (see Perron [7], pp. 252-253) :

Tietze’s Criterion:

Let $\{\,a_{n}\}_{n=1}^{\infty}$ be a sequence of integers and $\{\,b_{n}\}_{n=1}^{\infty}$ be a sequence of positive integers, with $\,a_{n}\,\neq\,0$  for any $n$. If there exists a positive integer $N_{0}$ such that

[TABLE]

for all  $n\,\geq N_{0}$ then   $K_{n=1}^{\infty}\displaystyle{a_{n}/b_{n}}\,$ converges and it’s limit is irrational.

It would seem from the literature that finding cases of equal degrees or even close degrees is difficult. If one picks a typical continued fraction from published tables, the degree of the numerator tends to be twice that of the denominator. One easy way in which this can arise is when the continued fraction is equal to a series after using the Euler transformation:

[TABLE]

If one side of this equality converges, then so does the other as the $n$th approximants are equal. The Euler transformation is easily proved by induction.

In this formula, if the terms of the series are rational functions of the index of fixed degree, then in the continued fraction after simplification, one will get the degrees of the numerators to be at least twice that of the denominators. The continued fraction for $\pi$ given by (1.1) is an example of this phenomenon. Another example of this is the series definition of Catalan’s constant:

[TABLE]

which, by (1.4), transforms into the continued fraction given by

[TABLE]

Here the degree of the numerator is four times that of the denominator. This continued fraction appears to be new.

Taking contractions of continued fractions (see, for example, Jones and Thron [5], pp. 38–43) also leads to a relative increase in the degree of the numerator over that of the denominator. For example, forming the even part of the continued fraction will cause a continued fraction with equal degrees to be transformed into one with twice the degree in the numerator as the denominator.

The even part of the continued fraction $K_{n=1}^{\infty}\displaystyle{a_{n}/b_{n}}$ is equal to

[TABLE]

Other work on polynomial continued fractions was done by Ramanujan [1], chapter 12. He gave several cases of equal degree in which the sum is rational. For example, Ramanujan gave the following: If $x$ is not a negative integer then

[TABLE]

Despite the simplicity of this formula, Ramanujan did not give a proof: the first proof seems to be have been given by Berndt [1], Page 112.

In this paper we examine a large number of infinite classes of polynomial continued fractions in which the degrees are equal, or close. Our results follow from a theorem of Pincherle and a variant of the Euler transformation discussed above. We obtain generalizations of Ramanujan’s results in which the degrees are equal and the values rational as well as cases of equal degree with irrational limits. Many of our theorems give infinite families of continued fractions. While we concentrate on polynomial continued fractions, many of the results hold in more general cases. Here are some special cases of our general results (proofs are given throughout the paper) :

[TABLE]

[TABLE]

where $J_{0}(x)$ is the Bessel function of the first kind of order [math].

[TABLE]

(Notice that the irrationality criterion mentioned above means that the last two quantities on the right are irrational)

[TABLE]

[TABLE]

[TABLE]

2. Infinite Polynomial Continued Fractions with Rational Limits

In this section we derive some general results about the convergence of polynomial continued fractions in some infinite families and give some examples of how these results can be used to find the limit of such continued fractions. Many of the results in this paper are consequences of the following theorem of Pincherle  [9] :

Theorem 1**.**

*(Pincherle) Let $\{\,a_{n}\}_{n=1}^{\infty}$, $\{\,b_{n}\}_{n=1}^{\infty}$ and $\{\,G_{n}\}_{n=-1}^{\infty}$ be

sequences of real or complex numbers satisfying $a_{n}\neq 0$ for $n\geq 1$ and for all    $n\geq 1$,*

[TABLE]

Let $\{B_{n}\}_{n=1}^{\infty}$ denote the denominator convergents of the continued fraction $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{a_{n}}{b_{n}}}$.

If $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$ then $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{a_{n}}{b_{n}}}$ converges and its limit is $-G_{0}/G_{-1}$.

Proof.

See, for example, Lorentzen and Waadeland [6], page 202. ∎

For many sequences it may be difficult to decide whether the condition $\lim_{n\to\infty}$ $\displaystyle{G_{n}/B_{n}}=0$  is satisfied. Below are some easily proven properties governing the growth of the $B_{n}$’s which will be useful later.

(i)  Let $a_{n}$  and   $b_{n}$ be non-constant polynomials in $n$ such that $a_{n}\geq 1$, $b_{n}\geq 1$, for $n\geq 1$ and suppose $b_{n}$ is a polynomial of degree $k$. If the leading coefficient of $b_{n}$ is $D$, then given $\epsilon>0$, there exists a positive constant $C_{1}=C_{1}(\epsilon)$ such that $B_{n}\geq C_{1}(|D|/(1+\epsilon))^{n}(n!)^{k}$.

(ii)  If $a_{n}$  and   $b_{n}$ are positive numbers $\geq 1$, then there exists a positive constant $C_{3}$ such that $B_{n}\geq C_{3}\phi^{n}$ for $n\geq 1$, where $\phi$ is the golden ratio  $\displaystyle{(1+\sqrt{5})/2}$.

Corollary 1**.**

If $m$ is a positive integer and $b_{n}$ is any polynomial of degree $\geq 1$ such that $b_{n}\geq 1$ for $n\geq 1$, then

[TABLE]

Proof.

With $a_{n}=mb_{n}+m^{2}$, for  $n\geq\,1$, and $G_{n}=(-1)^{n+1}m^{n+1}$, for  $n\geq-1$, $\{\,a_{n}\}_{n=1}^{\infty}$, $\{\,b_{n}\}_{n=1}^{\infty}$ and $\{\,G_{n}\}_{n=-1}^{\infty}$ satisfy equation (2.1). By (i) above

[TABLE]

∎

A special case is where $m=1$, in which case $|G_{n}|=1$, for all $n$ and all that is necessary is that $\lim_{n\to\infty}B_{n}=\infty$. The following generalization of the result (1.5) of Ramanujan, for positive numbers greater than 1 follows easily:

If $\{\,b_{n}\}_{n=1}^{\infty}$ is any sequence of positive numbers with $b_{n}\geq 1$ for $n\geq 1$ then

[TABLE]

Letting $b_{n}=n^{\alpha}$, $\alpha>0$, gives (1.6) in the introduction.111Lorentzen and Waadeland give an exercise [6, page 234], question 15(d) which effectively involves a similar result in the case where $b_{n}$ belongs to a certain family of quadratic polynomials in $n$ over the complex numbers.

Entry 12 from the chapter on continued fractions in Ramanujan’s second notebook [1] , page 118, follows as a consequence of the above theorem:

Corollary 2**.**

If $x$ and $a$ are complex numbers, where $a\neq 0$ and $x\neq-ka$, where $k$ is a positive integer, then

[TABLE]

Proof.

Note that

[TABLE]

Replace $\displaystyle{x/a}$ by $m$ to simplify notation; the result will follow if it can be shown that

[TABLE]

With  $G_{-1}=1$, $G_{n}=(-1)^{n+1}\prod_{i=0}^{n}(m+i)$, for  $n\geq 0$ and $b_{n}=1$, $a_{n}=(m+n-1)(m+n+1)$ for  $n\geq\,1$, $\{\,a_{n}\}_{n=1}^{\infty}$, $\{\,b_{n}\}_{n=1}^{\infty}$ and $\{\,G_{n}\}_{n=-1}^{\infty}$ satisfy equation (2.1) so that the result will follow from Theorem 1 if it can be shown that $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$, in which case the continued fraction will converge to $-G_{0}/G_{-1}=m$. However, an easy induction shows that for $k\geq\,1$,

[TABLE]

Thus $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$, and the result follows. ∎

Corollary 3**.**

Let $m$ be a positive integer and let $b_{n}$ be any polynomial of degree $\geq 1$ such that $b_{n}\geq 1,$ for $n\geq 1$ and either degree $b_{n}>1$ or if degree  $b_{n}=1$ then its leading coefficient is $D>m$. Then

[TABLE]

Proof.

Letting $G_{n}=(-1)^{n+1}m^{n+1}(n+1)!$ for  $n\geq-1$ and $a_{n}=mnb_{n}+m^{2}n(n+1)$, for  $n\geq\,1$, one has that $\{\,a_{n}\}_{n=1}^{\infty}$, $\{\,b_{n}\}_{n=1}^{\infty}$ and $\{\,G_{n}\}_{n=-1}^{\infty}$ satisfy equation (2.1). By (i), $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0\Longrightarrow\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{a_{n}}{b_{n}}}=-G_{0}/G_{-1}=m$. ∎

Theorem 1 does not say directly how to find the value of all polynomial continued fractions $K_{n=1}^{\infty}\displaystyle{a_{n}/b_{n}}$ as it does not say how the sequence ${G_{n}}$ can be found or even if such a sequence can be found. However, Algorithm Hyper (see [8]) can be used to determine if a hypergeometric solution ${G_{n}}$ exists to equation (2.1) and, if such a solution exists, the algorithm will out-put $G_{n}$, enabling the limit of the continued fraction to be found, if ${G_{n}}$ satisfies $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$.

Even if for the particular polynomial sequences ${a_{n}}$ and ${b_{n}}$ it turns out that the sequence ${G_{n}}$ found does not satisfy $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$, then these three sequences ${a_{n}}$, ${b_{n}}$ and ${G_{n}}$ may be used to find the value of infinitely many other continued fractions when ${G_{n}}$ is a polynomial or rational function in $n$.

The following proposition shows how, given any one solution of (2.1), one can find the value of infinitely many other polynomial continued fractions in an easy way.

Proposition 1**.**

*Suppose that there exists complex sequences $\{\,G_{n}\,\}_{n=-1}^{\infty}$,

$\{\,a_{n}\,\}_{n=1}^{\infty}$ and $\{\,b_{n}\,\}_{n=1}^{\infty}$ satisfying*

[TABLE]

Let $f_{n}$ be any sequence, let $s_{n}=f_{n}G_{n-1}+a_{n}$ be such that $s_{n}\neq 0$, for $n\geq 1,$  and let   $t_{n}=f_{n}G_{n-2}-b_{n}$ . Let $A_{n}/B_{n}$ denote the convergents to $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{s_{n}}{t_{n}}}$. If $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$ then $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{s_{n}}{t_{n}}}$ converges and its limit is $G_{0}/G_{-1}$.

Proof.

Let $G^{\prime}_{n}=(-1)^{n+1}G_{n}$. Then

[TABLE]

Thus $\{\,G^{\prime}_{n}\,\}_{n=-1}^{\infty}$, $\{\,s_{n}\,\}_{n=1}^{\infty}$ and $\{\,t_{n}\,\}_{n=1}^{\infty}$ satisfy the conditions of theorem 1 so $K_{n=1}^{\infty}\displaystyle{s_{n}/t_{n}}$ converges and its limit is $-G^{\prime}_{0}/G^{\prime}_{-1}=G_{0}/G_{-1}.$

∎

Entry 9 from the chapter on continued fractions in Ramanujan’s second notebook [1] , pages 114-115, follows in the case $a$ is real and positive and $x$ is real as a consequence of the above proposition:

Corollary 4**.**

Let $a$ be a real positive number and let $x$ be a real number such that $x\neq-ka$ where k is a positive integer. Then

[TABLE]

Proof.

It is enough to prove this for $x-1>0$ since for $n$ sufficiently large $x+(n-1)a-1>0$ and then the result will hold for a tail of the continued fraction and then resulting finite continued fraction will collapse from the bottom up to give the result. Let $G_{n}=\displaystyle{(x+(n+1)a+1)/(x+1)}$. Put $f_{n}=x+1,\,\,a_{n}=-1$    and    $b_{n}=2$ so that $a_{n}G_{n-2}+b_{n}G_{n-1}-G_{n}=0$, $x+na=f_{n}G_{n-1}+a_{n}$ and $x+(n-1)a-1=f_{n}G_{n-2}-b_{n}$. Since $G_{n}$ is a degree 1 polynomial in $n$, $x+na,x+(n-1)a-1>0$   for $n\,\geq\,1$, it can easily be shown that $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$ and so by Proposition 1 the continued fraction converges to $G_{0}/G_{-1}=\displaystyle{(x+a+1)/(x+1)}$. ∎

Remarks: (1) In Proposition 1 any polynomial $G_{n}$ satisfying (2.2) can always be assumed to have positive leading coefficient (if necessary multiply (2.2) by $-1$.) If $f_{n}$ is then taken to be a polynomial of sufficiently high degree with leading positive coefficient then both $s_{n}$ and $t_{n}$ will be polynomials with positive leading coefficients so that there exists a positive integer $N_{0}$ so that for all $n\geq N_{0}$, $s_{n}$, $t_{n}>0$. If it happens that for some $m\geq N_{0}$ that both $B_{m}$ and $B_{m+1}$ are of the same sign then $B_{n}$ will go to $+\infty$ or $-\infty$ exponentially fast. In these circumstances $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0,$ since $G_{n}$ is only of polynomial growth.

In many of the following corollaries $f_{n}$ will be restricted so as to have $N_{0}$ small (typically in the range $1\leq N_{0}\leq 3$), but of course there are $f_{n}$ for which this is not the case but for which the results claimed in the corollaries hold.

(2) One approach is to take the polynomial $G_{n}$ as given and search for polynomials $a_{n}$ and $b_{n}$ satisfying equation (2.2). It can be assumed that degree($a_{n}$), degree($b_{n}$) $<$ degree($G_{n}$). This follows since if a solution exists with degree($a_{n}$) $\geq$ degree($G_{n}$) then the Euclidean algorithm can be used to write $a_{n}=p_{n}G_{n-1}+a_{n}^{\prime}$, $b_{n}=q_{n}G_{n-2}+b_{n}^{\prime}$, where $p_{n}$, $q_{n}$, $a_{n}^{\prime}$ and $b_{n}^{\prime}$ are polynomials in $n$. Substituting into (2.2) and comparing degrees gives that (2.2) holds with $a_{n}$ replaced with $a_{n}^{\prime}$ and $b_{n}$ replaced with $b_{n}^{\prime}$.

(3) In theory it is possible to find polynomials $G_{n}$ of arbitrarily high degree and polynomials $a_{n}$ and $b_{n}$ of lesser degree (with rational coefficients) satisfying (2.2), by using (2.2) to define equations expressing the coefficients of $a_{n}$ and $b_{n}$ in terms of those of $G_{n}$. If $G_{n}$ has degree $k$ and $a_{n}$ and $b_{n}$ both have degree $k-1$, then (2.2) is a polynomial identity of degree $2k-1$, giving $2k$ equations for the  $2k$ coefficients of $a_{n}$ and $b_{n}$.222Starting with $a_{n}$ and $b_{n}$, arbitrary polynomials of a certain degree, it is possible to look for solutions $G_{n}$ satisfying (2.2) with coefficients defined in terms of those of $a_{n}$ and $b_{n}$ using the the Hyper Algorithm (see [8]). However there is no certainty that the solutions (if they exist) will be polynomials or that they will have any particular desired degree.

In practice these equations and the requirement that the coefficients of $G_{n}$ be integers introduces conditions on the coefficients of $G_{n}$. For example, if there exists $G_{n}=an^{2}+bn+c$, $a_{n}=dn+e$, and $b_{n}=fn+g$, polynomials with integral coefficients, satisfying (2.2) , then

[TABLE]

giving restrictions on the allowable values of $a$, $b$ and $c$.

(Parts (ii) –(ix) of the following corollary correspond, respectively, to the solutions $\{a=b=m,c=1\}$, $\{a=1,b=4,c=4\}$, $\{a=m^{2},b=3m^{2}-2m,c=2m^{2}-2m+1\}$, $\{a=m^{2},b=m^{2}+2m,c=2m+1\}$, $\{a=m,b=3m,c=2m+1\}$, $\{a=m,b=m-2,c=-1\}$, $\{a=m,b=3m+2,c=2m+3\}$ and $\{a=4m,b=16m^{2}+8m+1,c=16m^{3}+16m^{2}+5m+1\}$ )

Proposition 1 is too general to easily calculate the limit of particular polynomial continued fractions. The following corollary enables these limits to be calculated explicitly in many particular cases.

Corollary 5**.**

Let $m$ be a positive integer, $k$ a positive integer greater than $m$ and $\{\,f_{n}\}_{n=1}^{\infty}$ a non-constant polynomial sequence such that $f_{n}\geq 1$, for $n\geq 1$. For each of continued fractions below assume that $f_{n}$ is such that no numerator partial quotient is equal to zero. (This holds automatically in cases (i) –(vi)).

(i)asdasdsadfdfdsf $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{(mn+k-m)f_{n}-1}{(mn+k-2m)f_{n}-2}}=\frac{k}{k-m}$.

(ii)asdasdsa $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{((n^{2}-n)m+1)f_{n}+nm-1}{((n^{2}-3n+2)m+1)f_{n}+mn-(2m+2)}}=1$.

(iii)asdasdsadfdfdsf $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{(n+1)^{2}f_{n}+4n+5}{n^{2}f_{n}+4n-4}}=4$.

(iv) $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{f_{n}(m^{2}n^{2}+n(m^{2}-2m)+1)+mn+m-2}{f_{n}(m^{2}n^{2}-n(m^{2}+2m)+2m+1)+mn-m-3}}=2m^{2}-2m+1$.

(v) $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{f_{n}(m^{2}n^{2}+n(2m-m^{2})+1)+mn}{f_{n}(m^{2}n^{2}-n(3m^{2}-2m)++2m^{2}+2m+1)+mn-2m-1}}~{}=~{}2m+~{}1$.

(vi)asdasdsadfdfdsf$\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{f_{n}(n(n+1)m+1)+mn+m-1}{f_{n}(n(n-1)m+1)+mn-m-2}}=2m+1$.

(vii) Let $\displaystyle{A_{n}/B_{n}}$ denote the convergents to the continued fraction below and suppose $\lim_{n\to\infty}\displaystyle{n^{2}/B_{n}}=0$ . Then

[TABLE]

(viii)asdasdsa $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{f_{n}(n(n+1)m+2n+1)-(m^{2}(n+1)+m+1)}{f_{n}(n(n-1)m+2n-1)-(m^{2}(n-1)+m+2)}}=2m+3$.

(ix) $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{\begin{matrix}-1-8\,m-32\,m^{2}-128\,m^{3}-64\,m^{2}\,n+\phantom{asddfdsfgsdfsdfsdfdgh}\\ \phantom{asddffdfdsfgsdfsdfdgh}\left(m+16\,m^{3}+\left(1+16\,m^{2}\right)\,n+4\,m\,n^{2}\right)\,f_{n}\end{matrix}}{\begin{matrix}&-2-8\,m+96\,m^{2}-128\,m^{3}-64\,m^{2}\,n+\phantom{asddgfdgfdgsdfdsfggdsfsdh}\\ &\phantom{asdggh}(-1+5\,m-16\,m^{2}+16\,m^{3}+\left(1-8\,m+16\,m^{2}\right)\,n+4\,m\,n^{2})\,f_{n}\end{matrix}}}$

[TABLE]

Proof.

In each case below an easy check shows that with the given choices for $\{\,G_{n}\,\}_{n=-1}^{\infty}$, $\{\,a_{n}\,\}_{n=1}^{\infty}$ and $\{\,b_{n}\,\}_{n=1}^{\infty}$ that equation (2.2) holds, that the continued fraction in question corresponds to the continued fraction $\displaystyle{\mathop{K}\limits_{n=1}^{\infty}}\displaystyle{\frac{s_{n}}{t_{n}}}$ of proposition (1), that if $\{A_{n}/B_{n}\}$ are the convergents to this continued fraction then $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$ and that (by fact or assumption) no $s_{n}=0$. Finally, the limit of the continued fraction is $G_{0}/G_{-1}$. The fact that some early partial quotients may be negative does affect any of the results - a tail of the continued fraction will have all terms positive so that $\lim_{n\to\infty}\displaystyle{G_{n}/B_{n}}=0$ will hold for the tail which will then converge and the continued fraction will then collapse from the bottom up to give the result.

Remark: In some cases the result holds if $f_{n}$ is a constant polynomial such that $f_{n}\geq 1$ for $n\geq 1$.

(i) Let $G_{n}=mn+k$, $a_{n}=-1$, and   $b_{n}=2$.

(ii) Let $G_{n}=n(n+1)m+1$, $a_{n}=mn-1$   and    $b_{n}=-mn+(2m+2)$.

(iii) Let $G_{n}=(n+2)^{2}$, $a_{n}=4n+5$ and $b_{n}=-4n+4$ .

(iv) Let $G_{n}=m^{2}n^{2}+n(3m^{2}-2m)+2m^{2}-2m+1$, a_{n}$$=mn+m-2 and $b_{n}=-mn+m+3$.

(v) Let $G_{n}=n(n+1)m^{2}+2m(n+1)+1$, $a_{n}=mn$ and $b_{n}=-mn+2m+1$.

(vi) Let $G_{n}=(n+2)(n+1)m+1$, $a_{n}=mn+m-1$ and $b_{n}=-mn+m+2$.

(vii) Let $G_{n}=mn^{2}+(m-2)n-1$, $a_{n}=-m^{2}n+m-1$ and $b_{n}=m^{2}n-2m^{2}-m+2$.

(viii) Let $G_{n}=(n+2)(n+1)m+2n+3$, $a_{n}=-(m^{2}n+m^{2}+m+1)$, and $b_{n}=m^{2}(n-1)+m+2$.

(ix) Let $G_{n}=1+5\,m+16\,m^{2}+16\,m^{3}+\left(1+8\,m+16\,m^{2}\right)\,n+4\,m\,n^{2}$, $a_{n}=-1-8\,m-32\,m^{2}-128\,m^{3}-64\,m^{2}\,n$ and $b_{n}=2+8\,m-96\,m^{2}+128\,m^{3}+64\,m^{2}\,n$.

∎

Examples:

1. Letting $m=5$ and $f_{n}=10n^{8}$ in (ii) above gives

[TABLE]

2. Also in (ii), letting $f_{n}=n^{8}$ and $m$ be an arbitrary positive integer,

[TABLE]

3. Letting $f_{n}=2n^{5}$ and $m=3$ in (vi) above gives (1.10) in the introduction. Similarly, letting $f_{n}=n^{10}$ gives (1.9) in the introduction.

4. In (vii) above a general class of examples may be obtained by choosing $m>1$ and $f_{n}>nm^{2}$ for $n\geq 1$. With the notation of the proposition it can easily be seen that $s_{n},t_{n}\geq 1$ for $n\geq 3$. If $f_{n}$ is such that $B_{2}$ and $B_{3}$ are negative, then $B_{n}$ will be negative for all $n\geq 2$ and by a similar argument to the reasoning behind condition (ii), it will follow that $\lim_{n\to\infty}\displaystyle{n^{2}/B_{n}}=0$ and the conditions of the corollary will be satisfied. For example, letting $f_{n}=16n$ and $m=3$ gives that

[TABLE]

All the examples in the last corollary were derived from solutions to equation (2.2) where $G_{n}$ had degree $2$. Table 1 below gives several families of solutions to equation (2.2), where $G_{n}$ is of degree 3 in $n$.

Considering the third and fourth row of entries in the Table 1, for example, there is the following corollary to Proposition 1:

Corollary 6**.**

Let $f_{n}$ be a polynomial in $n$ such that $f_{n}\geq 1$ for $n\geq 1$ and let $m$ be a positive integer.

(i) If $f(2)>2$ then

[TABLE]

(ii)

[TABLE]

Proof.

(i) In the light of the fact that $G_{n}$, $a_{n}$ and $b_{n}$ satisfy (2.2) simply note that the numerator of the continued fraction is $f_{n}G_{n-1}+a_{n}$ and that the denominator is $f_{n}G_{n-2}-b_{n}$. It is easily seen that $a_{n}\geq 1$, for all $n\geq 1$ and that $b_{n}\geq 1$, for all $n\geq 2$. It can also be shown that $B_{2}$ and $B_{3}$ are positive for all $m$ and $f_{n}$ satisfying the conditions of the corollary. In the light of what was said in an earlier remark this is sufficient to ensure the result.

(ii) The proof of this follows the same lines as that of (i) above. ∎

Taking $f_{n}$ to be $n^{3}$ and $m=3$ in part (i) gives (1.11) in the introduction.

One could continue to prove similar results by finding other solutions to equation (2.2) for degrees $2$ or $3$ or by going to higher degrees, but these corollaries should be sufficient to illustrate the principle at work.

3. Infinite Polynomial Continued Fractions with Irrational Limits

In this section we use a continued fraction-to-series transformation equivalent to Euler’s transformation to sum some polynomial continued fractions with irrational limits.

Theorem 2**.**

For $N$  $\geq 1$

[TABLE]

Thus, when $N\rightarrow\infty$, the continued fraction converges if and only if the series converges.

Proof.

See, for example, Chrystal [2], page 516, equation (14). ∎

Remark: The irrationality criterion mentioned in the introduction means that if $\{b_{n}\}_{i=0}^{\infty}$ is a sequence of integers, then $\sum_{n=0}^{\infty}\displaystyle{(-1)^{n}x^{n}/b_{0}b_{1}\cdots b_{n}}$ is not rational for $x=1/m$, $m$ being a non-zero integer, provided $|mb_{n}-1|\,\,\geq\,\,|mb_{n-1}|+1$, for all $n$ sufficiently large.

Corollary 7**.**

For all non-zero integers $m$ (and indeed for all non-zero real numbers $m$ )

[TABLE]

Remarks:

(1) Glaisher, [4] states continued fraction expansions essentially equivalent to this one and the one in the next corollary .

(2) The irrationality criterion gives that $\sin(\frac{1}{m})$ is irrational for $m$ either a non-zero integers or the square-root of a positive integer.

Proof.

(i) In Theorem 2 let $b_{n}=\displaystyle{(2n+2)(2n+3)m^{2}}$ and $x=1$.

∎

Corollary 8**.**

For all non-zero integers $m$ (and indeed for all non-zero real numbers $m$ )

[TABLE]

Note that the irrationality criterion gives that $\cos(\frac{1}{m})$ is irrational for $m$ either a non-zero integers or the square-root of a positive integer.

Proof.

(i) In Theorem 2 let $b_{n}=\displaystyle{(2n+1)(2n+2)m^{2}}$  and  $x=1$.

∎

Corollary 9**.**

For all positive integers $\nu$ and all non-zero integers $m$   (and indeed for all non-zero real numbers $m$ )

[TABLE]

where $J_{\nu}(x)$ is the bessel function of the first kind of order $\nu$.

The Tietze irrationality criterion shows that if $\nu$ is a non-negative integer and $m$ is a non-zero integer or the squareroot of a positive integer then $J_{\nu}(\pm\frac{1}{m})$ is irrational.

Proof.

(1) In Theorem 2 letting $b_{n}=\displaystyle{4(n+1)(\nu+n+1)m^{2}}$  and   $x=1$   gives

[TABLE]

from the power series expansion for $J_{\nu}(x)$. ∎

Taking $\nu$ to be [math] and $m=2$ gives (1.7) in the introduction. (1.8) in the introduction follows by letting $m=\frac{1}{\sqrt{2}}$ in Corollary  7.

Corollary 10**.**

For all non-zero integers $m$ (and indeed for all non-zero real numbers $m$)

[TABLE]

Proof.

In Theorem 2 let  $b_{0}=1$, $b_{n}=\displaystyle{(3n-2)(3n-1)(3n)}$,  for  $n\geq 1$ and $x=1/m^{3}$. Then

[TABLE]

Simplifying the continued fraction gives the left side and finally the right side equals $\left(\frac{1}{3}\exp{(\displaystyle{-1/m})}+\frac{2}{3}\exp{(\displaystyle{1/2m})}\cos{\left(\sqrt{3}/2m\right)}\right)^{-1}$. ∎

By the Tietze criterion the irrationality of this last function follows when $m$ is a non-zero integer or the real cube-root of a non-zero integer.