Śleszyński–Pringsheim theorem

Short description: Criterion for convergence of continued fractions

In mathematics, the Śleszyński–Pringsheim theorem is a statement about convergence of certain continued fractions. It was discovered by Ivan Śleszyński and Alfred Pringsheim in the late 19th century.

It states that if $n$ is a positive integer and $(a_n)$, $(b_n)$ are sequence real numbers such that $|b_n|\ge |a_n|+1$ for all $n$, then

$$\frac{a_1}{b_1+\frac{a_2}{b_2+\frac{a_3}{b_3+\ddots }}}$$

converges absolutely to a number $x$ satisfying $|x|\le 1$, meaning that the series

$$x=\sum _n\{\frac{A_n}{B_n}-\frac{A_{n-1}}{B_{n-1}}\},$$

where $A_n/B_n$ are the convergents of the continued fraction, converges absolutely.

Proof

Recall that the $n$th convergents of $x$, which will be denoted by $\frac{A_n}{B_n}$ in this article, can be computed from the following recurrence relation:

$$\begin{array}{rl}{A}_n& :=b_n{A}_{n-1}+a_n{A}_{n-2}\\ B_n& :=b_n{B}_{n-1}+a_n{B}_{n-2},n\ge 2\end{array}$$

where $A_0=0$, $A_1=a_1$, $B_0=1$, and $B_1=b_1$. See this article for more detail.

nth convergent as a series

First, we will prove the following claim via mathematical induction

Claim — For all positive integers $n$, then $$A_n{B}_{n-1}-A_{n-1}{B}_n=(-1{)}^{n-1}{a}_1{a}_2\cdots a_n$$

Proof

The case $n=1$ is trivial.

Suppose the claim is true for $n=k$. By using the two recurrence relation above, it follows that

$$\begin{array}{rl}{A}_{k+1}{B}_k-A_k{B}_{k+1}& =(b_{k+1}{A}_k+a_{k+1}{A}_{k-1})B_k-A_k(b_{k+1}{B}_k+a_{k+1}{B}_{k-1})\\ & =b_{k+1}{A}_k{B}_k+a_{k+1}{A}_{k-1}{B}_k-b_{k+1}{A}_k{B}_k-a_{k+1}{A}_k{B}_{k-1}\\ & =a_{k+1}{A}_{k-1}{B}_k-a_{k+1}{A}_k{B}_{k-1}\\ & =-a_{k+1}(A_k{B}_{k-1}-A_{k-1}{B}_k)\\ & =-a_{k+1}\cdot (-1{)}^{k-1}{a}_1{a}_2\cdots a_k\\ & =(-1{)}^k{a}_1{a}_2\cdots a_k{a}_{k+1}\end{array}$$

which finishes the induction step.

By dividing both sides of the claim by $B_n\cdot B_{n-1}$, the equation becomes

$$\frac{A_n}{B_n}-\frac{A_{n-1}}{B_{n-1}}=(-1{)}^{n-1}\frac{a_1{a}_2\cdots a_n}{B_{n-1}\cdot B_n}$$

Thus,

$$\begin{array}{rl}\sum _{i=1}^n\frac{A_i}{B_i}-\frac{A_{i-1}}{B_{i-1}}& =\sum _{i=1}^n(-1{)}^{i-1}\frac{a_1{a}_2\cdots a_i}{B_{i-1}{B}_i}\\ \frac{A_n}{B_n}-\frac{A_0}{B_0}& =\frac{a_1}{B_0{B}_1}-\frac{a_1{a}_2}{B_1{B}_2}+\frac{a_1{a}_2{a}_3}{B_2{B}_3}-\dots +(-1{)}^{n-1}\frac{a_1{a}_2\cdots a_n}{B_{n-1}{B}_n}\\ \frac{A_n}{B_n}& =\frac{a_1}{B_0{B}_1}-\frac{a_1{a}_2}{B_1{B}_2}+\frac{a_1{a}_2{a}_3}{B_2{B}_3}-\dots +(-1{)}^{n-1}\frac{a_1{a}_2\cdots a_n}{B_{n-1}{B}_n}\end{array}$$

Absolute value of nth convergent as a series

Now suppose that $|b_n|\ge |a_n|+1$ for all $n$. Using the recurrence relation of $(B_n)$, note that

$$\left|b_n{B}_{n-1}\right|=|B_n-a_n{B}_{n-2}|\le \left|B_n\right|+\left|a_n{B}_{n-2}\right|.$$

Thus,

$$\begin{array}{rl}\left|B_n\right|& \ge \left|b_n\right|\left|B_{n-1}\right|-\left|a_n\right|\left|B_{n-2}\right|\\ & \ge (\left|a_n\right|+1)\left|B_{n-1}\right|-\left|a_n\right|\left|B_{n-2}\right|\\ \left|B_n\right|-\left|B_{n-1}\right|& \ge \left|a_n\right|(\left|B_{n-1}\right|-\left|B_{n-2}\right|)\end{array}$$

Since $\left|B_1\right|-\left|B_0\right|=\left|b_1\right|-1\ge \left|a_1\right|$ by assumption, then by using mathematical induction, one can show that

$$\left|B_n\right|-\left|B_{n-1}\right|\ge \prod _{i=1}^n\left|a_i\right|.$$

Consequently, the sequence of $\left|B_n\right|$ is monotone nondecreasing and bounded from below by $\left|B_0\right|=1$. Moreover,

$$\frac{1}{\left|B_n{B}_{n-1}\right|}\prod _{i=1}^n\left|a_i\right|\le \frac{\left|B_n\right|-\left|B_{n-1}\right|}{\left|B_n{B}_{n-1}\right|}=\frac{1}{\left|B_{n-1}\right|}-\frac{1}{\left|B_n\right|}$$

Since the right-hand side forms a telescoping series, it is easy to see that

$$\frac{\left|a_1\right|}{\left|B_0{B}_1\right|}+\frac{\left|a_1{a}_2\right|}{\left|B_1{B}_2\right|}+\dots +\frac{|a_1{a}_2\cdots a_n|}{\left|B_{n-1}{B}_n\right|}\le \frac{1}{\left|B_0\right|}-\frac{1}{\left|B_n\right|}=1-\frac{1}{\left|B_n\right|}<1$$

for all values of $n$. Furthermore, the nondecreasing property of the sequence $\left|B_n\right|$ also implies the nondecreasing property of the sequence $1-\frac{1}{\left|B_n\right|}$. Thus, the sequence $1-\frac{1}{\left|B_n\right|}$ must converge, by the monotone convergence theorem.

Note that the left-hand side is the upper bound of series representation of $\left|\frac{A_n}{B_n}\right|$ after applying triangle inequality, which completes the proof.

See also

• Convergence problem