Overview
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 \geq a_n + 1 for all n , then \cfrac a_1 b_1 + \cfrac a_2 b_2 + \cfrac a_3 b_3 + \ddots converges absolutely to a number x satisfying x \leq 1 , meaning that the series x = \sum_n \left\ \frac A_n B_n - \frac A_ n - 1 B_ n - 1 \right\ , 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 \tfrac A_n B_n in this article, can be computed from the following recurrence relation: \begin align A_n &:= b_n A_ n - 1 + a_n A_ n - 2 \\ B_n &:= b_n B_ n - 1 + a_n B_ n - 2 , \qquad n \geq 2 \end align where A_0 = 0 , A_1 = a_1 , B_0 = 1 , and B_1 = b_1 . See this article for more detail.
th convergent as a series
First, we will prove the following claim via mathematical induction By dividing both sides of the claim by B_n \cdot B_ n - 1 , the equation becomes \dfrac A_n B_n - \dfrac A_ n - 1 B_ n - 1 = (-1)^ n - 1 \dfrac a_1 a_2 \cdots a_n B_ n - 1 \cdot B_n Thus, \begin align \sum_ i \, = \, 1 ^ n \dfrac A_i B_i - \dfrac A_ i - 1 B_ i - 1 &= \sum_ i \, = \, 1 ^ n (-1)^ i - 1 \dfrac a_1 a_2 \cdots a_i B_ i - 1 B_i \\ \dfrac A_n B_n - \dfrac A_0 B_0 &= \dfrac a_1 B_0 B_1 - \dfrac a_1 a_2 B_1 B_2 + \dfrac a_1 a_2 a_3 B_2 B_3 - \ldots + (-1)^ n - 1 \dfrac a_1 a_2 \cdots a_n B_ n - 1 B_n \\ \dfrac A_n B_n &= \dfrac a_1 B_0 B_1 - \dfrac a_1 a_2 B_1 B_2 + \dfrac a_1 a_2 a_3 B_2 B_3 - \ldots + (-1)^ n - 1 \dfrac a_1 a_2 \cdots a_n B_ n - 1 B_n \end align
Absolute value of th convergent as a series
Now suppose that b_n \geq a_n + 1 for all n . Using the recurrence relation of (B_n) , note that \left b_n B_ n - 1 \right = \left B_n - a_n B_ n - 2 \right \leq \left B_n \right + \left a_n B_ n - 2 \right . Thus, \begin align \left B_n \right &\geq \left b_n \right \left B_ n - 1 \right - \left a_n \right \left B_ n - 2 \right \\ &\geq \left( \left a_n \right + 1 \right) \left B_ n - 1 \right - \left a_n \right \left B_ n - 2 \right \\ \left B_n \right - \left B_ n - 1 \right &\geq \left a_n \right \left( \left B_ n - 1 \right - \left B_ n - 2 \right \right) \end align Since \left B_1 \right - \left B_0 \right = \left b_1 \right - 1 \geq \left a_1 \right by assumption, then by using mathematical induction, one can show that \left B_n \right - \left B_ n - 1 \right \geq \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, \dfrac 1 \left B_n B_ n - 1 \right \prod_ i \, = \, 1 ^ n \left a_i \right \leq \dfrac \left B_n \right - \left B_ n - 1 \right \left B_n B_ n - 1 \right = \dfrac 1 \left B_ n - 1 \right - \dfrac 1 \left B_n \right Since the right-hand side forms a telescoping series, it is easy to see that \dfrac \left a_1 \right \left B_0 B_1 \right + \dfrac \left a_1 a_2 \right \left B_1 B_2 \right + \ldots + \dfrac \left a_1 a_2 \cdots a_n \right \left B_ n - 1 B_n \right \leq \dfrac 1 \left B_0 \right - \dfrac 1 \left B_n \right = 1 - \dfrac 1 \left B_n \right 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 - \tfrac 1 \left B_n \right .
Thus, the sequence 1 - \tfrac 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 \tfrac A_n B_n \right after applying triangle inequality, which completes the proof.
See also
* Convergence problem