Śleszyński–Pringsheim theorem

nth convergent as a series

First, we will prove the following claim via mathematical induction

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

$${\displaystyle {\dfrac {A_{n}}{B_{n}}}-{\dfrac {A_{n-1}}{B_{n-1}}}=(-1)^{n-1}{\dfrac {a_{1}a_{2}\cdots a_{n}}{B_{n-1}\cdot B_{n}}}}$$

Thus,

$${\displaystyle {\begin{aligned}\sum _{i\,=\,1}^{n}{\dfrac {A_{i}}{B_{i}}}-{\dfrac {A_{i-1}}{B_{i-1}}}&=\sum _{i\,=\,1}^{n}(-1)^{i-1}{\dfrac {a_{1}a_{2}\cdots a_{i}}{B_{i-1}B_{i}}}\\{\dfrac {A_{n}}{B_{n}}}-{\dfrac {A_{0}}{B_{0}}}&={\dfrac {a_{1}}{B_{0}B_{1}}}-{\dfrac {a_{1}a_{2}}{B_{1}B_{2}}}+{\dfrac {a_{1}a_{2}a_{3}}{B_{2}B_{3}}}-\ldots +(-1)^{n-1}{\dfrac {a_{1}a_{2}\cdots a_{n}}{B_{n-1}B_{n}}}\\{\dfrac {A_{n}}{B_{n}}}&={\dfrac {a_{1}}{B_{0}B_{1}}}-{\dfrac {a_{1}a_{2}}{B_{1}B_{2}}}+{\dfrac {a_{1}a_{2}a_{3}}{B_{2}B_{3}}}-\ldots +(-1)^{n-1}{\dfrac {a_{1}a_{2}\cdots a_{n}}{B_{n-1}B_{n}}}\end{aligned}}}$$

Absolute value of nth convergent as a series

Now suppose that ${\displaystyle |b_{n}|\geq |a_{n}|+1}$ for all ${\displaystyle n}$. Using the recurrence relation of ${\displaystyle (B_{n})}$, note that

$${\displaystyle \left|b_{n}B_{n-1}\right|=\left|B_{n}-a_{n}B_{n-2}\right|\leq \left|B_{n}\right|+\left|a_{n}B_{n-2}\right|.}$$

Thus,

$${\displaystyle {\begin{aligned}\left|B_{n}\right|&\geq \left|b_{n}\right|\left|B_{n-1}\right|-\left|a_{n}\right|\left|B_{n-2}\right|\\&\geq \left(\left|a_{n}\right|+1\right)\left|B_{n-1}\right|-\left|a_{n}\right|\left|B_{n-2}\right|\\\left|B_{n}\right|-\left|B_{n-1}\right|&\geq \left|a_{n}\right|\left(\left|B_{n-1}\right|-\left|B_{n-2}\right|\right)\end{aligned}}}$$

Since ${\displaystyle \left|B_{1}\right|-\left|B_{0}\right|=\left|b_{1}\right|-1\geq \left|a_{1}\right|}$ by assumption, then by using mathematical induction, one can show that

$${\displaystyle \left|B_{n}\right|-\left|B_{n-1}\right|\geq \prod _{i\,=\,1}^{n}\left|a_{i}\right|.}$$

Consequently, the sequence of ${\displaystyle \left|B_{n}\right|}$ is monotone nondecreasing and bounded from below by ${\displaystyle \left|B_{0}\right|=1}$. Moreover,

$${\displaystyle {\dfrac {1}{\left|B_{n}B_{n-1}\right|}}\prod _{i\,=\,1}^{n}\left|a_{i}\right|\leq {\dfrac {\left|B_{n}\right|-\left|B_{n-1}\right|}{\left|B_{n}B_{n-1}\right|}}={\dfrac {1}{\left|B_{n-1}\right|}}-{\dfrac {1}{\left|B_{n}\right|}}}$$

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

$${\displaystyle {\dfrac {\left|a_{1}\right|}{\left|B_{0}B_{1}\right|}}+{\dfrac {\left|a_{1}a_{2}\right|}{\left|B_{1}B_{2}\right|}}+\ldots +{\dfrac {\left|a_{1}a_{2}\cdots a_{n}\right|}{\left|B_{n-1}B_{n}\right|}}\leq {\dfrac {1}{\left|B_{0}\right|}}-{\dfrac {1}{\left|B_{n}\right|}}=1-{\dfrac {1}{\left|B_{n}\right|}}<1}$$

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

Note that the left-hand side is the upper bound of series representation of ${\displaystyle \left|{\tfrac {A_{n}}{B_{n}}}\right|}$ after applying triangle inequality, which completes the proof.