Irrationality and transcendence of alternating series via continued fractions (Q2097090)

Let \(\{ A_n\}_{n=0}^\infty\) be the sequence of integers greater than one. Then the author proves that \begin{itemize} \item[1.] For any sequence \(\{ x_n\}_{n=0}^\infty\) of positive real numbers we have the equivalence between an alternating series and a general continued fraction \[ \alpha =\sum_{n=0}^\infty \frac {(-1)^n}{\prod_{n=0}^\infty A_n}= \frac {x_0}{A_0x_0+\frac{A_0x_0x_1}{(A_1-1)x_1+\frac{A_1x_1x_2}{(A_2-1)x_2+\dots}}}. \] \item[2.] If \(A_{n+1}>A_n\) for all \(n\geq 0\) then \(\alpha\) is transcendental. \item[3.] If the continued fraction is simple for some \(\{ x_n\}_{n=0}^\infty\) then \(\alpha\) is transcendental number with irrationality measure exponent \(2.5\). \end{itemize} The author shows many consequences and examples of this result for a well-known constants and their alternatives including also the conjecture if the primorial constant \(\sum_{n=0}^\infty \frac {(-1)^n}{\prod_{n=0}^\infty p_n}\) is transcendental, where \(\{ p_n\}_{n=0}^\infty\) is the increasing sequence of all primes. For the entire collection see [Zbl 1478.11004].