For the continued fraction \(\xi = a_0+\frac{1\,|}{|\, a_1} + \frac{1\,|}{|\, a_2} + \dots\) the author proves the following theorems:
1. Suppose that \(\xi\) is not periodic, that (1) \(a_n=a_{n+1}=\dots = a_{n+\lambda(n)-1}\) for \(n=n_i\), \(i=1,2,\ldots\) and \(\lambda(n)\) is greater than a certain function of \(q(n)\), the denominator of the \(n\)-th approximant of \(\xi\), and that \((\log \lambda_i)(\log n_i)^{\frac12}/n_i \to\infty\) as \(i\to\infty\), where \(\lambda_i = \lambda(n_i)\). Then \(\xi\) is transcendental.
2. Suppose that \(\xi\) is not periodic, that \(a_n\leq A\) for all \(n\), that (1) holds for all \(n=n_i\), \(i = 1, 2, \ldots\), and that \(\limsup \lambda_i/n_i > B =B(A)\), where \(\lambda_i= \lambda(n_i)\) and \(B\) is defined by \(\left(\tfrac12(1+\sqrt 5)\right)^{\frac12(B+1)}=\tfrac12 \left(A + \sqrt{A^2 + 4)}\right)\). Then \(\xi\) is transcendental.
3. Consider a quasi-periodic continued fraction
\[ \xi = \left[a_0,a_1,\ldots, a_{n_s-1}, \overset{\lambda_i}{\overline{a_{n_i},\ldots, a_{n_i+k_i-1}}}, \overset{\lambda_i}{\overline{a_{n_i},\ldots, a_{n_i+k_i-1}}}, \ldots\right], \]
where the notation implies that \(n_i = n_{i-1} + \lambda_{i-1} k_{i-1}\) and the \(\lambda\)’s indicate the number of times a block of partial quotients is repeated. Suppose that the \(a_i\) and \(k_i\) are bounded. If \(\lambda_i/\lambda_{i-1} > 2 + \delta\) for all \(i\), where \(\delta > 0\) is independent of \(i\), then \(\xi\) is transcendental.
Slightly more general theorems are also proved.