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Let $<a_0;a_1,a_2,\dots>$ be an infinite sequence of integers such that $$0<n\implies a_n>0.$$

For any natural $n$ we know there exists a convergent: one rational number $r_n$ such that it is equal to the simple continued fraction $$r_n=a_0+\frac{1}{a_1 +\frac{1}{a_2 +\frac{1}{\ddots_{a_n+0}}}}$$ $r_n$ is the number given by the finite continued fraction of the original sequence truncated at $n$, $<a_0;a_1,a_2,\dotsm,a_n>$.

The question is: given any sequence of positive integers, does the simple continued fraction always converge to a real number?

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In fact, an even better theorem is available, the Seidel-Stern theorem. Let $\langle a_i\rangle$ be any sequence of positive real numbers. Then the continued fraction $[a_0; a_1, a_2, \ldots]$ converges if and only if the series $$\sum_{i=0}^\infty a_i$$ diverges!

By restricting the $a_i$ to be positive integers, we get the answer to your question: all such continued fractions converge.

(This is theorem 10 (p.10–12) in Continued Fractions, A. Ya. Khinchin, Phoenix Science Press 1964, or theorem 9.31 (p.185) in Neverending Fractions: An Introduction to Continued Fractions by Borwein et al., Cambridge University Press 2014.)

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I just found the Continued Fraction Limit (See Theorem 5.2).

The answer is: yes!

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