I was contemplating convergent sums, trying to think of very unusual or unorthodox sums that might be treatable recursively. Eventually, the following sum occurred to me:
$$ \xi = 1 + \frac{ \frac{1}{2} + \frac{ \frac{3}{4} + \frac{ \frac{7}{8} + \frac{\cdots}{\cdots} }{ \frac{8}{9} + \frac{\cdots}{\cdots}} }{ \frac{4}{5} + \frac{ \frac{9}{10} + \frac{\cdots}{\cdots}}{ \frac{10}{11} + \frac{\cdots}{\cdots}} } }{ \frac{2}{3} + \frac{ \frac{5}{6} + \frac{ \frac{11}{12} + \frac{\cdots}{\cdots}}{ \frac{12}{13} + \frac{\cdots}{\cdots}} }{ \frac{6}{7} + \frac{ \frac{13}{14} + \frac{\cdots}{\cdots}}{ \frac{14}{15} + \frac{\cdots}{\cdots}} } } \cdots $$
There is no especial motivation, but I endeavored to determine if I could show at least that it was bounded. I was not sure how to treat it however; it seemed like it might be possible to put it into some form of continued fraction, however I confess I am not familiar enough with continued fractions to see how this might be done (if it is possible). I calculated the first "convergents", or terms of the sequence $\left\{ \xi_n \right\}_{n \in \mathbb{N}^0}$ as I imagined it, if I were to write it as
\begin{aligned}[t] \xi_0 &= 1, \\ \xi_1 &= 1 + \frac{ \frac{1}{2} }{ \frac{2}{3} } = 1 + \frac{3}{4} = 1.75, \\ \xi_2 &= 1 + \frac{ \frac{1}{2} + \frac{ \frac{3}{4} }{ \frac{4}{5} } }{ \frac{2}{3} + \frac{ \frac{5}{6} }{ \frac{6}{7} } } = 1 + \frac{ \frac{1}{2} + \frac{15}{16} }{ \frac{2}{3} + \frac{35}{36} } = 1 + \frac{ \frac{23}{16} }{ \frac{59}{36} } \approx 1.87712, \\ \xi_3 &= 1 + \frac{ \frac{1}{2} + \frac{ \frac{3}{4} + \frac{ \frac{7}{8} }{ \frac{8}{9} } }{ \frac{4}{5} + \frac{ \frac{9}{10} }{ \frac{10}{11} } } }{ \frac{2}{3} + \frac{ \frac{5}{6} + \frac{ \frac{11}{12} }{ \frac{12}{13} } }{ \frac{6}{7} + \frac{ \frac{13}{14} }{ \frac{14}{15} } } } = \cdots \approx 1.8887. \end{aligned}
which reinforced my thought that it was potentially a convergent sum. If anyone has any advice as to what methods may be used to establish this, or any other insights, they would be a great assistance in sating my curiosity.
The perceived intractability of the aforementioned sum lead me to consider the "visually similar" but more simple sum
$$ 1 + \frac{ \frac{1}{2} }{ \frac{2}{3} } + \frac{ \frac{ \frac{3}{4} }{ \frac{4}{5} } }{ \frac{ \frac{5}{6} }{ \frac{6}{7} } } + \frac{ \frac{ \frac{\frac{7}{8}}{\frac{8}{9}} }{ \frac{\frac{9}{10}}{\frac{10}{11}} } }{ \frac{ \frac{\frac{11}{12}}{\frac{12}{13}} }{ \frac{\frac{13}{14}}{\frac{14}{15}} } } + \cdots $$
One can notice that the fraction ratios are of the form $\frac{ \frac{p-1}{p} }{ \frac{p}{p+1} } = \frac{p^2 - 1}{p^2} = 1 - \frac{1}{p^2} ,$ so that I could express this sum as
$$ 1 + \left( 1 - \frac{1}{2^2} \right) + \frac{ 1 - \frac{1}{4^2} }{ 1 - \frac{1}{6^2} } + \frac{ \left( 1 - \frac{1}{8^2} \right) \left( 1 - \frac{1}{14^2} \right) }{ \left( 1 - \frac{1}{10^2} \right) \left( 1 - \frac{1}{12^2} \right) } + \frac{ \left( 1 - \frac{1}{16^2} \right) \left( 1 - \frac{1}{22^2} \right) \left( 1 - \frac{1}{26^2} \right) \left( 1 - \frac{1}{28^2} \right) }{ \left( 1 - \frac{1}{18^2} \right) \left( 1 - \frac{1}{20^2} \right) \left( 1 - \frac{1}{24^2} \right) \left( 1 - \frac{1}{30^2} \right) } + \cdots $$
Unfortunately, I do not see a very obvious way to generalize this to a sum perhaps roughly of the form $$ \sum_{n \in 2 \mathbb{Z}^+} { \prod_{i = 0} ^{2^n} { \left( \text{ something... } \right) }} $$ which might be tackled more effectively.
If anyone can offer any assistance, or is familiar with similar problems, I am very interested in learning more regarding properties, especially convergence, of these types of "nested" sums.
