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.
It's bounded. Consider the columns you have formed. In each column, the section above the bar is less than the one below the bar, so the fractional part is guaranteed to be less than 1, so the whole sum is less than 2.