Consider the sequence: $J=\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\cdots$
For which values L does there exist a subsequence that this converges to.
My attempt(s):
Notice that each finite sequence $b^m_n=\frac{k}{10^m}$ is a finite subsequence of $J$, where $k$ ranges from $1$ to $(10^m-1)$, now notice that $\frac{b_1\cdots b_m}{10^m}=0.b_1 \cdots b_m$
Now for the proof that all real numbers in the interval [0,1] have a convergent subsequence, let $\psi=0.\psi_1\psi_2\psi_3\cdots$ be a real number $\in (0,1)$ define a sequence $b_n=\frac{\psi_1\cdots\psi_n}{10^n}$, notice that by our previous observation the resulting sequence $b_n$ is monotonically increasing, as each term is the same decimal expansion of $\psi$ with more digits appended.
$b_n$ is a subsequence of $a_n$ by the facts that the denominators of the $b_n$ increase as $n$ increase, meaning the terms come after eachother in the sequence $a_n$, and that no two terms share the same denominator.
Now, the proof that $b_n$ converges to
$|a_n-\psi|=|0.\psi_1\cdots\psi_n-0.\psi_1\psi_2\psi_3\cdots|=|0.0\cdots 0\psi_{n+1}\psi_{n+2}\cdots|<\frac{1}{10^n}$
$|a_n-\psi|$ being bound by $\frac{1}{10^n}$ means that for any $\epsilon$ chose $N$ to be any $N$ such that $\epsilon<\frac{1}{10^N}$, all of this together:
$\forall\epsilon\exists N:(|b_n-\psi|<\epsilon \ \forall n>N)$ meaning we have found a convergent subsequence for an arbitrary real number between 0 and 1.
Please let me know if this proof checks out, I have a proof that I am sure works for all rational numbers, but am not sure about the soundness of this one.
Yes your proof works. the sequence you found is clearly a subsequence of $J$ which converges to $L$ by construction, since each element gets closer to $L$ by the value of the next decimal. Below I've provided another proof that is somewhat less explicit in its construction. The core idea is that chunks of $J$ define finer and finer partitions of the interval $[0,1]$, and $L$ must fall into some part in each partition, so you can squeeze your way to $L$.
Fix $L \in [0,1]$ and consider sets $A_1, A_2, \dots$ made from the chunks of the sequence $a_n$ which have the same denominator, i.e. $A_2 = \{\frac{1}{2}\}$, $A_3 = \{\frac{1}{3}, \frac{2}{3}\}$, etc... (we start at $A_2$ just so the index of the set lines up with the denominator.) Now note that each of these sets defines a partition of $[0,1]$, e.g $A_2$ defines the partition into two parts given by $[0,\frac{1}{2})$ and $[\frac{1}{2}, 1]$, $A_3$ defines the partition $[0,\frac{1}{3}), [\frac{1}{3}, \frac{2}{3}), [\frac{2}{3},1]$, etc... The number $L$ falls into one part of the partition for each partition $A_i$. Let $b_i$ be the upper bound of the part that $L$ falls into in partition $A_i$, e.g if $L = \sqrt{2}/2 \approx 0.707$, then $b_1 = 1$, $b_2 = 1$, $b_3 = \frac{3}{4}$, $b_4 = \frac{4}{5}$, etc...
Remove all instances of the number $1$ in the sequence $b_i$ and reindex to start at the first fraction. The resulting sequence is a subsequence of $J$ (since each successive element comes from a later partition) and converges to $L$ since the partitions get smaller each time (unless $L = 1$ but that case is trivial).