Let $\{a_n\}_{n=0}^{\infty}$ be a sequence, such that $\forall n\in\mathbb{N}: 0 < a_{n+1} < a_n + \frac{1}{2^n}$. Prove that $\{a_n\}$ converges.
Now I've been investigating a couple of approaches:
My first idea was to search for monotonicity, since the sequence can easily be shown to be bounded based off of the fact that the given inequalities imply $\forall n\in\mathbb{N} ~\forall p\in\mathbb{N}_+$: $$0 < a_{n+p} < a_n + \frac{1 - \frac{1}{2^p}}{2^{n-1}}$$ and thus plugging in $n = 0$ in particular we can obtain an upper bound as well. Turned out, the sequence could very well be non-monotone, so I discarded that approach.
This generalization of the inequality brought me to my next idea: finding a way to prove that the sequence is Cauchy, but the problem is that reordering the inequality to give the difference between two indices $n$ and $n + p$ like that: $$-a_n < a_{n+p} - a_{n} < \frac{1 - \frac{1}{2^p}}{2^{n-1}}$$ unfortunately leaves this rather wild $-a_n$ in the LHS and thus makes bounding the absolute value of that difference a bit tricky... so I found no way around it.
My last strike of insight came in the form of a more theoretical approach, since both ideas above had an issue with the sequence being able to arbitrarily 'drop down' with $a_{n+1}$ anywhere between 0 and $a_n$. So I thought: according to the Bolzano-Weierstraß theorem, since the sequence is bounded, it must have at least one convergent subsequence with, say, a limit $L$. Let's fix an $\epsilon > 0$ and pick an entry $a_k\in U_\epsilon(L)$. Now I wanted to prove, using the general inequality above, that for $p$ large enough, if $a_{k + p}$ happens to drop down too low from $L$, the sequence will never be able to come back to $L$ thus contradicting the initial assumption. Furthermore, we already know that $a_{k + p}$ can't go up indefinitely, since we have a nice upper boundary and voilà - an only cluster point is proven. The problem is, I really don't know how to approach the rigorous proof based on this intuition.
Any help/ideas would be greatly appreciated! Thanks in advance!