Does $\frac{1}{2^{\frac{3}{4^{\frac{5}{6\ldots}}}}}$ converge? 
Does $$\cfrac{1}{2^{\cfrac{3}{4^{\cfrac{5}{6\ldots}}}}}$$ converge?

I calculated $1200$ terms, and the answer seems to be stuck at $0.99797135454244139755827322630312164...$
Formally, the sequence would be defined as: $a_n= 1/2^{3/4^{...(2n-2)^{(2n-1)/(2n)}}}$
There seem to be 2 limits, one of $a_{2n}$ (goes just below 1) and $a_{2n+1}$ (goes to just above 1/8) I'm looking for $a_{2n}$.
I can't find a recurrence relation, so I have no idea how to go about proving this thing might converge. Is there any info on this out there? This seems to be a fairly natural question.
$a_1=\frac{1}{2}$
$a_2 = \frac{1}{2^{\frac{3}{4}}} =\frac{1}{1.68179} \approx 0.594604$
$a_3 = \frac{1}{2^{\frac{3}{4^{\frac{5}{6}}}}} = \frac{1}{2^{\frac{3}{3.17}}} = \frac{1}{1.92702} \approx 0.519450872$
 A: (I'm going to post my proof for convergence, if anyone finds a way to compute the limits, I'll delete my answer.)
Claim $1$: $\forall n\geq 1, 0\leq a_n \leq 1$.
These bounds are obvious, just by expanding the beginning of the expression.
Claim $2$: By induction, $\forall n \geq 1, a_{2n} \leq a_{2(n+1)}$.
It's clearly true for $n=1$. To see why this is true in general, take the inequality above and expand the terms:
$$a_{2n}= 2^{-...^{-(4n-1)\times(4n)^{-1}}}\leq a_{2n+2}= 2^{-...^{-(4n-1)(4n)^{-(4n+1)(4n+2)^{-(4n+3)\times(4n+4)^{-1}}}}}$$
We then apply successively logarithms and multiply by $(-1)$ a total of $2n$ (even) times (each $a_n$ has $n$ negative signs in the power tower), to get:
$$1 \leq (4n+1)(4n+2)^{-(4n+3)\times(4n+4)^{-1}}$$
$$\iff \frac{4n+3}{4n+4}\leq \frac{\ln{(4n+1)}}{\ln{(4n+2)}}$$
Using some real analysis, one can see that this is always true for $n\geq 1$ (more specifically, $4n\geq2.86...$).
Claim $3$: By induction, $\forall n \geq 2, a_{2n-1} \geq a_{2n+1}$.
Also clearly true for $n=2$. The proof is basically the same, and you get:
$$1 \leq (4n-1)(4n)^{-(4n+1)\times(4n+2)^{-1}}$$
$$\iff \frac{4n+1}{4n+2}\leq \frac{\ln{(4n-11)}}{\ln{(4n)}}$$
as now you have flipped the inequality $2n-1$ times. By the same argument, this is true for $n\geq 2$ (more specifically, $4n\geq4.86...$).
Conclusion: The subsequences are monotonous and bounded hence they converge.
