The following question is related to my question here about behavior of this sequence $a_{n}=(1-\frac12)^{ \left( \frac12-\frac13 \right)^{...^{ \left( \frac{1}{n}-\frac{1}{n+1} \right)}}}$. Now I define a new sequence generated from the linked sequence by:
$A_{n}= \lfloor (2n!((1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}) \rfloor$ where $\lfloor \cdot \rfloor$ is the floor function. My question here is: Is $\lim\limits_{n\to\infty}(A_{n+1}-A_{n})$ finite?
I have calculated some terms for $(A_{n+1}-A_{n})$ and it seems the difference does not grow fast enough to lead to divergence, which leads me to guess that the overall limit is finite.
The terms can be written much more simply:
$$A_n = \Bigl\lfloor 2n! \left( \frac{1}{2} \right)^{ {{\left( \frac{1}{6} \right)}^{\cdots} }^{ \left( \frac{n}{n+1} \right)}} \Bigr\rfloor$$
and with a little effort you can replace $\left( \frac{1}{2} \right)^x$ to $2^{-x}$, better revealing the structure of the term:
$$\Bigl\lfloor 2n!\ 2^{{{\left(-6 \right)}^\cdots}^{\pm \left( 1 + \frac{1}{n} \right)}} \Bigr\rfloor$$
Given the alternating nature of your original sequence, it is easy to prove (from what has already been said in you previous post) that $\frac{1}{2}\leq a_n<1$ for all $n$. Therefore, $$A_{n+1}-A_n=$$ $$\lfloor (2n+2)!\cdot a_{n+1}\rfloor-\lfloor (2n)!\cdot a_n\rfloor>$$ $$(2n+2)!\cdot a_{n+1}-(2n)!\cdot a_n-1>$$ $$\frac{(2n+2)!}{2}-(2n)!-1=$$ $$(2n)!\left(2n^2+3n\right)-1.$$ This last value clearly diverges, as does the original.