I am unable to evaluate the following limit, nor is my computer algebra system able to produce a float approximation. But the graphical plot it produces shows a pattern that is indicative of convergence to a specific constant value, and I realise that this is by no means a guarantee that indeed the limit exists, but it does allow us to eliminate a few of the commonly known types of divergence.(see figure 1 below):

$$\lim_{n \rightarrow \infty}\Biggl(\frac{1}{2n}\Biggl\lfloor\ln(2) \Bigl(\Bigl\lfloor\frac{3^{1/3}n}{3^{1/3}-1}\Bigr\rfloor+1\Bigr)\Biggr\rfloor\Biggr)$$

Also for the sake of a sort of comparative analysis, see the function plotted in figure 2, which likewise is very suggestive of having a limit as $n$ is made infinite.

Figure 1

Figure 2

relevant question previously asked

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    $\begingroup$ Hint. $x-1 \leq \lfloor x \rfloor \leq x$ holds for any $x \in \mathbb{R}$. $\endgroup$ – Sangchul Lee Oct 28 '18 at 11:53
  • $\begingroup$ Ok so is this for the application of the squeeze theorem as a means of confirmation of the existence of the limit? I did have something similar in mind for which I was hoping to employ somehow, that is a preliminary lemma in the proof of Hermite's identity: $\forall x \in \mathbb R,$ there is exactly one $k \in {\{1,2,3,...,n}\}$ such that: $$1-{\frac {k}{n}}\leq {\{x}\} \lt 1-\frac{k-1}{n} $$ $\endgroup$ – Adam L Oct 28 '18 at 12:44

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