Can anyone help me to prove that for large $n$
whether $$\dfrac{n\choose \big[\frac{n}{2\log_2n}\big]-1}{(\sqrt{2})^n}\geq(1+\beta)^n$$ for some $\beta>0$ where and $\big[x\big]$ denotes the integer part of $x$, for example, letting $x=5.3222$, then $\big[x\big]=5$.
Which is equivalent to ask whether $ n\choose \big[\frac{n}{2\log_2n}\big]-1 $$>(\sqrt{2}+\beta)^n$ for some $\beta>0$.
Thanks very much for your help!!
This is not true however; $\sqrt[3m]{2m} = \left(e^{\ln (2m) }\right)^{\frac{1}{3m}}$ $=e^{\frac{\ln(2m)}{3m}} \rightarrow 1$ for $m \rightarrow \infty$ which is what is happening here for $m =\log n$. This gives $$\lim_{n \rightarrow \infty} \left(\frac{\sqrt[3\log n]{2\log n}}{\sqrt{2}}\right)^n < \left(\frac{1.1}{\sqrt{2}}\right)^n =0.$$
This still holds true even for the new question:
$${n \choose k} =\frac{n!}{(n-k)!k!}$$ $$\le \frac{n^k}{k!} \le 2ke^k\left(\frac{n}{k}\right)^k,$$ the latter a consequence of Stirling's Inequality.
For $k\in \theta\left(\frac{n}{\log n}\right)$ then: $${n \choose k} \le$$ $$ke^k\left(\frac{n}{k}\right)^k = ke^k(\theta(\log n))^{\theta\left(\frac{n}{\log n}\right)}$$ $$\le ke^k\left(2^{\theta(\log_2\log_2n)}\right)^{\theta\left(\frac{n}{2\log n}\right)}=2^{o(n)}.$$
Thus, $${n \choose k} = 2^{o(n)} \text{ for } k = \frac{n}{2\log n} \pm 1 $$ and dividing both sides of this by $2^{\frac{n}{2}}$ gives
$$2^{-\frac{n}{2}}{n \choose k} =2^{o(n)}2^{-\frac{n}{2}} < 1$$ for such $k$, which shows that the inequality in this question cannot be.
Something similar can be concluded for any $k\in o(n)$ infact.