What is $\lim_{n\to\infty} \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$ What is $$\lim_{n\to\infty} \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor}  \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$$

I have posted a longer explanation at mathoverflow.  In short, the sum diverges, because the $k = \lfloor n/4\rfloor$ summand increases without bound.  For this summand, the log of $\binom{n}{2k}$ is $n \log 2 + O(\log n)$, while the log of the other term is $-n \log 2 - \frac{n \log 2}{\log n}\log (\pi/4) + O(\log n)$.  The main contributions cancel, and the subleading contribution increases without bound because $\pi/4 < 1$.
