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

Question

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

Answer 1

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 $\pi/4 < 1$.

Answer 2

$$ \sum_{k=0}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large\frac{2n}{\log_2(n)}} =\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\left[4^{-k}\binom{2k}{k}^{\large\frac2{\log_2(n)}}\right]^{\large n} $$ and when $n\gt4$, $\frac2{\log_2(n)}\lt1$ and the term in the brackets decays exponentially since $$ \binom{2k}{k}\le\frac{4^k}{\sqrt{\pi k}} $$ This sum converges to $1$ much faster than the one in the linked question.