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

This case seems to be a border between converging and diverging as far as I can tell.

Related to http://math.stackexchange.com/questions/742510/limit-of-displaystyle-sum-k-0-lfloor-n-2-rfloor-2-2nk-binomn2k which has a great answer. I can't work out how to apply that answer to this case unfortunately.

(Also asked at http://math.stackexchange.com/questions/746895/what-is-lim-n-to-infty-displaystyle-sum-k-0-lfloor-n-2-rfloor-binom ).

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