What is $\lim_{n\to\infty} \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$ What is $$\lim_{n\to\infty} \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor}  \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$$
 A: First, we note that Stirling's series yields $4^{-k} \binom{2k}{k} = \frac{1}{\sqrt{\pi k}}e^{-1/8k + O(k^{-3})}$.
Let $c = k/n$.  Then


*

*$\log n!$ expands as $n \log n - n + O(\log n)$.

*$\log (2k)!$ expands as $2cn \log 2cn - 2cn + O(\log n)$.

*$\log(n-2k)!$ expands as $n(1-2c)(\log n + \log(1-2c)) - n(1-2c) + O(\log n)$.

*$\log \binom{n}{2k}$ expands as $-2cn\log 2c - n(1-2c) \log (1-2c) + O(\log n)$.  This is $n$ times the following function of $2c$: 

*$\frac{2n \log 2}{\log n} \log\left(\frac{1}{\sqrt{\pi k}}e^{-1/8k + O(k^{-3})}\negthinspace\right)$ expands as $\frac{n \log 2}{\log n}(-\log n -\log c -\log \pi) - O(\frac{1}{c\log n})$


We need to consider the asymptotics of the sum of the 4th and 5th terms.  The 5th term is dominated by $-n \log 2$, unless $c\sim 1/n$ (i.e., $k\sim 1$ - a short calculation shows we don't need to worry about this case).  From calculus (or brief examination of Pascal's triangle) the fourth term achieves its maximum value of $n \log 2$ at $c=1/4$.  The sum is then dominated by the term $\frac{n \log 2}{\log n}(-\log c -\log \pi)$.  Since $\pi c = \pi/4 < 1$, we have $(-\log c -\log \pi)>0$, so the sum $\frac{n \log 2}{\log n}(-\log c -\log \pi) + O(\log n)$ increases without bound.
In conclusion, the sum diverges, because the summand for $k = \lfloor \frac{n}{4} \rfloor$ increases without bound as $n$ increases.
