Limit of $\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right)$ I can see numerically that $$\lim_{n \to \infty} \sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right) = 1$$ but how can you prove this?  Using Stirling's approximation doesn't seem to be enough.
 A: $$
\sum_{k=1}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large n}
=\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}\left[4^{-k}\binom{2k}{k}\right]^{\large n}\tag{1}
$$
The $k=0$ term is $1$ so we need to determine how the rest of the terms go to $0$. Let's look at the ratio between consecutive terms in $(1)$
$$
\begin{align}
&\frac{(n-2k)(n-2k-1)}{(2k+1)(2k+2)}\left[\frac14\frac{(2k+2)(2k+1)}{(k+1)(k+1)}\right]^{\large n}\\
&=\left(\frac{n+1}{2k+1}-1\right)\left(\frac{n+1}{2k+2}-1\right)\left[1-\frac1{2k+2}\right]^{\large n}\\[6pt]
&=\left(\frac{2k+2}{2k+1}\frac{n+1}{2k+2}-1\right)\left(\frac{n+1}{2k+2}-1\right)\frac{2k+2}{2k+1}\left[1-\frac1{2k+2}\right]^{\large n+1}\\[6pt]
&\le\left(\frac43\frac{n+1}{2k+2}-1\right)\left(\frac{n+1}{2k+2}-1\right)\frac43\left[1-\frac1{2k+2}\right]^{\large n+1}\\[6pt]
&\le\frac43\left(\frac43x-1\right)(x-1)e^{-x}\qquad\text{where }x=\frac{n+1}{2k+2}\\[6pt]
&\lt\frac25\qquad\qquad\text{for all }x\ge1\tag{2}
\end{align}
$$
Thus, the error when truncating the sum is no more than $\frac53$ times the first truncated term.
Therefore,
$$
\sum_{k=0}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large n}
\sim1+\binom{n}{2}(1/2)^n+O\left(n^4(3/8)^n\right)\tag{3}
$$
since $\binom{n}{4}\sim\frac{n^4}{24}$.
