Sum involving the product of binomial coefficients I wonder whether it is possible to calculate the folowing sum that involves the Binomial coefficients
$$\sum_{k=0}^n \binom{n}{k}^2 \binom{2k}{k} .$$
 A: Let
$$
a_k=\binom{n}{k}^2\binom{2k}{k}
$$
Then, with $k=\frac23n+j$,
$$
\begin{align}
\log\left(\frac{a_{k+1}}{a_k}\right)
&=2\log\left(2\frac{n-k}{k+1}\right)+\log\left(\frac{k+1/2}{k+1}\right)\\
&=2\log\left(\frac{\frac23n-2j}{\frac23n+j+1}\right)
+\log\left(\frac{\frac23n+j+\frac12}{\frac23n+j+1}\right)\\
&=-\frac{9j}n+O\!\left(\frac1n\right)\\
\end{align}
$$
Therefore,
$$
a_k=\color{#C00000}{a_{\frac23n}}\,\color{#009000}{e^{-\frac{9j^2}{2n}+O\left(\frac jn\right)}}
$$
Thus, we can estimate $a_{\frac23n}$ using Stirling and sum the exponential using a Riemann Sum to get
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}^2\binom{2k}{k}
&=\color{#C00000}{\frac32\left(\frac3{2\pi n}\right)^{3/2}9^n}\color{#009000}{\sqrt{n}\int_{-\infty}^\infty e^{-9x^2/2}\,\mathrm{d}x}\left(1+O\!\left(\frac1n\right)\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{3\sqrt3}{4\pi n}\,9^n\left(1+O\!\left(\frac1n\right)\right)}
\end{align}
$$

A Note About The Error Estimate
Even though the $O\!\left(\frac jn\right)$ in the exponential would normally introduce an error of $O\!\left(\frac1{\sqrt{n}}\right)$, the only terms beyond $-\frac{9j^2}{2n}$ that would contribute an error of $O\!\left(\frac1{\sqrt{n}}\right)$ are the $\frac jn$ and $\frac{j^3}{n^2}$ terms and they are odd and so get cancelled upon integration. This leaves an error of $O\!\left(\frac1n\right)$ which combines with the error for Stirling's Formula.
