<s>[INST] Bound for $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$. In the book Complex analysis by Bak J. & Newman J., chapter 11, talks about Sums Involving Binomial coefficients and find a bound $\frac{16}{9}\sqrt{3}$ for $|(z-1)^2(z+1)|$ in the "Example 3" on the unit circle, my way was using lagrange multipliers in this form:
We want find $\max{|(z-1)^2(z+1)|}$ ahnd have that $|z-1|^2+|z+1|^2=4$. Let be $a=|z-1|$ and $b=|z+1|$ and then the exercise is: "Maximize $f(a,b)=a^2b$ subject to $a^2+b^2=4$" then $\nabla f=\lambda\nabla g$ so $\begin{cases}2ab=\lambda(2a)\\a^2=\lambda(2b)\end{cases}$, i.e., $ab=\lambda a$ if $a(b-\lambda)=0$ therefore
i)$a=0$ or $b=\lambda$, $b^2=4$ then $b=2$ then $|z+1|=2$ or $|z-1|=0$ then $z=1$ and $a^2b=0$.
ii)$b=\lambda$, $a^2=2b^2$ then $4=a^2+b^2=3b^2$ then $b^2=4/3$ then $b=\frac{2\sqrt{3}}{3}$ then $a^2=\frac{8}{3}$ then $a=\frac{2\sqrt{2}}{\sqrt{3}}$
So $a^2b=\frac{8}{3}\frac{2\sqrt{3}}{3}=\frac{16}{9}\sqrt{3}$.
After i want to use this idea for exercise 17.b with $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$ but, i don´t see what should be $a$ and $b$.
Edit: this exercise says that $|\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}|\leq4^n$.
 [/INST] The first term in the asymptotic expansion follows from the analysis given by 'Asymptotics of a Family of Binomial Sums' by R. Noble.  Putting this in a form for comparison one gets
$$ \sum_{k=0}^n (-1)^k\binom{3n}{k} \binom{n}{k} \sim (-4)^n \frac{2^{1/4}}{\sqrt{\pi n}} \cos(n \tan^{-1}(10\sqrt{2}/23) + \tan^{-1}((1-\sqrt{2})/(1+\sqrt{2})) $$
This is not a bound, of course.  However, when $n$ is large enough,
$$ \Big|\sum_{k=0}^n (-1)^k\binom{3n}{k} \binom{n}{k} \Big| < C\frac{4^n}{\sqrt{n}}.$$
Is $C=2^{1/4}/\sqrt{\pi} \approx 0.671$ ?  Checking up through $n=1000$ says it is.  To be rigorous the next term in the asymptotic expansion needs to be generated.
For comparison of the asymptotic formula and the exact: n=200, true=-1.2130 x $10^{119},$ appx=-1.2145 x $10^{119},$ and absolute % error between them is 0.12%.
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