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Could you help me to find an asymptotic for this sum?

$$ \sum_{k=0}^{n – 1} (-1)^k {n \choose k} {3n – k – 1 \choose 2n – k} = {n \choose 0} {3n – 1 \choose 2n} – {n \choose 1} {3n – 2 \choose 2n – 1} + … + (-1)^{n-1} {n \choose n-1} {2n \choose n + 1} $$

I have tried to write binomials through factorials and work with it, but it seems like not right way to evaluate an asymptotic.

Thank you for your help:)

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$$\begin{align}

\sum_{k=0}^{n-1}(-1)^k\binom nk \color{blue}{\binom {3n-k-1}{2n-k}}

&=\sum_{k=0}^{n-1}(-1)^k\binom nk\color{blue}{(-1)^{2n-k}\binom {-n}{2n-k}}

\qquad&\color{blue}{\text{(upper negation)}}\\

&=\color{green}{\left[\sum_{k=0}^{n}\binom nk\binom {-n}{2n-k}\right]}-\color{orange}{\binom nk\binom {-n}{2n-k}\Biggr|_{k=n}}\\

&=\color{green}{\binom 0{2n}}-\color{orange}{\binom nn\binom {-n}{n}}

\qquad&\color{green}{\text{(Vandermonde)}}\\

&=-\color{orange}{(-1)^n\binom{2n-1}n}

\qquad&\color{orange}{\text{(upper negation)}}\\

&=(-1)^{n-1}\binom{2n-1}n=(-1)^{n-1}\binom{2n-1}{n-1}\quad\blacksquare

\end{align}$$

Suppose we first seek to evaluate the somewhat more general

$$S_m(n) = \sum_{k=0}^n {n\choose k} (-1)^k

{mn-k-1\choose 2n-k}$$

with $m$ an integer parameter.

Now introduce

$${mn-k-1\choose 2n-k} =

\frac{1}{2\pi i}

\int_{|z|=\epsilon}

\frac{1}{z^{2n-k+1}} (1+z)^{mn-k-1} \; dz.$$

We thus get for the sum

$$\frac{1}{2\pi i}

\int_{|z|=\epsilon}

\frac{(1+z)^{mn-1}}{z^{2n+1}}

\sum_{k=0}^n {n\choose k} (-1)^k \frac{z^k}{(1+z)^k} \; dz

\\ = \frac{1}{2\pi i}

\int_{|z|=\epsilon}

\frac{(1+z)^{mn-1}}{z^{2n+1}}

\left(1-\frac{z}{1+z}\right)^n\; dz

\\ = \frac{1}{2\pi i}

\int_{|z|=\epsilon}

\frac{(1+z)^{mn-n-1}}{z^{2n+1}} \; dz

\\ = {mn-n-1\choose 2n}.$$

We see that this is zero when $m=2$ and $m=3$ and non-zero otherwise.

In the problem being asked we are computing

(the last term is not being included)

$$S_3(n) – (-1)^n {2n-1\choose n}$$

so we obtain

$$(-1)^{n+1} {2n-1\choose n}.$$

This can be treated with the asymptotic for the central binomial

coefficient. We get

$$(-1)^{n+1} \frac{n}{2n} {2n\choose n}

= \frac{1}{2} (-1)^{n+1} {2n\choose n}.$$

The central binomial coefficient is

OEIS A000984

and is asymptotic to

$$\frac{4^n}{\sqrt{\pi n}}.$$

**Hint:** Use $\displaystyle{a-1\choose b}=(-1)^b~{b-a\choose b}$ in conjunction with Vandermonde’s identity. Also, notice

that the upper summation limit is $n-1$ instead of *n*.

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