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Is it the case that, as $N\to\infty$, $$\binom{2N}{N+j}_q\to (-1)^j,$$ where convergence of the $q$-binomial coefficient is seen as a convergence of formal power series in the variable $q$?

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  • $\begingroup$ The coefficient of $q^1$ in any non-constant Gaussian binomial coefficient is $1$; how could the limit possibly have a zero coefficient there? Also the constant term is always $1$. $\endgroup$ Feb 23, 2015 at 15:17
  • $\begingroup$ Why is the coefficient of $q^1$ always $1$? $\endgroup$
    – Nishant
    Feb 23, 2015 at 16:31
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    $\begingroup$ Because of the combinatorial interpretation of those coefficients. At a given corner of a non-empty grid rectangle, there is exactly one way to remove one square. $\endgroup$ Feb 23, 2015 at 18:03

2 Answers 2

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$$A_{N,j}=\binom{2N}{N+j}_q=\frac{(1-q^{2N})(1-q^{2N-1})\ldots(1-q^{2N-N-j+1})}{(1-q)(1-q^2)\ldots(1-q^{N-j})}$$

Take $\log$ and expand using that $$\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}.$$

We obtain that $$\begin{align}\log(A_{N,j})&=\sum_{k=N-j+1}^{2N}\log(1-q^{k})-\sum_{k=1}^{N-j}\log(1-q^k)\\&=\sum_{k=N-j+1}^{2N}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}-\sum_{k=1}^{N-j}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}\end{align}$$

The first group with the double summation tends to $0$ as $N\to\infty$ because all exponents are above $N-j+1$.

For the second we have $$\sum_{k=1}^{N-j}\sum_{r=1}^{\infty}\frac{q^{kr}}{r}=\left(\frac{q^1}{1}+\frac{q^{2}}{2}+\cdots\right)+\left(\frac{q^2}{1}+\frac{q^{4}}{2}+\cdots\right)+\left(\frac{q^3}{1}+\frac{q^{6}}{2}+\cdots\right)+\cdots+\left(\frac{q^{N-j}}{1}+\frac{q^{2(N-j)}}{2}+\cdots\right)\\=a_1q+a_2q^2+\cdots$$

where $a_i$ is the sum of the reciprocals of the divisors of $i$. This series is not constant and its exponential is not either.

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Let ${\left( {a;q} \right)_n}=\prod\limits_{j = 0}^{n - 1} {(1-{q^j}a} ).$

Then $ {2n\brack {n+j}}=\frac{(q;q)_{2n}}{(q;q)_{n+j}(q;q)_{n-j}}$ converges to $\frac{1}{(q;q)_\infty}=\sum\limits_{n \ge 0} {p(n){q^n}}=1+q+2q^2+3q^3+5q^4+ \dots,$ where $p(n)$ is the number of partitions of $n$.

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