Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$ Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj} \binom{2n}{j}_{q^k}$$ for integers $m,r$ and positive integers $k$.

Note that a famous result of Gauss says that $f(n,0,0,1)=(q;q^2)_n$.

For $n=1$ we get $f(1,m,r,k)=1-q^{m+r}-q^{k+m+r}+q^{4m+2r}$

and therefore

$$\frac{f(1,m,r,k) } {1-q} ={\frac{1-q^{k+m+r}}{1-q}}-q^{m+r}{\frac{1-q^{3m+r}}{1-q}}$$

which for ${q\to1}$ converges to ${(k+m+r)-(3m+r)}={k-2m}.$

Computer experiments suggest that for each positive integer $n$

$$\lim_ {q\to1}\frac{f(n,m,r,k) } {{(q;q^2)_n} }=(k-2m)^n.$$

Till now I did not find a method which leads to a proof of this result. I asked the special case $ \lim_ {q\to1}\frac{f(n,m,m,1)}{f(n,1,1,1)}$ of this Problem already in Mathematics Stack Exchange 1359886.

  • $\begingroup$ This limit appears to diverge, and it doesn't match the limit in the Stack Exchange question. Could you please clarify? $\endgroup$ Aug 3, 2015 at 19:00
  • $\begingroup$ The Stack Exchange question is in fact a special case of this question. I have edited my question in order to clarify the Situation. $\endgroup$ Aug 4, 2015 at 8:22

2 Answers 2


I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider

$$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{(m-k/2)j^2+rj}}{ \partial q^a}(1) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

Now $\frac{\partial^a q^{(m-k/2) j^2+rj}}{ \partial q^a}(1)$ is a polynomial of degree $2a$ in $j$. So we can write this in terms of

$$ F(b,c) = \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}j^b \frac{ \partial ^{c} q^{(k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{c}}(1)$$

where $b+2c \leq 2i$.

We want to show that this vanishes for $i < n$ and evaluate it for $i=n$. For the first, it suffices to show that for all $b$ and $c$ with $b+2c < 2n$ we have:


But these terms all show up in computing the partial derivatives for the sum

$$\sum\limits_{j = 0}^{2n} {( - 1)}^{ j} q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}$$

$$ \frac{ \partial ^i \sum\limits_{j = 0}^{2n} {( - 1)}^{ j} q^{(k/2) j^2 + rj} \binom{2n}{j}_{q^k}}{ \partial q^i} (1) = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{\partial^a q^{rj}}{ \partial q^a}(1) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1) $$

$$ = \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j} (rj)(rj-1) \dots (rj+1-a) \frac{ \partial ^{i-a} q ^{ (k/2) j^2} \binom{2n}{j}_{q^k}}{ \partial q^{i-a}}(1)$$

$$ = \sum_{i=0}^a \binom{a}{i} \sum_{b=0}^a[j^b] (rj)(rj-1) \dots (rj+1-a) F(b,i-a) $$

Now suppose by induction that $F(b,c)=0$ for $b+c < i $ for some fixed $i \leq 2n$. Then this sum simplifies to only include the terms where $b=a$. Here

$$[j^a] (rj)(rj-1) \dots (rj+1-a) = r^a$$

so we obtain

$$= \sum_{i=0}^a \binom{a}{i} r^a F(a,i-a) $$

If $i<2n$ then this polynomial is identically $0$ by Max's computation, so all the coefficients vanish, so $F(a,i-a)=0$. This verifies the induction step.

Hence $F(b,c)=0$ for $b+c<2n$, which includes $b+2c<2n$, so the first $n-1$ partial derivatives vanish.

What about the $n$th partial derivative? Well the only $F(b,c)$ with $b + 2c \leq 2n$ that we don't know is $0$ is $F(2n,0)$. So dropping all the other terms we get

$$\frac{ \partial^n f(n,m,r,k)}{\partial q^n} (1)= F(2n,0) [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) $$


$$ [ j^{2n}]\frac{\partial^n q^{(m-k/2)j^2+rj}}{ \partial q^n}(1) = [j^{2n}] ( (m-k/2)j^2+rj) ((m-k/2)j^2+rj-1) \dots ((m-k/2)j^2+rj+1-n) = (m-k/2)^{n}$$

So we get a constant depending only on $n$ times $(2k-m)^n$. To get the correct constant, you can compute it directly, or just note that it is sufficient to handle one special case for each $n$ and note that you did the $f(n,0,1,2)$ case in the comments to Max's answer.

  • $\begingroup$ Very nice! We just happen to be lucky that $F(b,c)=0$ for $b+c<2n$, which was not obvious at all. $\endgroup$ Aug 12, 2015 at 2:30
  • $\begingroup$ This is a beautiful proof and deserves the bounty. $\endgroup$ Aug 12, 2015 at 10:54
  • $\begingroup$ @MaxAlekseyev Well it's not purely luck, because once you find evidence for a nice formula, it gets a lot more likely that there is some nice reason that a nice formula holds. $\endgroup$
    – Will Sawin
    Aug 12, 2015 at 12:59
  • $\begingroup$ This answer is cited in this paper of Johann Cigler's arxiv.org/abs/1602.07850 $\endgroup$
    – j.c.
    Sep 27, 2017 at 15:54

A couple of remarks.

First, it easy to see that if $q=1-t$, then $$(q;q^2)_n = \prod_{j=0}^{n-1} (1-q^{2j+1}) = \prod_{j=0}^{n-1} ((2j+1)t+O(t^2)) = (2n-1)!!\cdot t^n + O(t^{n+1}).$$ The limit in question can therefore be restated as $$\lim_{q\to 1} \frac{f(n,m,r,k)}{(1-q)^n} = (k-2m)^n\cdot (2n-1)!!.$$

Second, the limit can be easily proved for $k=2m$. In fact, we can even get a closed form for $f(n,m,r,2m)$. Using generating function for $q$-binomial coefficients, we have $$f(n,m,r,2m) = \sum_{j=0}^{2n} (-1)^j q^{mj^2+rj} \binom{2n}{j}_{q^{2m}}$$ $$=\sum_{j=0}^{2n} (-1)^j q^{(r+m)j} q^{2mj(j-1)/2} \binom{2n}{j}_{q^{2m}} =\sum_{j=0}^{2n} (-1)^j q^{(r+m)j} [z^j]\ (-z,q^{2m})_{2n}$$ $$=\sum_{j=0}^{2n} (-q^{r+m})^j [z^j]\ (-z;q^{2m})_{2n} = (q^{r+m};q^{2m})_{2n}.$$ Here $[z^j]$ is the operator of extracting the coefficient of $z^j$.

Again letting $q=1-t$, we trivially get $(q^{r+m};q^{2m})_{2n} = O(t^{2n})$, which implies the zero limit for any $n>0$.

  • $\begingroup$ Yes, this is one of the special cases, where I had a proof. The other one is $f(n,0,r,1)$. $\endgroup$ Aug 5, 2015 at 11:03
  • $\begingroup$ Another special case with a closed formula is $f(n,0,1,2)=(q;q^2)_n(-q^2;q^2)_n.$ This implies $\lim_ {q\to1}\frac{f(n,0,2r+1,2)}{{(q;q^2)_n} }=2^n.$ $\endgroup$ Aug 11, 2015 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.