# Asymptotic for combinatorial function

Let

$$F_q(k) = \sum_{n=1}^{\infty} \binom{n-1}{k} \binom{1/2}{n} q^n$$

be a function on $\mathbb{N}$. I am interested in the asymptotic behavior of $F$. Any ideas how to tackle it?

• 1) Did you try anything? 2) $\binom{x}{k} = 0 \ \forall \ 0 \leq x <k$
– Alex
May 18, 2016 at 22:59
• 1) I try to estimate $C_{n-1}^k$ like $n^k/k!$ then series converges to $\operatorname{Li}_{-k}(q)/k!$ but I dont find anything about asymptotic of this function when $k \to \infty$ 2) yes May 18, 2016 at 23:19
• $$\binom xk=\frac{x!}{k!(x-k)!}$$ where $n!=1\times2\times3\times\dots\times n$ May 18, 2016 at 23:20
• $$\binom{1/2}{k} = \frac{1/2 (1/2 - 1) (1/2 - 2) ... (1/2 - k + 1)}{k!}$$ if you about that. May 18, 2016 at 23:22
• So... I only have to tell you about how its asymptotic? Not actually find the closed form solution, if it exists? May 19, 2016 at 0:13

How about this: Let's assume $$|q|<1$$, then

\begin{align*} F_q(k) &= \sum_{n\geqslant 1} \binom{n-1}{k} \binom{1/2}{n} q^n\\ &= \sum_{n \geqslant 1} \binom{n-1}{k} \frac{1/2}{n} \binom {-1/2} {n-1}q^n\\ &= \frac 1 2 \sum_{n} \binom {n-1} k \binom{-1/2}{n-1} \frac{q^n}{n}\\ &= \frac 1 2 \sum_n \binom {-1/2} k \binom{-1/2-k}{n-1-k} \frac {q^n} n\\ \frac d {dq} F_q(k) &= \frac 1 2 \binom {-1/2} k \sum_n \binom {-1/2 -k} {n-1-k}q^{n-1} \\ &= \frac 1 2 \binom {-1/2} k q^k \sum_n \binom {-1/2 - k} n q^n \\ &= \frac 1 2 \binom {-1/2} k q^k (1+q)^{-k-1/2}. \end{align*}

I'm not good with integrals but this one can apparently be expressed in terms of a hypergeometric function. Note $$F_0 (k)$$ = 0; we get

$$F_q(k) = \frac{q^{k+1}}{2k+2} \binom{-1/2}{k} {_2F_1}(k+\tfrac 1 2,k+1;k+2;-q).$$

This appears to be correct for all integer $$k$$ and all $$|q|<1$$ (I checked a few small values numerically.) Note that it is not valid for noninteger $$k$$ because then line 4 is invalid. Of course, we need to verify that $$F_q(k)$$ actually converges: the radius of convergence seems to be 1, but could somebody prove it?

Update (2016-05-19):

Here's an idea for where to go from here. Apply Euler's identity

$${_2F_1}(a,b;c;z) = (1-z)^{c-a-b} {_2F_1}(c-a,c-b;c;z),$$

and now all that remains is to find an approximation for

$${_2F_1}(3/2,1;k+2;-q).$$

But now that the upper parameters are fixed, provided that also $$\mathrm{Re}(q) \geqslant -\tfrac 1 2$$, we can use a result from the NIST DLMF, which simplifies slightly to

$$_2F_1(3/2,1;k+2;-q) = \sum_{s=0}^{m-1} \frac{(3/2)_s}{(k+2)_s}(-q)^s + O((k+2)^{-m}),$$

where $$(x)_s$$ is a rising factorial. Pick some small $$m$$, and you're done.

Update 2 (2016-05-19):

In fact we only need to take the first term of the last sum above! Unless of course you want a better approximation. Here's the result with $$m=3$$:

$$F_q(k) = (-1)^k \frac{q^{k+1}(1+q)^{-k+1/2}}{2\sqrt{\pi k}(k+1)}\left(1 - \frac{3q}{2(k+2)} + \frac{15q^2}{4(k+2)(k+3)}\right) + O(k^{-9/2}).$$

Or, if you want something a bit cleaner, though this isn't great for small $$k$$,

$$F_q(k) \sim \left(\frac{q(1+q)^{1/2}}{2\sqrt{\pi}}\right) \left(\frac{-q}{1+q}\right)^k k^{-3/2}.$$

• Thank you! Wolfram gets me representation in ${}_pF_q$ functions too, but it doesn't get me any information about asymptotic. May 19, 2016 at 15:15
• Hmm, I'm not sure how to go about finding an asymptotic for that thing in closed form. But at least now it seems a bit more tractable. The binomial coefficient is not too hard to deal with, so you just need the behavior of the well-known $_2F_1$. Maybe you'll have better luck if you post a new question specifically about that. May 19, 2016 at 17:37
• Aw dang, you beat me to it. However, I do think you meant $(1+q)^{k-1/2}$. May 19, 2016 at 21:06
• SimpleArt, I did not mean $(1+q)^{k-1/2}$. How did you get that? Also, what justifies your approximation $x \approx x+1$? Since $|q|$ is less than $1$, this seems like a very poor approximation. Sorry, I can't post comments on your answer directly (I have no "reputation.") May 19, 2016 at 21:42
• @YakovShklarov Sorry, it seems like I need estimate, not asymptotic. Maybe you have ideas how to get it? May 21, 2016 at 9:12

$$\sum_{n=1}^\infty\binom{n-1}k\binom{1/2}nq^n=\sum_{n=1}^\infty\frac{(n-1)!(1/2)!}{k!(n-k)!n!(1/2-n)!}q^n$$

$$=\sum_{n=1}^\infty\frac{(1/2)!}{n(k)!(n-k)!(1/2-n)!}q^n$$

$$=\frac1{k!}\sum_{n=1}^\infty\frac{(1/2)!}{n(n-k)!(1/2-n)!}q^n$$

$$F_q(k)=\int\frac d{dq}F_q(k)dq=\int\frac d{dq}\frac1{k!}\sum_{n=1}^\infty\frac{(1/2)!}{n(n-k)!(1/2-n)!}q^ndq$$

$$=\int\frac1{k!}\sum_{n=1}^\infty\frac{(1/2)!}{(n-k)!(1/2-n)!}q^{n+k-1}dq$$

$$\sum_{n=1}^\infty\frac{(1/2)!}{(n-k)!(1/2-n)!}q^{n-1}=\sum_{n=0}^\infty\frac{(1/2)!}{n!(1/2-k-n)!}q^{n+k-1}$$

To get to that step, I assumed $k\in\mathbb{N}$.

$$=\frac{(1/2)!}{(1/2-k)!}\sum_{n=0}^\infty\frac{(1/2-k)!}{n!(1/2-k-n)!}q^{n+k-1}$$

$$=\frac{(1/2)!}{(1/2-k)!}q^{k-1}(q+1)^{1/2-k}$$

$$F_q(k)=\int_0^q\frac{(1/2)!}{(1/2-k)!}x^{k-1}(x+1)^{1/2-k}dx$$

In words, you want to graph that, and the area under the function from $0$ to $q$ is equal to $F_q(k)$.

As for asymptotes, we can use $x+1\approx x$.

$$F_q(k)\approx\int_0^q\frac{(1/2)!}{(1/2-k)!}x^{k-1}x^{1/2-k}dx$$

$$=\int_0^q\frac{(1/2)!}{(1/2-k)!}x^{-1/2}dx$$

$$=\frac{(-1/2)!}{(1/2-k)!}q^{1/2}$$

So

$$F_q(k)\sim\frac{(-1/2)!}{(1/2-k)!}q^{1/2}$$

It is difficult for me to make conclusions about $k\to\infty$, but I do believe that it diverges with fixed $q$.

$$\lim_{k\to\infty}(-k)!=0$$

That is a usually excepted statement.

• Wow, it seems very impressive for me, thanks! May 18, 2016 at 23:47
• How did you get $\frac{k!}{(n-k)!}=\frac{1}{n!}$ after fifth "equality" sign? May 18, 2016 at 23:54
• @kp9r4d Aw, appears I goofed up. Will see if I can fix my answer May 18, 2016 at 23:56
• @kp9r4d You get the hang of it after a few months of practice... ;D May 18, 2016 at 23:56
• @kp9r4d Man, redoing this problem, its a lot harder than I anticipated. May 19, 2016 at 0:11