# Inequalities involving polynomials with combinatorial coefficients

For all non-negative integers $i$ and $j$ such that $j\leq i$, define the array of polynomials $$p_{ij}(z):=\sum_{h=(j-1)_+}^{i-1} {i\choose h}{i-j\choose{i-h-1}}z^h,$$ where $(a)_+=\max\{a,0\}$ (we need this lower bound for the sum in order for the second binomial coefficient to exist).

With these polynomials, I wish to prove the following inequalities: for all $i>2$ and for $0<z<1$, $$\frac{p_{i0}(z)}{p_{i1}(z)}<1+\frac1{\sqrt{z}}<\frac{p_{i2}(z)}{p_{i3}(z)}.$$

I've graphed these functions for various values of $i$, and not only do the inequalities appear to be true, but they also appear to be tight, in that the upper and lower bounds converge pointwise to $1+\frac1{\sqrt{z}}$ as $i \rightarrow \infty$. I'm not aware of any well known rational function sequences that converge to square root functions besides the binomial series, but I don't think these rational functions can reduce down to something of that form.

Any help I receive will be appreciated.

Fortunately, $p_{i j}(z)$ can be summed formally, using Gauss hypergeometric functions. With the help of Wolfram Alpha, one obtains for $j = 0$ $$p_{i 0}(z) = i \;\; {_2{F_1}}(1-i, -i, 2, z)$$ and for $j \geq 1$ $$p_{i j}(z) = z^{-1+j} {i \choose {-1+j}} \;\; {_2{F_1}}(j-i, -1+j-i, j, z)$$ Hence the inverse of the first fraction can be written $$\frac{i \; p_{i1}(z)}{p_{i0}(z)} = \frac{ {_2{F_1}}(1-i, -i, 1, z)}{{_2{F_1}}(1-i, -i, 2, z)} > \frac{i}{1+\frac1{\sqrt{z}}}$$ Now there exist continued fraction expressions for exactly that type of quotients in the 1956 work of Evelyn Frank, A New Class of Continued Fraction Expansions for the Ratios of Hypergeometric Functions. We have from her formula (2.7.) that
$$\frac{ {_2{F_1}}(a, b, c, z)}{{_2{F_1}}(a, b, c+1, z)} = 1 + \frac{ \frac{a b z}{c (c+1)} }{ \frac{- b z}{(c+1)} + 1 - } \quad \frac{ \frac{(a+1)(c+1 -b) z}{(c+1) (c+2)} }{ \frac{(a+1- b) z}{(c+2)} + 1 - } \quad \frac{ \frac{(a+2)(c+2 -b) z}{(c+2) (c+3)} }{ \frac{(a+2- b) z}{(c+3)} + 1 - } \cdots$$ Indeed this is not an infinite expansion but a finite expression since it terminates for negative integer $a,b$, as soon as the first coeffient is zero, with the preceding term.

Inserting the values for $a,b,c,$ gives $$\frac{ {_2{F_1}}(1-i, -i, 1, z)}{{_2{F_1}}(1-i, -i, 2, z)} = 1 + \frac{ \frac{(1-i) (-i) z}{2} }{ \frac{i z}{2} + 1 - } \quad \frac{ \frac{(2-i)(2+i) z}{2 \cdot 3} }{ \frac{2 z}{3} + 1 - } \quad \frac{ \frac{(3-i)(3+i) z}{3 \cdot 4} }{ \frac{3 z}{4} + 1 - } \cdots$$ This will have $i$ terms.

Starting as requested by the OP with $i=3$ we have the condition for $z \in (0,1)$: $$1 + \frac{ \frac{2 \cdot 3 z}{2} }{ \frac{3 z}{2} + 1 - } \quad \frac{ \frac{-5 z}{2 \cdot 3} }{ \frac{2 z}{3} + 1 } > \frac{3}{1+\frac1{\sqrt{z}}}$$ It is easy to show that this holds true.

Now for larger $i$ one needs to proceed to show that the inequality holds. (induction? continued fraction expansion of the root term?)

The paper of Evelyn Frank also has other quotients, so it may be used for the OP's upper bound as well.

However I'm not an expert in continued fractions so I'll stop here. I'd be curious to read some continuation of this approach.

Hope it helped anyway.