# Limit of Product of Monomials, Specifically $\prod_{i=1}^{n}(x+i)$

If $$p_n(x)=\prod_{i=1}^{n}(x+i)$$ then what is the order of the power of $$x$$ (with respect to $$n$$) with the greatest coefficient for the following polynomial? $$\lim_{n \to \infty} p_n(x)$$

I plugged in some small values of $$n$$ and saw that the desired power tends to hang within the smaller powers of $$x$$ (for example, $$p_4(x)=x^4 + 10 x^3 + 35 x^2 + 50 x + 24$$, so the power of $$x$$ with the greatest coefficient is $$1$$), but I don't know enough about infinite polynomials to make any sort of generalization.

Also, I know that the power goes to infinity as $$n$$ goes to infinity, what I want to know is the order of the power relative to $$n$$- is it $$\sqrt{n}$$? $$\ln(n)$$? $$n^{\frac{1}{e}}$$? That sort of thing.

• Not sure you are aware, but formally you can write $$a_k(n)=\frac{1}{k!} \, \frac{{\rm d}^k}{{\rm d}x^k} \frac{\Gamma(x+n+1)}{\Gamma(x+1)} \Bigg|_{x=0}$$ where $a_k(n)$ is defined by $p_n(x)=\sum_{k=0}^n a_k(n) x^k$. Nov 20, 2019 at 18:08
• Nope, definitely was not aware of that. Now I need to look into this and justify it. Nov 20, 2019 at 19:54

$$\displaystyle \prod\limits_{k=1}^n (x+k) = \sum\limits_{v=0}^n a_{n,v}x^v \approx\frac{n!n^x}{\Gamma(1+x)}=\frac{1}{\Gamma(1+x)}\sum\limits_{v=0}^\infty\frac{n!(\ln n)^v}{v!}x^v$$

For $$~v\ll n~$$ we can write $$\displaystyle~a_{n,v}\approx \frac{n!(\ln n)^v}{v!}~$$ .

So, the question is about $$\displaystyle~\max\frac{(\ln n)^v}{v!}~$$ .

With the Stirling formula $$\displaystyle~v!\approx \left(\frac{v}{e}\right)^v\sqrt{2\pi v}~$$ we get $$\displaystyle~a_{n,v}\approx \frac{n!}{\sqrt{2\pi v}}\left(\frac{e\ln n}{v}\right)^v~$$ .

We have $$\displaystyle~\max\left(\frac{e\ln n}{v}\right)^v \leq\left(e^{1/e}\right)^{e\ln n} = n =\left(\frac{e\ln n}{v}\right)^v|_{v=\ln n}~$$ .

It follows $$~v\approx \ln n~$$ .

• Why do you identify $a_{n,v}\approx \frac{n!(\ln n)^v}{v!}$? What about the denominator $\Gamma(x+1)$? Nov 20, 2019 at 22:38
• @Diger : It's an approximation for $n\to\infty$, where $x$ doesn't play a role $(a_{n,v}$ is independend of $x$). And of course: It's not exact. It must be $v\ll n$, that's why we can write $a_{n,v}\approx\frac{n!(\ln n)^v}{v!}$ by comparing the coefficients of $x^v$ . Nov 21, 2019 at 6:06

In terms of the unsigned Stirling numbers of the first kind $$p_n (x) = \sum\limits_{k = 0}^n {\left[ {\begin{array}{*{20}c} {n + 1} \\ {k + 1} \\ \end{array}} \right]x^k } .$$ By a result of Hammersley, the index $$k(n)$$ of the maximal coefficient (which is unique) is then given by $$k(n) = \left\lfloor {\log (n + 2) + \gamma - 1 + \frac{{\zeta (2) - \zeta (3)}}{{\log (n + 2) + \gamma - 3/2}} + \frac{{\theta _n }}{{(\log (n + 2) + \gamma - 3/2)^2 }}} \right\rfloor - 1,$$ with a suitable number $$\theta_n$$ satisfying $$-1.1 < \theta _n < 1.5$$. Here $$\zeta$$ is the Riemann zeta function and $$\gamma = 0.5772\ldots$$ is the Euler–Mascheroni constant. For more information, see, e.g., this paper.

Let $$p_n(x) = x^n + a_1x^{n-1}+a_2x^{n-2}+\ldots +a_{n-1}x+a_n$$

Then, $$-a_1 = 1+2+3+\ldots +n =\displaystyle\sum_{i=1}^n i = {n(n+1)\over 2}$$ $$a_2 = 1\cdot 2 + 1\cdot 3 + 1\cdot 4 +\ldots +n\cdot n-1 = 1(2+3+ \ldots + n) + 2(3+ 4+ \ldots n) +\ldots+ n$$ $$= \displaystyle\sum_{i=1}^n\left[ i\displaystyle\sum_{j=i+1}^n j\right] =\displaystyle\sum_{i=1}^n\left[ i\left(\displaystyle\sum_{j=1}^n j - \sum_{k=1}^i k\right)\right] =\displaystyle\sum_{i=1}^n\left[i\left({n(n+1)\over 2}-{i(i+1)\over 2}\right)\right]$$ $$= {n(n+1)\over 2}\sum i -\frac 12\sum(i^3+i) = {n^2(n+1)^2\over 4}+\frac 12 \left({n^2(n+1)^2\over 4} + {n(n+1)\over 2}\right)$$

$$=\frac 38 n^2(n+1)^2 + \frac 12 n(n+1)$$

To be continued...

• I see what you’ve done so far, but I don’t understand how what’s been written generalizes. Could you please continue your solution? Dec 26, 2019 at 18:23
• I will do it if i get time. I was trying to go the inductive route. Dec 27, 2019 at 4:14

Since this is the product of n linear polynomial in x, the degree is n. The coefficients are, I believe, unsigned Stirling numbers of either the first or second kind. One of my answers gave an asymptotic expansion in powers of $$x+(k+1)/2$$.