Warning: full-frontal mathematics.

From time to time, I'll try one of the problems from the American Mathematical Monthly; every month, they run about half a dozen problems, inviting readers to send in solutions. They publish reader solutions, and list names of people who submitted correct solutions. I haven't had the luck to get a solution of mine published, but I have gotten my name listed as a solver a few times.

Anyway, last May I submitted a solution to a problem that proved to be interesting to solve. The problem (number 12029, proposed by Hideyuki Ohtsuka) is to evaluate \[\lim_{n\to\infty}\prod_{k=1}^n\left(a+\frac kn\right)\] for \(a\gt0\).

At first, I thought this would be fairly straight-forward: this limit of a product resembles a Riemann sum for an integral. The fact that it is a product not a sum is not a challenge; there's such a thing as a product integral, based on products, not sums.

But there is no differential in the expression where it would be expected. I then realized that if the parameter \(a\) is smaller than a certain value, the product diverges to zero in the limit, while if the parameter is larger than that value, the product diverges to infinity in the limit.

Numerical experiments (direct attempts to calculate the limit for various values of \(a\)) suggested the magic value was about \(a=0.542\), and the value of the limit itself is about \(1.686\). Neither of these numbers looked familiar. It took a fair amount of work until my solution crystallized, and I knew what the exact values of these numbers are.

Anyway, here's the gist of the solution I submitted to the Monthly:

First, verify the following (this can be proved using the Trapezoid Rule): Suppose \(f\) satisfies certain reasonable conditions. Let \(\Delta x = (b-a)/n\) and for \(0\le k\le n\) let \(x_k=a+k\Delta x\). Then we have \[ \lim_{n\to\infty}\left( \sum_{k=1}^nf(x_k) - \frac1{\Delta x}\int_a^bf(x)\,dx\right)= \frac{f(b)-f(a)}2. \]

Now we use this to solve the problem. Let \(a\gt 0\). Let \(f(x)=\ln(a+x)\) on the interval \([0,1]\). So \(\Delta x=1/n\) and \(x_k=k/n\). We have \[ \lim_{n\to\infty} \left( \sum_{k=1}^n\ln\left(a+\frac kn\right) - n\int_0^1\ln(a+x)\, dx\right) = \frac{\ln(a+1)-\ln a}2. \] Now if \(\int_0^1\ln(a+x)\, dx \lt 0\), this implies that \[ \lim_{n\to\infty}\sum_{k=1}^n\ln\left(a+\frac kn\right) = -\infty, \] while if \(\int_0^1\ln(a+x)\, dx \gt 0\), we have \[ \lim_{n\to\infty}\sum_{k=1}^n\ln\left(a+\frac kn\right) = \infty. \] But if \(\int_0^1\ln(a+x)\, dx = 0\), equation (2) implies \[ \lim_{n\to\infty}\sum_{k=1}^n\ln\left(a+\frac kn\right) = \frac{\ln(a+1)-\ln a}2. \] Now \[\int_0^1\ln(a+x)\, dx = (a+1)\ln(a+1)-a\ln a-1.\] So if we take exponentials, we have the following solution to the problem: Let \(b\) be the unique positive number such that \(f(b)=0\), where \[f(x)=(x+1)\ln(x+1)-x\ln x - 1.\] It turns out that \(b\approx0.5422114\). Now let \[c = \exp\left(\frac{\ln(b+1)-\ln b}2\right) = \sqrt{\frac{b+1}{b}}.\] It turns out that \(c\approx 1.686505\). Then we have \[ \lim_{n\to\infty}\prod_{k=1}^n\left(a+\frac kn\right) = \left\{ \begin{array}{cl} 0 &\text{ if }0\le a \lt b \\ c & \text{ if }a = b\\ \infty & \text{ if }a \gt b \end{array}\right. \]

Update (2020-02-11): I was disappointed that my solution was not selected for publication, but my name did appear as one of the solvers of the problem. (The published solution was crisp; it used Stirling's formula.)