Infinitely many integrals

Note that the given integral can be written as the multiple integral $I(n,m) = \idotsint_{1\leq a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq n/m} 1\, d\mathbf{a}.$ Also, by symmetry, we may see that $I(n,m) = \idotsint_{1\leq a_{\sigma(0)} \leq a_{\sigma(1)} \leq \cdots \leq a_{\sigma(n-1)} \leq n/m} 1\, d\mathbf{a}$ for any of the $n!$ permutations $\sigma$ of $\{0,1,\ldots,n-1\}$ . Therefore, $\begin{aligned} I(n,m) &= \frac{1}{n!} \sum_{\sigma} \idotsint_{1\leq a_{\sigma(0)} \leq a_{\sigma(1)} \leq \cdots \leq a_{\sigma(n-1)} \leq n/m} 1\, d\mathbf{a} \\ &= \frac{1}{n!} \idotsint_{1\leq a_0, a_1, \ldots, a_{n-1} \leq n/m} 1\, d\mathbf{a} \\ &= \frac{1}{n!} \left(\int_1^{n/m} 1\, da\right)^n \\ &= \frac{1}{n!} \left(\frac{n}{m}-1\right)^n. \end{aligned}$

Now, using Stirling's approximation for the factorial, we have $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ and therefore, $I(n,m) = \frac{1}{n!} \left(\frac{n}{m}-1\right)^n \sim \frac{1}{\sqrt{2\pi n}(n/e)^n} \left(\frac{n}{m}\right)^n \cdot \left(1-\frac{m}{n}\right)^n \sim \frac{1}{\sqrt{2\pi n}}\left(\frac{e}{m}\right)^n \cdot e^{-m}.$

From this, we see that if $|m| < e$ , then $\displaystyle \lim_{n\rightarrow\infty} I(n,m)$ will diverge, and if $|m|\geq e$ , then $\displaystyle \lim_{n\rightarrow\infty} I(n,m) = 0$ .