Why is the following limit not 1?

Question

Consider $$\lim_{n\to\infty} f(n)=\lim_{n\to\infty}\frac{(n!)!}{(n!-n)!}\tag{1}$$

For large $n$, one can ignore $n$ wrt. $n!$ in the denominator. The limiting value should therefore approach $1$. However, as the plot shows, the limit blows up. Why is that?

Answer 1

If $n\in\Bbb N$, then$$\frac{(n!)!}{(n!-n)!}=n!\times(n!-1)\times\cdots\times\bigl(n!-(n-1)\bigr)\geqslant n!,$$and therefore$$\lim_{n\to\infty}\frac{(n!)!}{(n!-n)!}=\infty.$$You cannot simply ignore that second $n$ in the denominator since its presence makes the denominator much smaller than the numerator. And increasingly so as $n$ increases.

Answer 2

If you use the gamma function, taking logarithms and using carefully Stirling approximation $$\log\left(\frac{(n!)!}{(n!-n)!}\right)\sim-n+\frac{n}{12 (T+1) (n-T-1)}+$$ $$\left(n-T-\frac{1}{2}\right) \log (-n+T+1)+\left(T+\frac{1}{2}\right) \log (T+1)$$ where $$T=\sqrt{2 \pi }\, e^{-n}\, n^{n+\frac{1}{2}}$$

A few values of the logarithms (which for sure confirm your observations and your plot)

$$\left( \begin{array}{cccc} n & \text{approximation} & \text{exact value} & \Delta_n \\ 3 & 4.68466 & 4.78749 & 0.1028338 \\ 4 & 12.3601 & 12.4491 & 0.0889940 \\ 5 & 23.7684 & 23.8531 & 0.0846584 \\ 6 & 39.3711 & 39.4546 & 0.0835494 \\ 7 & 59.5886 & 59.6719 & 0.0833269 \\ 8 & 84.7528 & 84.8361 & 0.0832974 \\ 9 & 115.133 & 115.216 & 0.0833001 \\ 10 & 150.961 & 151.044 & 0.0833057 \\ \end{array} \right)$$

Edit

Searching the $ISC$, it seems that the asymptotic error is close to $$\Delta_\infty \sim 100\left(\frac{2}{3}\right)^{\pi \, \Gamma \left(\frac{1}{6}\right)}=0.0833058$$

Answer 3

You have to work it out in each case. $(n+\frac1n)^3-n^3$ grows but $(n+\frac1n)^2-n^2$ approaches a constant.

Let $g(n)=n!$ and $h(n)=\ln n!$.
Take the log of your ratio $f(n)$, and get $$h(g(n)) -h(g(n)-n)\\ \approx nh '(g(n))$$ Your 'minus n' term has to be so small it cancels the growth of $h'(g(n))$. Look up Stirling's formula, and work out how big $h'(g(n))$ is.