The ratio $\iota(D_n)/\iota(n!)$ computed for small $n$ as a quotient of two counts, with no probability space claimed

For $n \in \mathbb{N}$ consider the real number

$r_n \;:=\; \frac{\iota(D_n)}{\iota(n!)},$

the quotient of the count of derangements of an $n$-element set (The derangement number $D_n$: the number of bijections of an $n$-element set with no fixed point) by the count of all its bijections, $\lvert\operatorname{Bij}(n)\rvert = n!$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$, defined by recursion in $\mathbb{N}$). The quotient is legitimate because $n! \ne 0$. Dividing the derangement formula by $\iota(n!)$ gives

$r_n \;=\; \sum_{i<n+1}\frac{(-1)^{i}}{\iota(i!)} ,$

so $r_n$ is the truncated alternating sum itself. Its first values, obtained from $D_0 = 1$, $D_1 = 0$, $D_2 = 1$ and the first recurrence $\iota(D_n) = \iota(n)\iota(D_{n-1}) + (-1)^{n}$ ($\iota(D_n) = \iota(n)\,\iota(D_{n-1}) + (-1)^{n}$ for $n \ge 1$, and $D_n = (n-1)(D_{n-1} + D_{n-2})$ for $n \ge 2$), are

This is a ratio of two counts and nothing else. Nothing among this page's declared prerequisites defines a probability space, a measure or an expectation, so $r_n$ is not called a probability here and no statement about random behaviour is made. What is asserted is exactly that the numerator counts the fixed-point-free bijections, that the denominator counts all of them, and that the quotient is the displayed alternating sum.

• The alternation is visible in the table. $r_1 = 0$ is below $r_2 = 1/2$, which is above $r_3 = 1/3$, which is below $r_4 = 3/8$; each successive value differs from the previous one by the single term $(-1)^{n}/\iota(n!)$, whose sign alternates and whose size decreases.

• No limit is claimed. The quotient is computed at each $n$ from two counts, and nothing here asserts convergence or names a limiting value; the exponential function that would be needed to state such a limit is not among this page's declared prerequisites.

• Why the division is legitimate at every $n$, including $n = 0$. The denominator is $\iota(n!)$ and $n!$ is never $0$, its recursion starting at $0! = 1$ and multiplying by nonzero successors.