The portion of permutations that do not leave any number in its place is $\frac1{0!}-\frac1{1!}+\frac1{2!}-\frac1{3!}+\cdots+(-1)^n\left(\frac1{n!}\right)$. As $n$ grows, this sum gets closer to a certain number, which is approximately $0.367879$. That is, in general more than one third of the permutations leave no number in its place.
The exact value of this number is $\frac1e$, where $e$ is the famous Euler constant.
This is an extract from a book I've been reading, and it says the sequence converges to the reciprocal of Euler's number as n tends to large values.
What do they mean by convergence of a discrete, not continuous, sequence?
Could someone please show how and why the derangement sequences converges to $\frac1e$? (Treating the expression mathematically, using theorems of limits etc.)
Thanks!
1) All sequences are discrete- the elements are indexed by $\mathbb{N}$ (or $\mathbb{Z}^{+}$, if you prefer). An infinite series is the limit of a sequence of finite sums.
2) Recall that the Taylor expansion of $e^{x}$ is: $$e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$
Now $1/e = e^{-1}$. So:
$$e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!}$$
Edit: The derangement number: $$D_{n} = n! \sum_{i=0}^{n} \frac{(-1)^{i}}{i!}$$
This number has combinatorial interpretation via the principle of inclusion-exclusion. We count the number of permutations which fix no points. For each $i \in \{0, \ldots, n\}$, we count the number of permutations which fix at least $i$ points. We select the $i$ targeted fixed points in $\binom{n}{i}$ ways. The remaining $n-i$ elements can be freely varied in $(n-i)!$ ways. By rule of product, this yields $\binom{n}{i}(n-i)! = \frac{n!}{i!}$. So by the Principle of Inclusion-Exclusion, we have that:
$$D_{n} = \sum_{i=0}^{n} (-1)^{i} \binom{n}{i} (n-i)! = n! \sum_{i=0}^{n} \frac{(-1)^{i}}{i!}$$
As desired.