There are n letters for which each has a specific envelope. If each letter has randomly been put into an envelope, what is the probability of choosing at least one correct envelope as $n \rightarrow \infty$?
I want to use the inclusion-exclusion formula.
Let A be the event that at least one letter is in its correct envelope and let $A_i$ be the event that letter i is in its correct envelope, then:
$$P(A)=P(\bigcup^n_{i=1} A_i)=\sum_{I\in \{1,...,n\}, I \neq \emptyset} (-1)^{|I|-1}P(\bigcap_{i\in I}A_i)$$
Since for $k \in \{1,...,n\}: P(\bigcap_{i \in I} A_i) = \frac{(n-k)!}{n!}$, where $|i|=k$ and since there are $\binom{n}{k}$ combinations for such an i, I get the following formula:
$$P(A)=P(\bigcup^n_{i=1} A_i)=n\frac{1}{n}-\frac{1}{2!}+\frac{1}{3!}-....(-1)^{n+1}\frac{1}{n!}$$
Is that correct up until this point?
Now $e=\sum^{\infty}_{i=0}\frac{1^k}{k!}$. Is there some way of how to express $P(\bigcup^{\infty}_{i=1} A_i)$ in terms of e? My problem is the alternating sign. Can anybody help please?
$$\sum_{i=0}^\infty\frac{(-1)^i}{i!}=\frac1{\mathrm e}$$