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Write $N={\sum }_{i=1}^n{1}_{E_i}$, and put $S_0=1$, $S_j={\sum }_{|I|=j}\mathrm{Pr}({\bigcap }_{i\in I}{E}_i)$ for $j\ge 1$. Counting the $j$-subsets of the events that occur gives $S_j=\mathbb{E}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Set $S_j=0$ for $j>n$. The Jordan-Bonferroni exact-occurrence inequalities areHere $k\ge 0$ is an integer. To prove both bounds, evaluate the truncated expression before taking its expectation. At an integer value $N=j\ge k$ it equalsFor $j=k$ this is one. For $j>k$, Pascal's identity, or an induction on $m$, givesFor $j<k$ the original expression is zero. Thus its difference from $1_{\{N=k\}}$ is everywhere nonnegative for even $m$ and nonpositive for odd $m$. Taking expectations proves the inequalities, including $k=0$.

For a general nonnegative integer-valued random variable, define its factorial moment by ${\mathbb{E}}_r(X)=\mathbb{E}(X{)}_r$, where $(X{)}_r$ is the falling factorial and ${\mathbb{E}}_0(X)=1$. The same pointwise calculation applies to $X$, so putEvery required moment is finite under the given limit assumption: eventual finiteness of higher factorial moments implies finiteness of the lower ones. The neighbouring upper and lower bounds differ in absolute value byThe last ratio tends to one, and the first factor involving the moment tends to zero. Since $\mathrm{Pr}(X=k)$ lies between every two such neighbouring bounds, each $T_m$ tends to it. Consequently the factorial-moment inversion isThis proves convergence of the signed series without an unjustified interchange with an expectation.

For any prescribed $r\ge 1$, take $\mathrm{Pr}(X=j)=C_r{j}^{-(r+2)}$ for $j=1,2,\dots$, with $C_r=[{\sum }_{j\ge 1}{j}^{-(r+2)}{]}^{-1}$. Since $(j{)}_r\sim j^r$ and $(j{)}_{r+1}\sim j^{r+1}$, the two moment sums have tails comparable respectively to $\sum j^{-2}$ and $\sum j^{-1}$. Thus ${\mathbb{E}}_r(X)<\infty$ but ${\mathbb{E}}_{r+1}(X)=\infty$, as required.