= Solution
Write $N=\sum_{i=1}^n\mathbf1_{E_i}$, and put $S_0=1$, $S_j=\sum_{|I|=j}\Pr(\bigcap_{i\in I}E_i)$ for $j\ge1$. Counting the $j$-subsets of the events that occur gives $S_j=\mathbb E\binom Nj$. Set $S_j=0$ for $j>n$. The <Jordan-Bonferroni exact-occurrence inequalities> are
$$
\boxed{\sum_{r=0}^{2s+1}(-1)^r\binom{k+r}kS_{k+r}\le\Pr(N=k)\le\sum_{r=0}^{2s}(-1)^r\binom{k+r}kS_{k+r}\quad(s\ge0).}
$$
Here $k\ge0$ is an integer. To prove both bounds, evaluate the truncated expression before taking its <expectation>. At an integer value $N=j\ge k$ it equals
$$
\binom jk\sum_{r=0}^m(-1)^r\binom{j-k}r.
$$
For $j=k$ this is one. For $j>k$, <Pascal's identity>, or an induction on $m$, gives
$$
\sum_{r=0}^m(-1)^r\binom{j-k}r=\begin{cases}(-1)^m\binom{j-k-1}m,&m<j-k,\\0,&m\ge j-k.\end{cases}
$$
For $j<k$ the original expression is zero. Thus its difference from $\mathbf1_{\{N=k\}}$ 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 put
$$
T_m=\frac1{k!}\sum_{r=0}^m\frac{(-1)^r\mathbb E_{k+r}(X)}{r!}.
$$
Every 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 by
$$
|T_{m+1}-T_m|=\frac{\mathbb E_{k+m+1}(X)}{k!(m+1)!}=\frac1{k!}\frac{\mathbb E_s(X)s^k}{s!}\frac{(s)_k}{s^k},\qquad s=k+m+1.
$$
The last ratio tends to one, and the first factor involving the moment tends to zero. Since $\Pr(X=k)$ lies between every two such neighbouring bounds, each $T_m$ tends to it. Consequently the <factorial-moment inversion> is
$$
\boxed{\Pr(X=k)=\frac1{k!}\sum_{r=0}^\infty\frac{(-1)^r\mathbb E_{k+r}(X)}{r!}.}
$$
This proves convergence of the signed <series> without an unjustified interchange with an <expectation>.
For any prescribed $r\ge1$, take $\Pr(X=j)=C_rj^{-(r+2)}$ for $j=1,2,\ldots$, with $C_r=[\sum_{j\ge1}j^{-(r+2)}]^{-1}$. 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 \b[$\mathbb E_r(X)<\infty$ but $\mathbb E_{r+1}(X)=\infty$], as required.
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