I hope I understood your question correctly.
For any sequence of real numbers $(\alpha_n)_{n\geq 0}$ and $p\in\mathbb N$ let us denote by $(\alpha_n^{*p})$ (awkward notation) the sequence defined by
$$\alpha_n^{*p}=\sum_{k=0}^n \left(\frac{n^{\underline k}}{n^k}\right)^p \alpha_k\, . $$
Then, the following holds true: for any absolutely convergent series $\sum\alpha_k$, the sequence $(\alpha_n^{*p})_{n\geq 0}$ is convergent with limit $\sum_0^\infty\alpha_k$. In particular, with $\alpha_k=\frac{(-1)^k}{k!}$ and $p=2$ you get that $c_n\to1/e$.
To see this, note that one can write
$$\alpha_n^{*p}=\sum_{k=0}^\infty q_{n,k}\, \alpha_k\, ,$$
where
$$q_{n,k}=\left\{ \begin{matrix}
\left(\frac{n^{\underline k}}{n^k}\right)^p&k\leq n\\0&k>n
\end{matrix}\right. $$
For each fixed $k$ we have $\lim_{n\to\infty} q_{n,k}=1$ by the formula for $\frac{n^{\underline k}}{n^k}$ you give in your question. Moreover, since $0\leq q_{n,k}\leq 1$ by the same formula, we also have $\vert q_{n,k}\alpha_k\vert\leq \vert\alpha_k\vert$ for all $n$ and every $k\geq 0$. By the dominated convergence theorem (for series), it follows that
$$\lim_{n\to\infty} \sum_{k=0}^\infty q_{n,k}\alpha_k=\sum_{k=0}^\infty\alpha_k\, , $$
which is the required result.
Best Answer
Note that $$\frac{A(n)}{B(n)}=\left(1-\frac1{n^2}\right)^n\to 1 $$ because $$ 1\ge \left(1-\frac1{n^2}\right)^n\ge 1-\frac1n$$ by Bernoulli's inequality. Therefore, if either of $\lim A(n)$, $\lim B(n)$ exists, so does the other and is equal.