Short version: can we prove that
$$\sum_{k=0}^n (-1)^k \binom{n}{k}^2 \frac{k!}{n^{2k}} \to \frac1e$$
as $n \to \infty$?
Long version: First, consider
$$a_n = \sum_{k=0}^n \frac{(-1)^k}{k!}$$
It is well-known that $a_n \to \dfrac1e$ as $n \to \infty$; indeed $a_n = \dfrac{!n}{n!}$ is the truncation to the first $n$ terms of the power series for $e^x$, evaluated at $x = -1$ (where $!n$ denotes the subfactorial; it is also equal to the number of derangements on $n$ elements). There is also a simple expression for the generating function $A(z) = \sum_{n=0}^{\infty} a_n z^n$, which is $A(z) = \dfrac{e^{-z}}{1-z}$. (See Exponential Generating Functions For Derangements.)
Next, consider
$$b_n = \sum_{k=0}^n \frac{n^{\underline k}}{n^k} \frac{(-1)^k}{k!}$$
where $n^{\underline k} = \binom{n}{k}k!$ denotes a falling factorial, so the extra factor $\frac{n^{\underline k}}{n^k}$ is $\frac{n(n-1)\cdots(n-k+1)}{n(n)\cdots(n)} = 1\cdot\left(1 – \frac1n\right)\cdot\left(1 – \frac2n\right)\cdots\left(1 – \frac{k-1}n\right)$ which for large $n$ (and fixed $k$) is close to $1$.
I don't know if $b_n$ has a simple form for its generating function too, but it is easy to see that $b_n \to \dfrac1e$ as well; indeed by the binomial theorem we have $b_n = \sum_{k=0}^n \binom{n}{k} (\frac{-1}{n})^k = \left(1 – \frac1n \right)^n$ which is well-known to converge to $\frac1e$ (indeed such a limit is sometimes taken to be the definition of $e^x$).
Finally, consider
$$c_n = \sum_{k=0}^n \frac{n^{\underline k}}{n^k} \frac{n^{\underline k}}{n^k} \frac{(-1)^k}{k!}$$
This is the same transformation going from $b_n$ to $c_n$ as from $a_n$ to $b_n$. But can we prove that $c_n \to \frac1e$ too? (And can we write down its generating function compactly, perchance?) More generally, what techniques exist that help in proving something about $\sum t_n s_n$, given $\sum s_n$?
This question arose from an attempt to answer this question, where I arrived at the expression $c_n$ above (there I called it $P_{n, n, 0}$; next I'll try to understand $P_{m, w, k}$).
[Note: I'm tagging this special-functions too, as I understand $c_n$ has something to do with hypergeometric functions / Bessel functions / something like that.]
Best Answer
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.