Finding the limit of a sequence with an undesirable $\ln k$

EDITED 2. Here is another approach: The starting point is the following identity

$$ \int_{0}^{\infty} \frac{(1 - e^{-x})^{n}}{x^{m}} \, dx = \frac{(-1)^{m}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m-1} \log k. \tag{1} $$

which I proved here. Plugging $m = n$ to $\text{(1)}$, we have

$$ A_{n} := \frac{(-1)^{n}n^{2}}{n!} \sum_{k=2}^{n} \binom{n}{k}(-1)^{k} k^{n-1} \log k = \int_{0}^{\infty} n \left( \frac{1 - e^{-x}}{x} \right)^{n} \, dx. $$

Now let $f(x) : [0, \infty) \to (0, 1]$ be defined by $f(x) = (1 - e^{-x})/x$. This function is monotone decreasing, and let us denote its inverse by $g = f^{-1}$. Then it follows from the substitution $x = g(y)$ that

$$ A_{n} = \int_{0}^{1} n y^{n} \left| g'(y) \right| \, dy. $$

By observing that $y \mapsto ny^{n}$ behaves as an approximation to the identity of the unit mass $\delta_{1}$, it follows that

$$ \lim_{n\to\infty} A_{n} = \left| g'(1) \right| = \frac{1}{\left| f'(0) \right|} = 2. $$

EDITED. I finally came out with a complete solution:

To avoid taking account of the cancellation between large quantities, we find a more tractable representation. Let

$$ A_{n} = \frac{(-1)^{n}n^{2}}{n!} \sum_{k=2}^{n} \binom{n}{k}(-1)^{k} k^{n-1} \log k $$

and introduce

$$ f_{n}(x) = \frac{n^{2} x^{n-2} \log x}{(x-1)\cdots(x-n)}. $$

Then we have

$$ \operatorname{Res}\limits_{z=k} \, f_{n}(z) = \frac{(-1)^{n-k} n^{2} k^{n-1} \log k}{k!(n-k)!} $$

and hence for any simple closed contour $C \subset \Bbb{C} \setminus (-\infty, 0]$ winding $\{1, \cdots ,n\}$ in counter-clockwise direction it follows that

$$ \int_{C} f_{n}(z) \, dz = 2\pi i A_{n}. $$

But since $f_{n}(z) = O(|z|^{-2}\log|z|)$ as $|z| \to \infty$, we can modify contour so that $C$ winds $(-\infty, 0]$, and a simple calculation shows that

$$ A_{n} = \int_{0}^{\infty} \frac{n^{2} x^{n-2}}{(x+1)\cdots(x+n)} \, dx. \tag{1} $$

Using the substitution $x \mapsto n^{2}/x$, it follows from $\text{(1)}$ that

$$ A_{n} = \int_{0}^{\infty} \prod_{k=1}^{n} \left( 1 + \frac{kx}{n^{2}} \right)^{-1} \, dx. \tag{2} $$

Now notice that when $n \geq 2$, expanding the denominator and discarding higher order term, we have

\begin{align*} \prod_{k=1}^{n} \left( 1 + \frac{kx}{n^{2}} \right) &\geq 1 + \sum_{1\leq k \leq n} \frac{kx}{n^{2}} + \sum_{1 \leq l < k \leq n} \frac{kl x^{2}}{n^{4}} \\ &= 1 + \frac{n+1}{2n} x + \frac{(3n+2)(n^{2}-1)}{24n^{3}} x^{2} \\ &\geq 1 + \frac{x}{2} + \frac{x^{2}}{8}. \end{align*}

This means that the integrands of $\text{(2)}$ are bounded by an integrable function $1/(x^{2}/8 + x/2 + 1)$ and thus we can apply the dominated convergence theorem, once we prove that the integrand converges pointwise. But it is immediate that for any fixed $x$,

$$ \log \prod_{k=1}^{n} \left( 1 + \frac{kx}{n^{2}} \right) = \sum_{k=1}^{n} \log \left( 1 + \frac{kx}{n^{2}} \right) = \frac{x}{2} + \mathcal{O}\left(\frac{1}{n}\right) \longrightarrow \frac{x}{2} $$

and therefore

$$ \lim_{n\to\infty} A_{n} = \int_{0}^{\infty} e^{-x/2} \, dx = 2. $$