Let $$S_n=\lim_{n\to \infty} \sum_{k=2}^{n} 2^k {n \choose k} (n-2)^{n-k} n^{-n}, \quad L=\lim_{n \to \infty} S_n$$ $$S_n=\left(1-\frac{2}{n}\right)^n\sum_{k=2}^{n} {n \choose k}\left(\frac{2}{n-2}\right)^k =\left(1-\frac{2}{n}\right)^n \left[\left(\frac{n}{n-2}\right)^n-1-\frac{2n}{n-2}\right].$$ $$\implies S_n=\left(1-\frac{2}{n}\right)^n\left[\left(\frac{1}{1-2/n}\right)^n-1-2\left(\frac{1}{1-2/n}\right)\right].$$ When $n$ is large,neglecting $2/n$ in the third term we can write $$S_n\sim1-3\left(1-\frac{2}{n}\right)^{n}\implies L= \lim_{n \to \infty} S_n=1-3e^{-2}.$$
The question: Is this calculation rigorous, please comment.