Convergence of $\lim _{n \rightarrow \infty} n\left(1+\frac{1}{n}\right)^{-n^{2}}$ Assume we have $\lim _{n \rightarrow \infty} n\left(1+\frac{1}{n}\right)^{-n^{2}}$. My idea was $\lim _{n \rightarrow \infty} n({\left(1+\frac{1}{n}\right)^{n}})^{-n^{2}/n}$. So the inner part converges to $e$ for $n \to \infty$ and we have $\lim _{n \rightarrow \infty} ne^n \neq 0 $. Why is that wrong ?
 A: You have $\frac{-n^2}{n} = -n$ so the limit converges to $ne^{-n} \to 0$ as $n \to \infty$
A: In the same spirit as Daniel Schepler in his comment, consider
$$A_n=\left(1+\frac{1}{n}\right)^{-n^{2}}\implies \log(A_n)=-n^2 \log\left(1+\frac{1}{n}\right)$$
Using the Taylor expansion of the logarithm,
$$\log(A_n)=\sum_0^\infty \frac{(-1)^k}{k+2} n^k=-n+\frac{1}{2}-\frac{1}{3 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
$$A_n=e^{\log(A_n)}=\sqrt e\, e^{-n} \exp\left(-\frac{1}{3 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ Continuing with Taylor series for the exponential
$$A_n=\sqrt e\, e^{-n}\left(1-\frac{1}{3 n}+\frac{11}{36 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
$$ n\left(1+\frac{1}{n}\right)^{-n^{2}}=\sqrt e\,n\, e^{-n}\left(1-\frac{1}{3 n}+\frac{11}{36 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ which, for sure, shows the limit when $n \to \infty$.
But it also give a shortcut evaluation of the $n^{th}$ term. For example, using $n=5$, the exact value is
$$\frac{1490116119384765625}{28430288029929701376}\approx 0.052413$$ while the above truncated expansion gives
$$\frac{851}{180\, e^{9/2}}\approx 0.052521$$
