Show that $(1+\frac 1n)^{n^2} \mathrm e^{-n}$ is not a zero sequence without logarithms How can I show that
$z_n = \left(1+\dfrac 1n \right)^{n^2} \mathrm e^{-n}$
is not a zero sequence? WolframAlpha says, it converges to $\frac{1}{\sqrt{e}}$ but how can I prove? I'd appreciate a solution without logarithms.

Not a Zero-Sequence
In this answer, it is shown, using only an extension of Bernoulli's Inequality (which is proven there by induction), that $\left(1+\frac1n\right)^{n+1/2}$ is decreasing. Since the limit is $e$, this means
$$
\left(1+\frac1n\right)^{n+1/2}\ge e\tag1
$$
Raising both sides to the $n-1/2$ power, we get
$$
\left(1+\frac1n\right)^{n^2-1/4}\ge e^{n-1/2}\tag2
$$
Thus,
$$
\left(1+\frac1n\right)^{n^2}e^{-n}\ge e^{-1/2}\left(1+\frac1n\right)^{1/4}\tag3
$$
Therefore,
$$
\liminf_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}\ge e^{-1/2}\tag4
$$
which is definitely non-zero.

The Limit is $\boldsymbol{e^{-1/2}}$
To show the limit is $e^{-1/2}$ without logarithms requires a bit more work, but can be done using the same extension to Bernoulli's Inequality (case $m=2$ of Theorem $1$).
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{2n+1-2a}}{\left(1+\frac1{n-1}\right)^{2n-1-2a}}
&=\frac{n+1}{n-1}\left(1-\frac1{n^2}\right)^{2n}\left(1+\frac1{n^2-1}\right)^{2a}\\
&\ge\scriptsize\frac{n+1}{n-1}\left(1-\frac2n+\frac{2n-1}{n^3}-\frac{(2n-1)(2n-2)}{3n^5}\right)\left(1+\frac1{n^2-1}\right)^{2a}\\
&=\left(1-\frac{n^2+2n-2}{3n^5}\right)\left(1+\frac1{n^2-1}\right)^{2a}\\
&\ge\left(1-\frac{n^2+2n-2}{3n^5}\right)\left(1+\frac{2a}{n^2}\right)\\
&=1+\frac{2a}{n^2}-\frac1{3n^3}-\frac2{3n^4}+\frac{2-2a}{3n^5}-\frac{4a}{3n^6}+\frac{4a}{3n^7}\tag5
\end{align}
$$
Therefore, for any $a\gt0$, there is an $n_a$ so that for $n\ge n_a$,
$$
\left(1+\frac1n\right)^{n+\frac12-a}\ge\left(1+\frac1{n-1}\right)^{n-\frac12-a}\tag6
$$
Since it is increasing to $e$, we have
$$
\left(1+\frac1n\right)^{n+\frac12-a}\le e\tag7
$$
Raising both sides to the $n-\frac12+a$ power, we get
$$
\left(1+\frac1n\right)^{n^2-\left(\frac12-a\right)^2}\le e^{n-\frac12+a}\tag8
$$
That is, for $n\ge n_a$,
$$
\left(1+\frac1n\right)^{n^2}e^{-n}\le e^{-\frac12+a}\left(1+\frac1n\right)^{\left(\frac12-a\right)^2}\tag9
$$
Thus, for any $a\gt0$,
$$
\limsup_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}\le e^{-\frac12+a}\tag{10}
$$
Therefore, $(4)$ and $(10)$ show that
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}=e^{-1/2}\tag{11}
$$
