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.

We will first argue that $z_n \ge e^{-1/2}$, by showing that that for $x \in [0,1], e^{x - x^2/2} \le 1 + x$. 
TO do this, let 
$$ f(x) := e^{x - x^2/2} - 1 - x.$$ Note that, $$ f'(x) = (1-x) e^{x  -x^2/2} - 1, \\ f''(x) = ((1-x)^2 -1)e^{x- x^2/2}$$
Note that the second derivative is non-positive for $x \in [0,1]$. Thus, the derivative is nonincreasing in ths domain. Since $f'(0) = 0,$ the derivative is non-positive in $[0,1]$ - i.e., $f$ is nonincreasing on $[0,1]$. Finally, since $f(0) = 0,$ this tells us that for $x \in [0,1], f(x) \le 0 \iff e^{x- x^2/2} \le 1 + x$.
Using this for $x = 1/n,$ we find that $ e^{1/n - 1/2n^2} \le 1 + 1/n,$ for $n \ge 1,$ and thus 
$$(1 + 1/n)^{n^2} e^{-n} \ge (e^{1/n - 1/2n^2})^{n^2} e^{-n} = e^{-1/2}.$$
We will now get an upper bound on the $z_n$. By truncating the series, we get that $e^{u} \ge 1 + u + u^2/2$ for $u \ge 0$. Now, let $x_n = \sqrt{1+2/n} - 1$. Note that $x_n + x_n^2/2 = 1/n$. So, we find that $(1+1/n) \le e^{x_n},$ and thus $$ (1 + 1/n)^{n^2} e^{-n} \le e^{n^2 x_n - n}. $$
Since $\exp(\cdot)$ is continuous, if we can argue that $ n^2 x_n - n \to -1/2,$ then we'd be done by the sandwich theorem. We can show this via a Taylor expansion: $\sqrt{1 + u} = 1 + u/2 - u^2/8 + O(u^3).$
Thus, $$ n^2 x_n - n = n^2 \left( 1 + \frac{(2/n)}{2} - \frac{(4/n^2)}{8} + O(n^{-3}) -1\right) - n = - \frac{1}{2} + O(n^{-1}),$$ and we're done.
