I would like to prove $$(1-1/n)^{n \;-\; 1/2} < 1/e$$ where $n \geq 2$.
I saw the post Proving $(1 + 1/n)^{n+1} \gt e$, but it is not the same, and this answer (which is close to my question) involves some theorems I don't know.
I tried to take $\;\ln\;$ of both sides: $$g(n) := (n - 1/2) \ln(1-1/n) < -1$$ Proving that $g$ is increasing (by computing $g'$) seems difficult and painful. Proving that $g(n) \to -1$ when $n \to\infty$ shouldn't be hard, however.
Proof via differentiation is feasible. Put, for all $t>1$ :
$$u(t)=\left(t-\frac12\right)\ln \left(1-\frac1t\right)$$
Then :
$$u'(t)=\ln\left(1-\frac1t\right)+\left(t-\frac12\right)\frac{\frac1{t^2}}{1-\frac1t}=\ln\left(1-\frac1t\right)+\frac{t-\frac12}{t^2-t}$$
$$u''(t)=\frac1{t^2-t}+\frac{t^2-t-(2t-1)(t-\frac12)}{(t^2-t)^2}=\frac{-1}{2(t^2-t)^2}<0$$
Hence $u'$ is decreasing. But $\displaystyle{\lim_{+\infty}u'=0}$ and so $u'>0$, so that $u$ is increasing. Finally, since $\displaystyle{\lim_{+\infty}u}=-1$, we conclude that $\forall t>1,u(t)<-1$, hence :
$$\boxed{\forall t>1,\left(1-\frac1t\right)^{t-1/2}<\frac1e}$$
It can be seen as a consequence of the Hermite-Hadamard inequality, or just convexity.
$$-\left(n-\frac{1}{2}\right)\log\left(1-\frac{1}{n}\right) = \left(n-\frac{1}{2}\right)\int_{n-1}^{n}\frac{dx}{x}=\left(n-\frac{1}{2}\right)\int_{-1/2}^{1/2}\frac{dx}{\left(n-\frac{1}{2}\right)+x} $$ can be written as $$ \int_{-1/2}^{1/2}\frac{dx}{1+\frac{x}{2n-1}} = 2\int_{0}^{1/2}\frac{dx}{1-\left(\frac{x}{2n-1}\right)^2}> 2\int_{0}^{1/2}\,dx=1$$ hence by exponentiation $$ \left(1-\frac{1}{n}\right)^{n-1/2}<\frac{1}{e}$$ follows.
