I need to prove that for every $l \geq 3$, the $\Big( 1- \dfrac{1}{2 \cdot l}\Big)^{2 \cdot l} < \dfrac{1}{e}$ holds. ($l$ is integer)
This is what I tried so far.
$$ \begin{align} x &= \dfrac{1}{2\cdot l}\\ &\Rightarrow \Big(1-x \Big)^\dfrac{1}{x} < \dfrac{1}{e}\\ &\Rightarrow \dfrac{1}{x} \ln \Big(1-x \Big) < -1\\ &\Rightarrow \ln \Big(1-x \Big) < -x \end{align} $$
and I failed to prove the rest of the proof.
Let $$f(x)=\ln(1-x)+x$$ We know, $f(1/6)<0$. Moreover, $$f'(x)=1-\frac1{1-x}<0$$ for all $0<x<1$. So, $f$ is decreasing. Thus, $f(x)<f(1/6)<0$ for all $x<1/6$, i.e. $l>3$.