Show that $\forall n \in\Bbb N: e < \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right)$

The first sketch works.

We need to prove that $$\frac{x_n}{x_{n+1}}>1$$ or $$\frac{\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{2n}\right)}{\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{2n+2}\right)}>1$$ or $$\frac{(n+1)^n(2n+1)(n+1)^{n+1}(2n+2)}{n^n(n+2)^{n+1}2n(2n+3)}>1$$ or $$\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}>\frac{2n+3}{2n+1}$$ or$$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}>\frac{2n+3}{2n+1},$$ which is true because $$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}=\left(1+\frac{1}{n^2+2n}\right)^{n+1}>1+\frac{n+1}{n^2+2n}+\frac{(n+1)n}{2(n^2+2n)^2}>\frac{2n+3}{2n+1}.$$

$$\frac{\log(1+x)}{x}+\log\left(1+\frac{x}{2}\right)=1+\frac{5x^2}{24}-\frac{5x^3}{24}+\frac{59 x^4}{320}-\ldots=1+\sum_{m\geq 2}a_m x^m $$ for any $x\in(0,1)$, where the sequence $\{a_m\}_{m\geq 2}$ can be proved to have alternating signs and the sequence $\{|a_m|\}_{m\geq 2}$ can be checked to be weakly decreasing by induction on $m$. It follows that the LHS is $>1$ for any $x\in(0,1)$: your inequality follows by exponentiating both sides (with $x$ being chosen as $\frac{1}{n}$) and by checking $n=1$ as a separate case.