Monotonicity of (1 + 1/n)^n

1. $(a_n)$ increases. Apply AM-GM to the $n+1$ numbers $\underbrace{1+\frac1n,\dots,1+\frac1n}_{n},\,1$. Their mean is $\frac{n+1+1}{n+1}=1+\frac1{n+1}$, so
$$\left(1+\frac1n\right)^{n/(n+1)}\le1+\frac1{n+1}\ \Longrightarrow\ a_n\le a_{n+1}.$$

$(b_n)$ decreases. Apply AM-GM to the $n+2$ numbers $\underbrace{\frac n{n+1},\dots,\frac n{n+1}}_{n+1},\,1$. Their mean is $\frac{n+1}{n+2}$, so
$$\left(\frac n{n+1}\right)^{(n+1)/(n+2)}\le\frac{n+1}{n+2}\ \Longrightarrow\ \left(\frac{n}{n+1}\right)^{n+1}\le\left(\frac{n+1}{n+2}\right)^{n+2}.$$
Taking inverses, $\left(\frac{n+1}n\right)^{n+1}\ge\left(\frac{n+2}{n+1}\right)^{n+2}$, i.e. $b_n\ge b_{n+1}$.

2. We have $a_n\le b_n$ for all $n$, and
$$b_n-a_n=\frac1n a_n\le\frac{b_1}n\to0$$
since $a_n\le b_n\le b_1$. Therefore $a_n$ is increasing and bounded above by $b_1$, so it converges. Also $b_n$ is decreasing and bounded below by $a_1$, so it converges. they have the same limit, which is called $e$.