Let $n\in \mathbb N$ and: $$ x_n = \left(1 + {1\over 2n}\right)^n $$ Show that $\{x_n\}$ is an increasing sequence.
$\Box$ Consider ratio test of two consequent terms $x_n$ and $x_{n+1}$: $$ \frac{x_{n+1}}{x_n} = \frac{\left(1 + {1\over 2n + 2}\right)^{n+1}}{\left(1 + {1\over 2n}\right)^n} = \frac{\left(1 + {1\over 2n + 2}\right)^{n}}{\left(1 + {1\over 2n}\right)^n} \cdot\left(1 + {1\over 2n + 2}\right) = \\ = \left(\frac{2n(2n+3)}{(2n+1)(2n+2)}\right)^n \cdot\left(1 + {1\over 2n + 2}\right) $$
We are done in case this product is greater than $1$.
Denote: $$ P^n = \left(\frac{2n(2n+3)}{(2n+1)(2n+2)}\right)^n = \left(\frac{4n^2 + 6n}{4n^2 + 6n+2}\right)^n $$
Split $P$ into partial fractions: $$ P^n = \left(1 + \frac{1}{n+1} - \frac{2}{2n+1}\right)^n = \left(1 - \frac{1}{(n+1)(2n+1)}\right)^n $$
Since $\frac{-1}{(n+1)(2n+1)} > -1$ we may apply Bernoulli's: $$ P^n \ge 1 - \frac{n}{(n+1)(2n+1)} $$
Thus:
$$ \frac{x_{n+1}}{x_n} \ge \left(1 - \frac{n}{(n+1)(2n+1)}\right)\cdot\left(1+ \frac{1}{2n+2}\right) = \frac{2n^2 + 2n +1}{2n^2+3n+1} \cdot \frac{2n+3}{2n+2} = \\ = \frac{4 n^3 + 10 n^2 + 8 n + 3}{4 n^3 + 10 n^2 + 8 n + 2} $$
From here it's clear that:
$$ {x_{n+1}\over x_n} > 1 $$
This completes the proof that $x_n$ is monotonically increasing. ${\blacksquare}$
Is this a valid proof? Also I would appreciate any simpler methods to show that.
$b_n=(1+1/n)^n$ is increasing. Your sequence is $x_n=\sqrt{b_{2n}}$, so it is increasing.