<s>[INST] Proof verification of $\{x_n\} = \left(1 + {1\over 2n}\right)^n$ is an increasing sequence. 
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.
 [/INST] $b_n=(1+1/n)^n$ is increasing. Your sequence is $x_n=\sqrt{b_{2n}}$, so it is increasing.
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