<s>[INST] Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that 
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
 [/INST] We start from $\left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right)$:
$ \left( 1 - \frac{n}{(n+1)^2}\right)\left(\frac{n+2}{n+1}\right) =  \frac{((n+1)^2 - n)(n+2)}{(n+1)^3}$
We then set up the inequality and simplify in order to see if we can get a trivial inequality.
$ \frac{((n+1)^2 - n)(n+2)}{(n+1)^3} \geq 1 \Leftrightarrow (n+2)((n+1)^2 - n) \geq (n+1)^3$
$(n+2)((n+1)^2 - n) = (n+1 + 1)((n+1)^2 - n) = (n+1)^3 - (n)(n+1) + (n+1)^2 - n$
$(n+1)^3 - (n)(n+1) + (n+1)^2 - n \geq (n+1)^3 \Leftrightarrow -n(n+1) + (n+1)^2 - n \geq 0$
$-n(n+1) + (n+1)^2 - n \geq 0 \Leftrightarrow (n+1)^2  = n^2 + 2n + 1\geq n + n(n+1) = n^2 + 2n$
$n^2 + 2n + 1\geq n^2 + 2n \Leftrightarrow 1 \geq 0$
Since $1 \geq 0$ the inequality holds. 
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