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.
 A: The left hand side is
$$\frac1{(n+1)^3}\cdot((n+1)^2-n)(n+2)=\frac{n^3+3n^2+3n+2}{(n+1)^3} $$
A: A tricky way to verify the monotonicity is by applying AM-GM inequality as follows:
\begin{align*}
a_n & = \left(1 + \frac{1}{n}\right)^n \\
& = 1\times \left(1 + \frac{1}{n}\right) \times \cdots \times \left(1 + \frac{1}{n}\right)\\
& \leq \left[\frac{1 + n\left(1 + \frac{1}{n}\right)}{n + 1}\right]^{n + 1} \\
& = \left(1 + \frac{1}{n + 1}\right)^{n + 1} = a_{n + 1}.
\end{align*}
A: $$\frac { a_{ n+1 } }{ a_{ n } } =\frac { \left( 1+\frac { 1 }{ n+1 }  \right)  }{ { \left( 1+\frac { 1 }{ n }  \right)  }^{ n } } ^{ n+1 }=\left( 1-\frac { 1 }{ (n+1)^{ 2 } }  \right) ^{ n+1 }\frac { n+1 }{ n } >\left( 1-\frac { 1 }{ n+1 }  \right) \frac { n+1 }{ n } =1$$
A: 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. 
A: It can be proved using function. We have $f(x) = \left(1+\dfrac{1}{x}\right)^x, x \in [1,\infty)$ and taking log: $\log f(x) = x\log(x+1) - x\log x\implies f'(x) = f(x) \left(\log(x+1)+ \dfrac{x}{x+1} - \log x - 1\right)= f(x)\left(\log(x+1) - \log x - \dfrac{1}{x+1}\right)= f(x)g(x)\implies g'(x) = \dfrac{1}{x+1}-\dfrac{1}{x}+\dfrac{1}{(1+x)^2}= \dfrac{x(x+1)-(x+1)^2+x}{x(1+x)^2}= -\dfrac{1}{x(1+x)^2} < 0\implies g(x) > \displaystyle \lim_{x \to \infty} g(x)=0\implies f'(x) > 0\implies f(n+1) > f(n)$.
