The sequence $a_n = \left(\frac{n}{n+1}\right)^{n+1}$ is increasing I just wanna some help showing that the sequence $a_n = \left(\frac{n}{n+1}\right)^{n+1}$ is increasing, since it came up in some stuff I was doing and I'm not finding a quick solution. 
I actually just need it to be bounded below by a positive number, but I tested a bunch of values and it's increasing so I'm pretty sure it is.
 A: We need only use Bernoulli's inequality and some straightforward arithmetic.  To that end, we have
$$\begin{align}
\frac{a_{n+1}}{a_{n}}&=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\,\frac{n}{n+1}\\\\
&= \left(1+\frac{1}{n(n+2)}\right)^{n+2}\frac{n}{n+1}\\\\
&\ge \left(1+\frac1n\right)\frac{1}{1+\frac1n}\\\\
&=1
\end{align}$$
And we are done.
A: There is indeed an easy way to prove this by using AM-GM inequality for the following $n+2$ numbers:
$$1,\frac{n}{n+1},\dots,\frac{n}{n+1},$$
where the fraction $\frac{n}{n+1}$ is repeated $n+1$ times. These numbers have as sum $n+1$. By AM-GM we get:
$$\frac{n+1}{n+2}>\sqrt[n+2]{\left(\frac{n}{n+1}\right)^{n+1}},$$
which is obviously equivalent with the sequence $(a_n)$ being increasing. The inequality is strict since $1\ne \frac{n}{n+1}$.
A: I don't know if the process is appropriate in this case or if my methods are rigorus, but I think this can be done via Induction. If you can prove each term is greater than the previous than it follows that the sequence is increasing.
Let $P(n) : a_n \geq a_{n-1}$ or equivalently $P(n): (\frac{n}{n + 1})^{n+1} \geq (\frac{n - 1}{n})^n $ for $n \in \mathbb{Z}^+ | n \geq 1$. 
First check the base case $P(1)$
\begin{align}
(\frac{1}{2})^{2} > (\frac{0}{1})^1 
\end{align} 
Therefore $P(1)$ holds. Then assume $P(k)$ holds. Now we must show that $P(k + 1)$ holds. Now from the inductive step it follows that
\begin{align}
(\frac{k}{k + 1})^{k+1} \geq (\frac{k - 1}{k})^k
\end{align}
Recall that if $a \geq b$ and $c \geq d$ then $ac \geq bd$. If we can prove that $\frac{k}{k+1} > \frac{k - 1}{k}$  $\forall k \geq 1$ then the proof is complete. I will prove it from the fact that $\frac{1}{x-1} > \frac{1}{x} \forall x \geq 2$
\begin{align}
\frac{1}{x-1} &> \frac{1}{x}\\
1-\frac{1}{x - 1} &< 1 - \frac{1}{x}\\
1-\frac{1}{x-1} &< \frac{x - 1}{x}\\
\end{align}
Let $k = x - 1$, therefore $k \geq 1$:
\begin{align}
1 - \frac{1}{k} &< \frac{k}{k+1}\\
\frac{k - 1}{k} &< \frac{k}{k+1}\\
\end{align}
Now that this has been proved, we can use the multiplicative principle and the inductive step
\begin{align}
(\frac{k}{k + 1})^{k+1} \cdot \frac{k}{k+1} &\geq (\frac{k - 1}{k})^k \cdot \frac{k - 1}{k}\\
(\frac{k}{k + 1})^{k+2} &\geq (\frac{k - 1}{k})^{k + 1}\\
\end{align}
Therefore $P(k+1)$ holds if $P(k)$ does. From the principle of mathematical induction it follows that $P(n)$ holds for $n \in \mathbb{Z}^+ | n \geq 1$ holds as well. From this it follows that the sequence must be increasing.
