How to show a sequence of functions is increasing How does one show that for all $x\geq 0$ that the following sequence of functions is increasing where $f_n$ is defined by
$$f_n(x)= x\left(1+\frac{x^2}{n}\right)^n$$
using the fact that for all $y\geq 0$
$$\frac{y}{y+1}\leq \ln(1+y) \leq y$$
I have been trying different ways including trying to show $\frac{f_{n+1}}{f_n}$ is positive but I’m obviously missing the trick.
 A: Proof only using AM$-$GM: if $x\geq 0$ then proving $f_{n+1}(x)\geq f_{n}(x)$ is equivalent to prove $$\left(1+\frac{x^2}{n}\right)^{n}$$
is an incresing sequence. Using AM$-$GM inequality we have that
\begin{align*}
\left(1+\frac{x^2}{n}\right)^{\frac{n}{n+1}}&=\sqrt[n+1]{1\cdot\underbrace{\left(1+\frac{x^2}{n}\right)\cdot\left(1+\frac{x^2}{n}\right)\cdots\left(1+\frac{x^2}{n}\right)}_{n~\text{times}}}\\
\\
&\leq\frac{1+\overbrace{\left(1+\frac{x^2}{n}\right)+\left(1+\frac{x^2}{n}\right)+\cdots+\left(1+\frac{x^2}{n}\right)}^{n~\text{times}}}{n+1}\\
&=1+\frac{x^2}{n+1}
\end{align*}
and raising both sides to $n+1$ gives
$$\left(1+\frac{x^2}{n}\right)^{n}\leq\left(1+\frac{x^2}{n+1}\right)^{n+1}$$
In fact we have the strict inequality, since $1\neq 1+\frac{x^2}{n}$.
A: To show $\{f_n(x)\}$ is increasing is equivalent to showing $\{\ln f_n\}$ is increasing. For fixed $x\ge0$,
$$ \frac{\partial \ln f_n(x)}{\partial n}=\ln(1+\frac{x^2}{n})+n\frac{1}{1+\frac{x^2}{n}}(-\frac{x^2}{n^2})=\ln(1+\frac{x^2}{n})-\frac{x^2}{n+x^2}\ge \frac{x^2}{n}-\frac{x^2}{n+x^2}\ge0$$
implies that $\{\ln f_n\}$ is increasing.
