Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence Given 
$$
y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.
$$
Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ?  I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.

Note that
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}}
&=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\
&=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\
&=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\
&\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\
&=1
\end{align}
$$
Therefore,
$$
\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}
$$

Similarly,
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\left(\frac{n+2}{n+1}\right)^{n+1}\left(\frac{n}{n+1}\right)^n\\
&=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\
&=\frac{n+1}{n}\left(1-\frac1{(n+1)^2}\right)^{n+1}\\
&\ge\frac{n+1}{n}\left(1-\frac{n+1}{(n+1)^2}\right)\\
&=1
\end{align}
$$
Therefore,
$$
\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n
$$

Bernoulli's Inequality
In the preceding, we used Bernoulli's Inequality: for all $x\ge-1$ and non-negative integer $n$,
$$
(1+x)^n\ge1+nx
$$
This can be proven by induction:
Note that the inequality above is true for $n=0$.
Suppose that $x\ge-1$ and for a non-negative integer $n$, we have
$$
(1+x)^n-nx\ge1
$$
Then
$$
\begin{align}
(1+x)^{n+1}-(n+1)x
&=(1+x)^n-nx+x(1+x)^n-x\\
&\ge1+x((1+x)^n-1)\\
&\ge1
\end{align}
$$
If $-1\le x\le0$, then both $x$ and $(1+x)^n-1$ are negative. If $x\ge0$, then both both $x$ and $(1+x)^n-1$ are positive. Therefore, if $x\ge-1$, $x((1+x)^n-1)\ge0$. This justifies the last inequality above.
Note that if $x\ne0$ and $n\ge1$, the last inequality is strict. Thus, for $x\ne0$ and $n\ge2$, we have
$$
(1+x)^n\gt1+nx
$$

Negative Exponents
Bernoulli's Inequality is also true for negative integer exponents. That is, for $x\gt-1$ and non-negative $n\in\mathbb{Z}$,
$$
1-nx\le(1+x)^{-n}
$$
Suppose that
$$
(1-nx)(1+x)^n\le1
$$
which is trivially true for $n=0$, and strictly true for $x\ne0$ and $n=1$. Then
$$
\begin{align}
(1-(n+1)x)(1+x)^{n+1}
&=(1-nx)(1+x)^n-(n+1)x^2(1+x)^n\\
&\le1
\end{align}
$$
Therefore, for all non-negative $n\in\mathbb{Z}$,
$$
(1-nx)\le(1+x)^{-n}
$$
where the inequality is strict for $x\ne0$ and $n\ge1$.
