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.
 A: I think it needs just some basic algebraic manipulations: $$\frac{y_{n}}{y_{n+1}}=\frac{(1+\frac{1}n)^{n+1}}{(1+\frac{1}{n+1})^{n+2}}$$ You can show that the latter fraction is equal to $$(1+\frac{1}{n^2+2n})^{n+1}\times\frac{1}{1+1/(n+1)}$$ But $$(1+\frac{1}{n^2+2n})^{n+1}\ge {1+1/(n+1)}$$ Note that if $x\ge-1$ then $(1+x)^n\ge 1+nx.$
A: 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 $n\ge1$.
A: If you don't care about overkilling the Problem you can even do this with calculus, showing the derivative in n is negative. 
The derivative is 
$$ \frac{\left(\frac{1}{n}+1\right)^n (n+1) \left(n \log \left(\frac{1}{n}+1\right)-1\right)}{n^2}$$ and so we only need to show that 
$$1> n \log(1+\frac{1}{n})$$
By substitution $n=\frac{1}{x}$ we have 
$$\frac{\log(1+x)}{x}=\frac{\ln(1+x)-\ln(1)}{(x+1)-1}=\frac{1}{1+\xi}$$
with $\xi \in (0,x)$ (this is granted by the mean value theorem), and the last expression is less than 1.
A: Use the first derivative test and prove the function
$$ f(x)=\left(1+\frac{1}{x}\right)^{x+1}\hspace{-6 mm},\qquad   \quad x \geq 1, $$
is decreasing on $[1,\infty]$. That is prove $f'(x)<0$ on $[1,\infty]$. 
A: Lemma(1):
Let be $a,b\in\mathbb R^+$
$$\sqrt[n+1]{ab^n}\le \dfrac{a+bn}{n+1}$$
Proof:
$$\sqrt[n+1]{a\underbrace{bbb...b}_{n\;times}}\le \dfrac{a+b+b+b+...+b}{n+1}=\dfrac{a+bn}{n+1}\\ \Box.$$
Lemma(2):
Let be $x_n=\left(1+\dfrac1n \right)^{n}$ and $z_n=\left(1-\dfrac1n\right)^{n},\quad \forall n\neq 0\in\mathbb N,\quad x_n<x_{n+1}\quad and\quad z_n<z_{n+1}$
Proof:
Use "Lemma(1)" and choose $a=1$ and $b=\left(1\pm\dfrac1n\right)$
$$\Longrightarrow$$
$$\sqrt[n+1]{\left(1\pm\dfrac1n\right)^{n}}<\dfrac{1+n\left(1\pm\frac1n\right)}{n+1}=1\pm\dfrac1n$$
$$\Longrightarrow$$
$$\left(1\pm\dfrac1n\right)^{n}<\left(1\pm\dfrac1n\right)^{n+1}$$$$\Box.$$
Theorem:
$y_n=\left(1+\dfrac1n\right)^{n+1},\quad \forall n\neq 0\in\mathbb N,\quad y_{n+1}<y_n$
Proof:
We know that $z_{n+1}<z_{n+2}\Longleftrightarrow \dfrac1{z_{n+1}}>\dfrac1{z_{n+2}}$ from "Lemma(2)"
$$y_n=\left(1+\dfrac1n\right)^{n+1}=\left(\dfrac{n+1}{n}\right)^{n+1}=\dfrac{1}{\left(\dfrac{n}{n+1}\right)^{n+1}}=\dfrac{1}{\left(1-\dfrac1{n+1}\right)^{n+1}}=\dfrac1{z_{n+1}}$$
$$y_n>y_{n+1}\Box.$$
A: Hint: Try to use the A.M.- G.M. inequality for positive numbers.
A: the derivative of $(1+1/n)^{(n+1)}$ is $(ln(1+1/n)-1/n)*(1+1/n)^{(n+1)}$ so if that is less than $0$ then $(1+1/n)^{(n+1)}$ is decreasing. $(1+1/n)^{(n+1)}$ is greater than $0$. if $ln(1+1/n)-1/n<0$ and $(1+1/n)^{(n+1)}>0$ then $(ln(1+1/n)-1/n)*(1+1/n)^{(n+1)}<0$. to show that $ln(1+1/n)-1/n<0$ you add 1/n to both sides and get $ln(1+1/n)<1/n$ let $x=1/n$ so then I am proving $ln(1+n)<n$ so $n+1<e^n$ by bernellies inequaility this is true for x is not 0, but the original function is undefined at x=0, so that doesn't matter. I am left with proving $(1+1/n)^{(n+1)}>0$ and this is all that I was able to do. can somebody finish that from there please?
