Prove that the sequence $b_n=\left(1+\frac{1}{n}\right)^{n+1}$ is decreasing Prove that the sequence
$$b_n=\left(1+\frac{1}{n}\right)^{n+1}$$
Is decreasing.
I have calculated $b_n/b_{n-1}$ but it is obtain:
$$\left(1-\frac{1}{n^2}\right)^n \left(1+\frac{1}{n}\right)^n$$
But I can't go on.
Any suggestions please?
 A: $$y=\left(1+\frac{1}{x}\right)^{x+1}$$
$$\ln y=({x+1})\cdot\ln\left(1+\frac{1}{x}\right)$$
$$y'\frac{1}{y}=\ln\left(1+\frac{1}{x}\right)+(x+1)\cdot \frac{1}{1+\frac{1}{x}}\cdot\left(-\frac{1}{x^2}\right)$$
$$y'=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{x}\right)\cdot\left(1+\frac{1}{x}\right)^{x+1}$$
$$\Rightarrow y'<0$$
Hence $y$ is decreasing
A: Hint: using the tools of differential calculus, study the function
$$x\longmapsto \left(1+\frac{1}{x}\right)^{x+1}$$
for $x>0$.
A: Consider $f(x) = (x+1)\ln (x+1) - (x+1)\ln x$ on $x \geq 1$, we have:
$f'(x) = \ln\left(1 + \dfrac{1}{x}\right) - \dfrac{1}{x} < 0$ because it is true that $e^r \geq 1 + r$, and apply this for $r = \dfrac{1}{x} > 0$. From this the answer follows.
A: Why don't you change your goal to prove that $b_n-b_{n+1}>0$?
$b_n-b_{n+1}=(1+1/n)^n-(1+1/n)^n(1+1/n)=(1+1/n)^n(1-1-1/n)=(1+1/n)^n(-1/n)$
Since $n$ is a natural number, $-1/n < 0$. Since $1>0, 1/n>0$, we have $1+1/n>0$. It would be nice if you can use $(1+1/n)^n >0$ directly based on $1+1/n>0$. If not, just do a simple proof by induction.
Then you will have your $b_n-b_{n+1}>0$, i.e., $b_n> b_{n+1}$, meaning that $b_n$ is decreasing.
A: Since the inequality holds for any positive numbers:$$\sqrt[n]{{a_1}{a_2}\cdots {a_n}}\ge \frac{n}{a_1^{-1}a_2^{-1}\cdots a_n^{-1}},$$we have:$$\sqrt[n+2]{\underbrace{\frac{n+1}n\frac{n+1}n\cdots \frac{n+1}n}_{n+1\text{ times}}\cdot 1}\ge \frac{n+2}{\frac n{n+1}+\frac n{n+1}+\cdots +\frac n{n+1}+1}.$$Hence,$$\left(1+\frac 1n\right)^{n+1} \ge \left(1+\frac 1{n+1}\right)^{n+2}.$$
A: My 2¢: consider the function defined apriori for  $x>0$
$$f(x)=\log(1+x)\cdot (\frac{1}{x}+1)= \frac{\log(1+x)\cdot (1+x)}{x}$$
$f$ extends analytically to $(-1, \infty)$, and continuously to $[-1, \infty)$. We have $f(-1)=0$ and $f(0)=1$. 
We calculate: $$f'(x)= \frac{x - \log(1+x)}{x^2}$$
so $f'(x)>0$ for $x> -1$, $x \ne 0$ and so $f$ strictly increasing on $(-1, \infty)$. Now consider the decreasing sequence of values $\frac{1}{n}$ for $x$.
