Prove $(\frac{n+1}{n})^{n+1}$ is decreasing I managed to prove a similar fact: the following sequence is increasing: $\left( \dfrac{n+1}{n}\right)^n$, which means $\left( \dfrac{n+1}{n}\right)^n < \left( \dfrac{(n+1)+1}{(n+1)}\right)^{n+1}$.
All this took was some simple algebra and bernoulli's inequality. For the one in the title, I am not sure how.
My attempt was:
$$\left(\frac{n+1}{n}\right)^n \left(\frac{n+1}{n}\right)\stackrel{?}{>}\left(\frac{(n+1)+1}{(n+1)}\right)^{n+1} \left(\frac{(n+1)+1}{(n+1)}\right)$$
But this doesn't help me. Do you have other ideas?
 A: $$\frac{\left(\frac{n+1}{n}\right)^{n+1}}{\left(\frac{n}{n-1}\right)^n}=\frac{(n+1)^{n+1}(n-1)^n}{n^{2n+1}}=\underbrace{\left(1+\frac{1}{n}\right)}_{\leqslant\left(1+\frac{1}{n^2}\right)^n}\left(1-\frac{1}{n^2}\right)^n\leqslant\left(1-\frac{1}{n^4}\right)^n<1.$$
A: Easy approach with A.M$\gt $G.M
Let $u_n=(\frac{n+1}{n})^{n+1}=\left(1+\frac{1}{n}\right)^{n+1}$
Let us consider n+2 positive numbers $(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),(1-\frac{1}{n+1}),••••,(1-\frac{1}{n+1}),[(n+1)times] $and $1$.
Applying A.M$\gt $G.M.,we have,
$$\frac{(n+1)(1-\frac{1}{n+1})+1}{n+2}\gt (1-\frac{1}{n+1})^{\frac{n+1}{n+2}}$$
Or,$(\frac{n+1}{n+2})^{n+2}\gt (\frac{n}{n+1})^{n+1}$
Or,$(\frac{n+1}{n})^{n+1}\gt (\frac{n+2}{n+1})^{n+2}$
That is $ u_n\gt u_{n+1}$ for all $n$
A: Although this question is labeled precalculus, there is a calculus approach to show that this is decreasing.
Let's begin by considering the function
$$
f(x)=\left(\frac{x+1}{x}\right)^{x+1}.
$$
If we can show that this function is decreasing, then we've shown that the original sequence is decreasing.  Since the natural logarithm is an increasing function, we know that $f(x)$ is decreasing if and only if $\ln(f(x))$ is a decreasing function.  So, we consider
$$
g(x)=\ln(f(x))=(x+1)\ln\left(\frac{x+1}{x}\right).
$$
Now, we take the derivative of $g$ to get
$$
g'(x)=\ln\left(1+\frac{1}{x}\right)-\frac{1}{x}.
$$
Now, it is enough to show that $g'(x)<0$ for all $x$.  Consider the function $h(y)=\ln(1+y)-y$.  If we can show that this function is negative for $y>0$, then, since $g'(x)=h\left(\frac{1}{x}\right)$, the result has been shown.  Observe that $h(0)=0$ and consider 
$$
h'(y)=\frac{1}{1+y}-1=\frac{-y}{1+y}.
$$
For $y>0$, the derivative $h'(y)<0$, so $h(y)$ is decreasing, i.e., $h(y)<h(0)=0$.  Therefore, we know that $g'(x)<0$ and so $f(x)$ is decreasing.
