Prove that $f(n) =\frac{(n+1)^n}{n^{n+1}}$ is Monotonic 
Prove that for each $n \in N$,
  $$f(n) =\frac{(n+1)^n}{n^{n+1}}$$
  is monotonic.

First, I can tell that the function is decreasing. If I take $\frac{1}{n}$, the function looks like $\frac{1}{n}(1+\frac{1}{n})^n$. Can this help?
 A: Consider $$\frac{f(n)}{f(n+1)} = \frac{(n+1)^n}{n^{n+1}}\cdot \frac{(n+1)^{n+2}}{(n+2)^{n+1}}$$
$$ = \frac{(n+1)^n}{n^n \cdot n}\cdot \frac{(n+1)^n \cdot (n+1)^2}{(n+2)^n \cdot (n+1)}$$
$$ = \bigg(\frac{(n+1)^2}{n(n+2)}\bigg)^n \cdot\frac{(n+1)}{n}$$
Now for $n>0$ we have $$\frac{(n+1)^2}{n(n+2)}>1$$ And clearly $$\frac{n+1}{n} >1$$
So $$\frac{f(n)}{f(n+1)} >1 \Rightarrow f(n)>f(n+1)$$
A: $$f(n)=\frac{1}{n}\bigg(1+\frac{1}{n}\bigg)^n\longrightarrow f(n+1)=\frac{1}{n+1}\bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$
$$\frac{f(n+1)}{f(n)}=\frac{n}{n+1}\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^{n}}=\frac{n}{n+1}\bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)=\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\bigg(1+\frac{1}{n+1}\bigg)\frac{n}{n+1}$$
$$ =\bigg(\frac{n(n+2)}{(n+1)^2}\bigg)^n\frac{n(n+2)}{(n+1)^2}<1\cdot 1=1$$
Because $\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{n^2+2n+1} < 1$
A: Note that the function $g(x)=(x+1)^x/x^{x+1}$ satisfies
$$
\frac{g'(x)}{(x+1)^xx^{x+1}}=-\frac{2x+1}{x(x+1)}+\log\left(1+\frac1x\right)<0
$$
for $x>0$.
So your sequence is strictly decreasing.
