Demonstrate that the sequence is decreasing $$\left( 1-\frac{1}{n+1}\right)^n$$
How do I demonstrate that the sequence decreases? I tried using the newton binomial but I end up with a terribly complicated expression. 
 A: Use the GM-HM inequality:
$$\left(1-\frac1{n+1}\right)^n=\underbrace{\frac n{n+1}\cdot\ldots\cdot\frac n{n+1}}_n\cdot1\ge\left(\frac{n+1}{n\cdot\frac{n+1}n+1}\right)^{n+1}=\left(1-\frac1{n+2}\right)^{n+1}$$
A: This is a classic exercise.
Let 
$$a_n= \left( 1-\frac{1}{n+1}\right)^n=\left( \frac{n}{n+1}\right)^n$$
$$\frac{a_n}{a_{n+1}} =\left( \frac{n}{n+1}\right)^n\left( \frac{n+2}{n+1}\right)^{n+1}=\left( \frac{n}{n+1}\right)^{n+1}\left( \frac{n+2}{n+1}\right)^{n+1}\frac{n+1}{n}$$
Now write 
$$\left( \frac{n}{n+1}\right)^{n+1}\left( \frac{n+2}{n+1}\right)^{n+1}=\left( \frac{n^2+n}{(n+1)^2}\right)^{n+1}=\left( 1-\frac{1}{(n+1)^2}\right)^{n+1}$$
and use Bernoulli inequality. 
A: And because there's even more ways to do it:
Let $f(x)=\left(1-\frac{1}{x+1}\right)^x$ and show that $f(x)$ is decreasing by differentiating $\log f(x) = x \log\frac{x}{x+1}$.
We get
$$ \frac{d}{dx} \log f(x) = = \log\frac{x}{x+1} + x\left(\frac1x - \frac1{x+1}\right) = \log\left(1-\frac1{x+1}\right) + \frac1{x+1}  $$
which is negative for $x>0$ because $\log(1-\alpha)< -\alpha$ for $\alpha\in(0,1)$.
Therefore $\log f(x)$ decreases, and so does $f(x)$, and so does $f(n)$.
