Show a sequence is decreasing I'm stuck trying to show that the following sequence is decreasing
$$a_{n} = \left(\frac{n+x}{n+2x}\right)^{n}$$ where $x>0$.   I've tried treating $n$ as a real number and took derivatives but it didn't lead to anything promising.   Any hints would be appreciated.
 A: Fix $x>0$. Our aim is to show that $a'(n)<0$. Note that
$$
a(n)=\left(1-\frac{x}{n+2x}\right)^n>0
$$
$$
(\log a(n))'=\frac{a'(n)}{a(n)}
$$
Hence it is enough to show that $(\log a(n))'<0$. Well this is indeed true. Using inequality $\log(1-t)<-t$ we obtain
$$
(\log a(n))'=
\frac{\left(n^2+3 n x+2 x^2\right)\log\left(1-\frac{x}{n+2 x}\right)+n x}{(n+x)
(n+2 x)}\leq
$$
$$
\frac{\left(n^2+3 n x+2 x^2\right)\left(\frac{-x}{n+2 x}\right)+n x}{(n+x)
(n+2 x)}=
\frac{-x^2}{(n+x)
(n+2 x)}<0
$$
And now we are done
A: $$\frac{a_{n}}{a_{n+1}}=\frac{\left(\frac{n+x}{n+2x}\right)^{n}}{ \left(\frac{n+1+x}{n+1+2x}\right)^{n+1}}= \frac{n+1+2x}{n+1+x}\left( \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}\right)^n$$
Lets observe that
$$ \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}=\frac{n^2+n+3nx+x+2x^2}{n^2+n+3nx+2x+2x^2}=1-\frac{x}{n^2+n+3nx+2x+2x^2}$$
Then, by Bernoulli
$$\frac{a_{n}}{a_{n+1}} \geq (1+\frac{x}{n+1+x})(1-n\frac{x}{n^2+n+3nx+2x+2x^2}) $$
An easy computation shows that
$$(1+\frac{x}{n+1+x})(1-n\frac{x}{n^2+n+3nx+2x+2x^2}) \geq 1 \Leftrightarrow$$
$$\frac{x}{n+1+x} \geq \frac{nx}{(n+1+x)(n+2x)} +\frac{x}{n+1+x}\frac{nx}{(n+1+x)(n+2x)} \Leftrightarrow  $$
$$x(n+1+x)(n+2x) \geq nx(n+1+x)+nx^2 \Leftrightarrow $$
$$n^2x+nx+3nx^2+2x^2+2x^3 \geq n^2x+nx+nx^2+nx^2 \Leftrightarrow  $$
$$ nx^2+2x^2+2x^3 \geq 0 $$
Thus 
$$\frac{a_{n}}{a_{n+1}} \geq  1$$
