TITLE: Show a sequence is decreasing
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: $$\frac{a_{n}}{a_{n+1}}=\frac{\left(\frac{n+x}{n+2x}\right)^{n}}{ \left(\frac{n+1+x}{n+1+2x}\right)^{n+1}}= \frac{n+1+2x}{n+1+x}\left( \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}\right)^n$$
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 
$$\frac{a_{n}}{a_{n+1}} \geq  1$$