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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.

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    $\frac{n+x}{n+2x}=1-\frac{1}{an+2}$ where $a=\frac1x$, but I don't know if that leads anywhere...2012-06-16

2 Answers 2

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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

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$\frac{a_{n}}{a_{n+1}}=\frac{\left(\frac{n+x}{n+2x}\right)^{n}}{ \left(\frac{n+1+x}{n+1+2x}\right)^{n+1}}= \frac{n+1+2x}{n+1+x}\left( \frac{n+1+2x}{n+1+x} \frac{n+x}{n+2x}\right)^n$

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

$\frac{a_{n}}{a_{n+1}} \geq 1$