Show $\frac{x}{\left(1+x^2\right)^{n}}\underset{{n\to +\infty}}{\sim} \frac{1}{x^{2n-1}}$ for every $x>1$

Question

Show $$\dfrac{x}{\left(1+x^2\right)^{n}}\underset{{n\to +\infty}}{\sim} \dfrac{1}{x^{2n-1}}\quad \mbox{ with } x>1$$

My proof:

$$\dfrac{x}{\left(1+x^2\right)^{n}}\underset{{n\to +\infty}}{\sim} \dfrac{1}{x^{2n-1}} \iff \lim_{n\to +\infty}\dfrac{x}{\left(1+x^2\right)^{n}}\times x^{2n-1}=1$$

indeed,

For all $x\in \mathbb{R}^*$

• Note that : $$ q^{n}\underset{n\to +\infty}{\to +\infty}\quad \rm{ when}\quad q>1 $$

then :
$x^{2n}\underset{n\to +\infty}{\to +\infty}$ and $\left(1+x^2\right)^{n}\underset{n\to +\infty}{\to +\infty}$ since $1+x^{2}> 1$

$$\lim_{n\to +\infty}\dfrac{x}{\left(1+x^2\right)^{n}}\times x^{2n-1}=\lim_{n\to +\infty}\dfrac{x^{2n}}{\left(1+x^2\right)^{n}}=\lim_{n\to +\infty}\dfrac{x^{2n}}{\left(x\right)^{2n}}=1$$

Is my proof correct ? and could we use taylor expansion ?

Reference of my question:

https://books.google.com/books?id=Ihh2uOXnRQcC&hl=fr&pg=PA97&#v=onepage&q&f=false

Answer 1

After some trouble (!) I think we can conclude that the statement is not true. Take for example $x=2$: you would have $$ 2\cdot 5^{-n}\sim \frac{1}{2}\cdot 4^{-n}. $$ That's not true.