Prove that $f_n(x): x\in\mathbb{R}\mapsto \frac{x}{(1+x^2)^{n}}$ converges pointwise to null function

Question

Show that sequence of function :

$$f_n(x): x\in\mathbb{R}\longmapsto \dfrac{x}{\left(1+x^2\right)^{n}}$$

converges pointwise to null function

since we found from the following link that :

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

is false

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

• since the proof provided by the book Is there alternative proof

Answer 1

Of course $f_n(0)=0$ for any $n$ and there is nothing to prove. Suppose $x\neq 0$. Then $1+x^2=:q>1$ and $$ \lim_{n\to +\infty} \frac{1}{q^n}=0, $$ therefore $$ \lim_{n\to +\infty}f_n(x)=x\cdot \lim_{n\to +\infty} \frac{1}{q^n}=x\cdot 0=0, $$

Answer 2

Let us fix $x\in \Bbb R^*$. We see that $$f(x)=x\left(\dfrac{1}{1+x^2}\right)^n$$

Since $\left|\dfrac{1}{1+x^2}\right|<1$, we see that $\dfrac{f(x)}{x}$ is a geometric sequence converging to 0.