If $\displaystyle\lim_{n\to\infty}{\dfrac{a_n}{1+|a_n|}}=0$ then $\displaystyle\lim_{n\to\infty}{a_n}=0$

Question

Any hint to solve this?

It is obvious, by definition, that for every $\epsilon >0$ there exists $N_{\epsilon}$ such that $n\ge N_{\epsilon}$ implies $|a_n|<\epsilon(1+|a_n|)$. Why $a_n\to 0$?

Thanks.

Answer 1

Since $\frac{a_n}{1+|a_n|}$ is convergent there exist an integer $N$ such that $\frac{1}{2}\ge\frac{|a_n|}{1+|a_n|}$ for all $n\ge N.$ Then $1\ge |a_n|$ for all $n\ge N.$ Then for large $n$ values $$\frac{|a_n|}{1+|a_n|}\ge\frac{|a_n|}{2}\ge 0.$$ Now by comparison test the sequence $(\frac{|a_n|}{2})$ is convergent to $0.$.
I think now you can continue from here.

Answer 2

Hint: This is equivalent to showing that $$ \lim_{x \to \infty} \frac{|a_n|}{1+|a_n|} = 0 \implies \lim_{n \to \infty} |a_n| = 0 $$ It is useful to note that the function $f(x) = \frac{x}{1+x}$ has a continuous inverse in some neighborhood of $(0,0)$.

Answer 3

Divide your sequence into $a_n$ where $a_n \geq 0$ and $a_n$ with $a_n < 0$, it suffices to show that both sequences go to zero.

When $x \geq 0$, the function $x \over 1 + |x|$ = $x \over 1 + x$ has continuous inverse $y \over 1 - y$ and when $x < 0$ the function $x \over 1 + |x|$ = $x \over 1 - x$ has continuous inverse $y \over 1 + y$. So both cases $a_n = f^{-1} ({a_n \over 1 + |a_n|})$ for a continuous $f^{-1}$. So in either case, as ${a_n \over 1 + |a_n|}$ goes to zero, $a_n$ will go to $f^{-1}(0) = 0$ as needed.

Answer 4

Let $\epsilon>0$ be given.

Since $\displaystyle\lim_{n\to\infty}\frac{a_n}{1+|a_n|}=0$, there is an integer N such that $n\ge N\implies\displaystyle\left|\frac{a_n}{1+|a_n|}\right|=\frac{|a_n|}{1+|a_n|}<\frac{\epsilon}{1+\epsilon}$,

so $n\ge N\implies |a_n|<\epsilon$ and therefore $\displaystyle\lim_{n\to\infty}a_n=0$.