Prove that $inf\ \{|x_n|, n \in \mathbb{N}\}=0$

Solution 1:

The sequence $\{|x_{n}|\}$ is monotonic decreasing. as from the given condition $|x_{n+1}|<|x_{n}|$ . I would assume without loss of generality that $|x_{1}|>0$. If not then we could take the infimimum $n$ such that $|x_{n}|>0$ and repeat the proof which I did. If such an $n$ does not exist then we have a zero sequence and there is nothing to prove.

$|x_{2}|<c_{1}|x_{1}|$ .

$|x_{3}|<c_{2}|x_{2}|$

Now let $c'_{2}=\max\{c_{1}, c_{2}\}$.

Then $|x_{3}|<c'_{2}|x_{2}|<(c'_{2})^{2}|x_{1}|$

Call this $c'_{2}$ as $c_{2}$.

So proceeding inductively there is exist a $c_{n}\in(0,1)$ such that

$|x_{n+1}|<(c_{n})^{n}|x_{n}|<(c_{n})^{n}|x_{1}|$

Now consider the sequence $(c_{n})^{n}|x_{1}|$. This has a subsequence which converge to $0$. call it $\{(c_{n_{k}})^{n_{k}}|x_{1}|\}$

Let $\epsilon>0$ be arbitrary.

Then for $M\in\mathbb{N}$ we have $(c_{n_{k}})^{n_{k}}|x_{1}|<\epsilon\,\,\forall k\geq M$

So correspondingly we have $|x_{n_{k}+1}|<(c_{n_{k}})^{n_{k}}|x_{1}|<\epsilon$.

So $0$ is the greatest lower bound as any lower bound greater than $0$ fails to be an upper bound.

So $\inf\{|x_{n}|:n\in\mathbb{N}\}=0$