Assume $(x_n)$ converges and $x_n>0$ for all n. Prove $\lim_{n\rightarrow\infty}x_n>0$.

Question

Assume $(x_n)$ converges and $x_n>0$ for all n. Prove $\lim_{n\rightarrow\infty}x_n>0$.

My attempt so far:

I am trying to aim at contradiction by proving that $\lim_{n\rightarrow\infty}x_n<0$.

By Bolzano-Weiestrass, $(x_n)$ has a monotonic subsequence $(x_{n_{k}})$. Every subsequence of a converging sequence converges which mean that $x_{n_{k}}$ converges. By the definition of convergence, $\forall\epsilon>0\exists n_{\epsilon}$ such that $n>n_{\epsilon}\Rightarrow|x_{n_{k}}-L|<\epsilon$. Since $\epsilon$ is an arbitrary number, we can choose $\epsilon=1$. By the definition of triangle inequality, $|x_{n_{k}}-L|<\epsilon \Rightarrow |x_{n_{k}}|-|L|<\epsilon$. Now by absolute value,

$$|x_{n_{k}}|-|L|<1 \iff \\ |x_{n_{k}}|<1+|L| \iff \\ -1-|L|<x_{n_{k}}<1+|L|.$$

That is, $(x_{n_{k}})$ is bounded.

After this, I am trying to see the case when $L<0$ but I don't know how.

Answer 1

You can't prove it, since it is false. Just take $x_n=\frac1n$.

Answer 2

That’s not true, let consider

$$x_n=\frac1n$$

what is true is that $x_n\to L \ge0$.