Product of $\limsup_{x\to\infty} x_n $

Question

Given $x_n > 0$ and $$\limsup_{n\to\infty} x_n \cdot \limsup_{n\to\infty} \frac{1}{x_n} = 1$$ Does this mean that $x_n$ converges?

Answer 1

Yes. With $x_n>0,$ as stated. Let $r=\lim \sup_{n\to \infty}.$ Then $r>0.$

( because $r\geq 0$ and $\lim \sup_{n\to \infty}(1/x_n)\geq 0$ and $r\cdot \lim \sup_{n\to \infty}(1/x_n)=1.$).

Suppose $(x_n)_n$ does not converge to $r$. Then for some $s>0$ there are infinitely many $n$ for which $|x_n-r|>s.$ But there are only finitely many $n$ for which $x_n-r>s,$ else $r=\lim \sup_{n\to \infty} x_n\geq r+s>r,$ an absurdity.

So there are infinitely many $n$ for which $0<x_n<r-s .$ But that implies $\lim \sup_{n\to \infty}(1/x_n)\geq 1/(r-s)>1/r,$ a contradiction.