If $(\limsup_{n\to\infty} x_{n})\cdot(\limsup_{n\to\infty} 1/x_{n})=1$, then the sequence $x_n$ converges ${x_{n}}$ is a sequence in $\mathbb{R}$ where $n\geq 1$ such that all $x_{n} >0$. If 
$(\limsup_{n\to\infty} x_{n})\cdot(\limsup_{n\to\infty} 1/x_{n})=1$, how can I show that ${x_n}$ converges? I am thinking about using the contrapositive  of the statement but I'm not sure if that's the best way. Thank you. 
 A: If $x_n$ doesn’t converge then
$$\limsup x_n=L\neq \liminf x_n=l$$
then 
$$\limsup \frac1{x_n}=\frac 1 l$$
and
$$\limsup x_n\cdot \limsup \frac1{x_n}\neq1$$
Refer also to


*

*Proving $\limsup\frac 1 {a_n}=\frac 1 {\liminf a_n}$ and $\limsup a_n\cdot \limsup \frac 1 {a_n} \ge 1$
A: (Needs some explanation, but...)
$$
1 = (\limsup_{n\to\infty} x_{n})\cdot(\limsup_{n\to\infty} 1/x_{n})=
\frac{\limsup_{n\to\infty} x_{n}}{\liminf_{n\to\infty} x_{n}}
$$
so
$$
\limsup_{n\to\infty} x_{n} = \liminf_{n\to\infty} x_{n}
$$
and therefore $\lim_{n\to\infty} x_{n}$ exists.  
added
Note: Even if
$$
\frac{\limsup_{n\to\infty} x_{n}}{\liminf_{n\to\infty} x_{n}} = \frac{\infty}{\infty}\quad\text{or}\quad
\frac{\limsup_{n\to\infty} x_{n}}{\liminf_{n\to\infty} x_{n}} = \frac{0}{0}
$$
we may still conclude
$$
\limsup_{n\to\infty} x_{n} = \liminf_{n\to\infty} x_{n}
$$
and therefore that $\lim_{n\to\infty} x_n$ exists in $[0,\infty]$.
A: $x_n=\frac 1 {\frac 1 {x_n}}$. So $\lim \inf x_n =\lim \inf \frac 1 {\frac 1 {x_n}}=\frac 1 {\lim \sup \frac 1 {x_n}}=\lim \sup x_n$. Hence $(x_n)$ converges. 
