If $(\limsup_{n\to\infty} x_{n})\cdot(\limsup_{n\to\infty} 1/x_{n})=1$, then the sequence $x_n$ converges

Question

${x_{n}}$ 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.

Answer 1

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 $\limsup a_n\cdot \limsup \frac 1 {a_n} \ge 1$

Answer 2

(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 $[0,\infty]$.

Answer 3

$x_n=\frac 1 {\frac 1 {x_n}}$. 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.