prove-lim-sup-A-n-c-lim-inf-A-n-c-

Question Number 205054 by SANOGO last updated on 06/Mar/24

p r o v e

{(l i m s u p (A_{n}))}^{c} = l i m i n f (A_{n}^{c})

Commented by aleks041103 last updated on 11/Mar/24

B U T

A_{n} = (1 - {(- 1)}^{n}) \frac{n + 1}{4}

A_{n} : 1, 0, 2, 0, 3, 0, 4, \dots

A_{n} = {\begin{cases} 0, 2 ∣ n \\ k, n = 2 k - 1 \equiv 1 (m o d 2) \end{cases}

o b v .

l i m s u p A_{n} \to + \infty

i f c = 1 (f o r e x a m p l e) :

{(l i m s u p A_{n})}^{c} \to + \infty

l i m i n f A_{n}^{c} = l i m i n f A_{n} = 0

\Rightarrow {(l i m s u p A_{n})}^{c} \neq l i m i n f A_{n}^{c} f o r c = 1

\Rightarrow t h e s t a t e m e n t i s o b v i o u s l y f a l s e .

B u t i f A_{n} i s c o n v e r g e n t (o r d i v e r g e s t o \pm \infty) t h e n

l i m s u p A_{n} = l i m i n f A_{n} = l i m A_{n} = A

S i n c e A_{n} i s c o n v e r g e n t \Rightarrow A_{n}^{c} a l s o c o n v e r g e s

\Rightarrow l i m i n f A_{n}^{c} = l i m A_{n}^{c} = A^{c}

\Rightarrow l i m i n f A_{n}^{c} = A^{c} = {(l i m s u p A_{n})}^{c}