prove (lim sup(A_n ))^c = lim inf(A


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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
$BUT A_n =(1−(−1)^n )((n+1)/4) A_n : 1,0,2,0,3,0,4,... A_n = { ((0, 2∣n)),((k, n=2k−1≡1(mod 2))) :} obv. limsup A_n →+∞ if c=1(for example): (limsup A_n )^c →+∞ liminf A_n ^c = liminf A_n = 0 ⇒(limsup A_n )^c ≠ liminf A_n ^c for c=1 ⇒ the statement is obviously false. But if A_n is convergent(or diverges to ±∞) then limsup A_n = liminf A_n = lim A_n = A Since A_n is convergent ⇒ A_n ^c also converges ⇒liminf A_n ^c =lim A_n ^c =A^c ⇒liminf A_n ^c = A^c = (limsup A_n )^c$
$B U T$ $A_{n} = (1 - {(- 1)}^{n}) \frac{n + 1}{4}$ $A_{n} : 1, 0, 2, 0, 3, 0, 4, . . .$ $A_{n} = {\begin{cases} 0, 2 ∣ n \\ k, n = 2 k - 1 \equiv 1 (m o d 2) \end{cases}$ $o b v .$ $l i m s u p A_{n} \to + \infty$ $i f c = 1 (f o r e x a m p l e) :$ ${(l i m s u p A_{n})}^{c} \to + \infty$ $l i m i n f A_{n}^{c} = l i m i n f A_{n} = 0$ $\Rightarrow {(l i m s u p A_{n})}^{c} \neq l i m i n f A_{n}^{c} f o r c = 1$ $\Rightarrow t h e s t a t e m e n t i s o b v i o u s l y f a l s e .$ $B u t i f A_{n} i s c o n v e r g e n t (o r d i v e r g e s t o \pm \infty) t h e n$ $l i m s u p A_{n} = l i m i n f A_{n} = l i m A_{n} = A$ $S i n c e A_{n} i s c o n v e r g e n t \Rightarrow A_{n}^{c} a l s o c o n v e r g e s$ $\Rightarrow l i m i n f A_{n}^{c} = l i m A_{n}^{c} = A^{c}$ $\Rightarrow l i m i n f A_{n}^{c} = A^{c} = {(l i m s u p A_{n})}^{c}$
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