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#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

Hello everyone!

I'm trying to prove the following:

$a_n$ and $b_n$ are real sequences. If $a_n \rightarrow M \in R$, and $\lim \sup a_n b_n = L$ then $\lim \sup a_n b_n = \lim a_n \cdot \lim \sup b_n$.

This is what I got so far:

(1) Let $c_n = a_n b_n$, and let $C_n$ = $\sup \{c_k \; | \; k\geq n \}$. We know that $\lim C_n = L$.

Hence, for a given $\epsilon > 0, \; \exists N \in \mathbb{N} \; s.t. \; \forall n \geq N$, we have that

$|C_n - L| < \epsilon$ and thus $a_n b_n - L < \epsilon$

(2)$a_n \rightarrow M$. Let $\epsilon ' >0, \; \exists N_1 \in \mathbb{N} \; s.t. \; \forall n \geq N_1$, we have that

$M-\epsilon ' < a_n < M+\epsilon '$

(3) Combining (1) and (2), we get

$a_n b_n - L < (M+\epsilon ')b_n - L$

Now I'm stuck here, I want to show the RHS of the equation above is less the $\epsilon$, but I don't know how!

Any directions/suggestions are appreciated!

I'm trying to prove the following:

$a_n$ and $b_n$ are real sequences. If $a_n \rightarrow M \in R$, and $\lim \sup a_n b_n = L$ then $\lim \sup a_n b_n = \lim a_n \cdot \lim \sup b_n$.

This is what I got so far:

(1) Let $c_n = a_n b_n$, and let $C_n$ = $\sup \{c_k \; | \; k\geq n \}$. We know that $\lim C_n = L$.

Hence, for a given $\epsilon > 0, \; \exists N \in \mathbb{N} \; s.t. \; \forall n \geq N$, we have that

$|C_n - L| < \epsilon$ and thus $a_n b_n - L < \epsilon$

(2)$a_n \rightarrow M$. Let $\epsilon ' >0, \; \exists N_1 \in \mathbb{N} \; s.t. \; \forall n \geq N_1$, we have that

$M-\epsilon ' < a_n < M+\epsilon '$

(3) Combining (1) and (2), we get

$a_n b_n - L < (M+\epsilon ')b_n - L$

Now I'm stuck here, I want to show the RHS of the equation above is less the $\epsilon$, but I don't know how!

Any directions/suggestions are appreciated!

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