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This question already has an answer here:

Hello I want to make sure my work is correct!

Suppose that $\{a_n\}$ and $\{b_n\}$ are sequences such that $a_n \leq b_n$ for all $n $ and $b_n \to b$. Prove that $\lim \sup a_n \leq b.$


Let $\epsilon > 0$ be given. Choose an $N$ s.t. $|b_n-b|<\epsilon$ for $n\geq N$ where $b = \lim \sup b_n$.

$a = \lim \sup a_n$.

Proof by contradiction: If $a \geq b$, then, since $a_n\leq b$,

$b\leq a\leq b_n$, meaning $|b_n-a|<|b_n-b|<\epsilon $, implying $b_n \to a$. This is a contradiction,

therefore, $a\leq b$.

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marked as duplicate by Pedro Tamaroff, Sujaan Kunalan, Dominic Michaelis, Old John, azimut Nov 18 '13 at 8:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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No. It may happen that $b_n \to a$. That is not a problem.

What you can do is as follows : You know that $a_n \leq b_n$ for all $n \in \mathbb{N}$. Hence, for any $k \in \mathbb{N}$ $$ u_k := \sup\{ a_n : n\geq k \} \leq \sup\{ b_n : n\geq k\} =: v_k $$ Now, $v_k \to b$, and hence $$ \limsup a_n = \lim u_k \leq \lim v_k = b $$

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Another approach: Let $c_n:=b_n-a_n \geq 0$ . Since $c_n \geq 0$, $LimSup c_n >0$.

Claim: Let $c_n$ be a sequence of nonnegative terms. Then $LimSup c_n>0$.

Proof of Claim: $LimSup${$c_n$}$:=c$ is , by definition, the largest limit point of $c_n$ , so that, in particular, it is a limit point of the sequencee $c_n$. Assume, BWC , that $C_n$ has $c<0$ as a limit point. But then the ball $B(c, |c|/2)=(c-|c|/2,c+|c|/2)$ will not intersect {$c_n$} , since $c+|c|/2=c-c/2=c/2<0$ , i.e., the neighborhood $B(c, |c|/2)$ of $c$ that does not intersect the sequence, so that $c<0$ cannot be a limit point of {$c_n$}. It follows that $LimSup$ {$c_n$}$:=b_n-a_n \geq 0$, so that $LimSup a_n \geq LimSup $b_n$

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