For two sequences $\{a_n\}$ and $\{b_n\}$ with $a_n\le b_n,\forall n \in\mathbb{N}$, it is true that $\lim a_n \le \lim b_n$, right? I need this result to prove something but I can not find it anywhere.
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3Yes, it's true. But note that even if for every $n\in \Bbb{N}$ we have $a_n< b_n$, we might end up with $\lim a_n \leq \lim b_n$ i.e strict inequality could become a weak one in the limit. – peek-a-boo Sep 25 '20 at 19:04
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@peek-a-boo thansk! – Sorry Sep 25 '20 at 19:05
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To be more rigorous, you need to state both $a_n$ and $b_n$ have limits. – Zhanxiong Sep 25 '20 at 19:38
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Two cases, suppose that $\lim b_{n}= \infty$, then the result holds. So suppose that $\lim b_{n} \leq \infty$. Let $L_a$ and $L_b$ denote the two limits respectively.
Suppose that $L_a > L_b$. Let $0<\epsilon< L_a -L_b$. Now there exist an $N_{1}\in \mathbb{N}$ such that for all $n\geq N_1$, $ |a_n-L_a|< \frac{\epsilon}{2}$. Similarly an $N_{2}$ such that $|b_n - L_b|< \frac{\epsilon}{2}$ for all $n\geq N_2$. Thus by construction we have that for all $n \geq min(N_1,N_2)$ that $$ \begin{align} b_{n} &< L_b + \frac{\epsilon}{2} \\ &<L_a - \frac{\epsilon}{2}\\ & < a_n \end{align} $$ but that contradicts the assumption that $a_{n} \leq b_{n}$.
Daniel H. Hartman
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@Surb Because I was doing an assignment, and I assumed it was the case. Since I can not find it in the textbook I drop using that fact. – Sorry Sep 25 '20 at 19:48
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WLOG, $a_n=0$, because the difference of two limits is the limit of the differences (provided the limits exist). Now the limit of a non-negative sequence cannot be negative.