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Let $a_n \le b_n$ for all $n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = \infty$ . Can we conclude $\lim_{n \to \infty} b_n = \infty$ ? Intuitively , it seems right but I can't prove it or find a counterexample .

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  • $\begingroup$ How do you intuively see this as right? When you start explaining your intuitive thoughts about this problem in words you will automatically find a rigorous proof. Most simple proofs are intuitive and obvious, it is only a matter of putting your thoughts into words. $\endgroup$ – Paramanand Singh Jan 18 '18 at 13:49
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Yes, you can.

To see how, start with the definition: we say that $a_n\to\infty$ as $n\to\infty$ if given any $M\in\mathbb{R}$, we have $a_n\geq M$ for $n$ sufficiently large. (That is: for any $M$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $a_n\geq M$.)

Clearly, though, this implies that for $n\geq N$, $b_n\geq a_n\geq M$; therefore $b_n\to\infty$ as $n\to\infty$ as well.

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(If I didn't read the question wrong...) the proof should be a simple application of the definition of limits.

$$\lim_{n \rightarrow \infty} a_n = \infty$$ means that for each $N \in \mathbb{N}$ there exists some $n_{a} \in \mathbb{N}$ such that $\forall k \geq n_{a}$, $a_{k} \geq N$.

To show that the sequence $B = \{b_1, b_2, ...\}$ tends to infinity, given an $N$, we need to find a $n_{b}$ such that $\forall k \geq n_b$, $b_{k} \geq N$. Since $a_i \leq b_i$, our choice of $n_b$ is simply $n_a$. Then $\forall k \geq n_{a}$, $b_k \geq a_k \geq N$ as required.

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Given $M>0$, there exists $N$ such that $n\ge N$ implies that $$ M\leq a_n\leq b_n $$ as desired.

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