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Consider two sequences of positive real numbers $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ such that $\lim_{n\to\infty}\frac{a_n}{b_n}=+\infty$ and $\lim_{n\to\infty}a_n=+\infty$. Prove that $\lim_{n\to\infty}(a_n-b_n)=+\infty$.

The book I'm using to study sequences don't have a solution to this problem, so I don't think it is too difficult (since the book provides the solution to the difficult questions). Unfortunately I couldn't resolve this question and I wasn't even able to make any progress. So please help me.

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(Big) Hint: since $\lim_{n\to\infty}\frac{a_n}{b_n} =\infty$, there exists an $N\geq 0$ such that $$ \frac{a_n}{b_n} \geq 2 $$ for all $n\geq N$. Then $a_n - b_n \geq \frac{a_n}{2}$ for all $n\geq N$.

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$a_{n}-b_{n}=a_{n}\left(1-\dfrac{b_{n}}{a_{n}}\right)$, now $b_{n}/a_{n}\rightarrow 0$ so $1-\dfrac{b_{n}}{a_{n}}\rightarrow 1$, in particular, but $a_{n}\rightarrow\infty$, so $a_{n}\left(1-\dfrac{b_{n}}{a_{n}}\right)\rightarrow\infty$.

Also note that $1-b_{n}/a_{n}>1/2$ for large $n$, so $a_{n}-b_{n}>2^{-1}a_{n}$ for large $n$.

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    $\begingroup$ Not just bounded, but in particular, the sequence is bounded away from $0$ and is positive. $\endgroup$ – Clayton Nov 17 '17 at 17:08
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From $\lim a_n/b_n = \infty$ it follows that $a_n > 2 b_n$ for all large enough $n$ (which mean there is an $n_0$ such that the inequality holds for all $n\ge n_0$). Thus, $a_n-b_n \ge a_n/2,$ from which the claim follows.

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Hint: Prove that $\lim \frac{b_n}{a_n}=0$. Then try factoring $a_n$ out of the expression $a_n-b_n$ and consider the limit of the resulting product.

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$$\lim_{n\to\infty}(a_n-b_n)=\lim_{n\to\infty}a_n(1-\frac{b_n}{a_n})= \lim_{n\to\infty}a_n(1-\frac{1}{\frac{a_n}{b_n}})= \lim_{n\to\infty}a_n = \infty$$

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