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I have a question concerning a special limit

$$\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=1$$

Can I always conclude from this that

$$\lim_{n \rightarrow \infty}{(a_n-b_n)}=0$$

Even if $a_n$ and $b_n$ are divergent? I tried to find some counter examples but I couldn't.

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    $\begingroup$ Add condition $b_n$ bounded. No need for it to converge. $\endgroup$ – GEdgar Sep 13 '15 at 20:37
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False. Take $a_n = 1+n$ and $b_n = n$. Then $\frac{a_n}{b_n} \to 1$ and $a_n -b_n \to 1 \neq 0$.

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  • $\begingroup$ Are there conditions in which this might still be true? Even in the case of divergence? $\endgroup$ – MrYouMath Sep 13 '15 at 20:32
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    $\begingroup$ It seems to me that we must have $b_n \to b \in \Bbb R$. So we can do: $$a_n - b_n = b_n\left(\frac{a_n}{b_n}-1\right) \to b \cdot 0 = 0.$$If $b_n$ diverges then we'll have some $\infty \cdot 0$, and we can't conclude anything. $\endgroup$ – Ivo Terek Sep 13 '15 at 20:34
  • $\begingroup$ If $b_n$ is convergent also $a_n$ must be convergent to have a limit of 1 for the ratio. I think something like order of divergence of $b_n$ less than the order of Convergence of $\frac{a_n}{b_n}-1$ could used. $\endgroup$ – MrYouMath Sep 13 '15 at 20:41

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