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Let $a_n$ be a sequence that converges to $A$ with order of $n^\alpha$, that is $a_n = A + \mathcal{O}(n^\alpha)$ and $b_n$ is another sequence that converges to B with order of $n^\beta$; i.e. $b_n = B + \mathcal{O}(n^\beta)$. What is the order of convergence of $a_n \cdot b_n$?

My intuition tells me that:

$$ a_nb_n = AB + \mathcal{O}(n^{\max(\alpha,\beta)})$$

but could not formulate the proof. I used the trick:

$$|a_nb_n - a_nB + a_nB - AB| \leq |a_n||b_n-B| + |B| |a_n-A| \leq K n^{\alpha}n^{\beta}+ Bn^{\alpha}$$

Not sure how to go on. Could someone please tell me if my intuition is correct and how to go about proving it.

User1865345
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Morcus
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  • First, I think you meant to use the maximum rather than the minimum (presuming $n$ is growing without bound). Second, you can only estimate a bound for the order of convergence. To see why, let $a_n=n^\alpha$ when $n$ is odd and otherwise let $a_n=0;$ and let $b_n=n^\beta$ when $n$ is even and otherwise $b_n=0.$ Now both $a_n$ and $b_n$ converge to zero (at the specified rates) but $a_nb_n$ is identically zero and so has any order of convergence you care to name. There are many hints in this example concerning what restrictions to put on $A$ and $B$ and how to approach a proof. – whuber Jan 28 '21 at 22:51
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    Very helpful. Thanks! – Morcus Jan 29 '21 at 00:32

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