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I've run into some asymptotic analysis in research here and there and largely it feels pretty magical to me. My research has led me to consider the following question which I haven't the slightest idea how to address:

If we have two series $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $g(x) = \sum_{n=0}^{\infty} b_n x^n$ with $a_n,b_n\in\Bbb R$ which converge everywhere and $a_n \sim \frac{b_n}{n}$, can we conclude anything about the asymptotic behavior of $f$ in relation to $g$?

For example: if, say, $g(x) \sim x^{-\frac{1}{2}}$ at infinity, could we conclude that $f$ decays at least as fast as $x^{-\frac{1}{2}}$ at infinity? It seems reasonable to me to think that this would be the case but asymptotic analysis is pretty counter-intuitive at times, especially when the coefficients are allowed to oscillate.

Cameron Williams
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  • Can you restrict your problem to $b_n\geq0$? Otherwise I would search for a counterexample with oscillating $b_n$ and $a_n = |b_n|/n$. – AD - Stop Putin - Jun 07 '15 at 20:50
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    What precisely do you mean by $\sim$ here? If $a_n \sim b_n$, do you mean that $\frac{a_n}{b_n} = 1 + o(n)$? Similarly, what precisely do you mean by $g(x) \sim x^{-1/2}$ at infinity? – davidlowryduda Jun 07 '15 at 20:51
  • @mixedmath Here, I mean that $a_n = \frac{b_n}{n+\varepsilon}$ where $\varepsilon > 0$ is small so in this context $\sim$ means approximately equal, I suppose. The statement at infinity is that $\lim_{x\to\infty}\frac{g(x)}{x^{-\frac{1}{2}}}=C$ for some constant $C$. – Cameron Williams Jun 07 '15 at 21:08
  • @AD. They both have the same sign so no issues like that. I did think of that issue, though. – Cameron Williams Jun 07 '15 at 21:14
  • Do you really mean "there exists $\epsilon>0$ such that $a_n=b_n/(n+\epsilon)$ for every $n$", or is it another approximate thing? Because if this relation is true, then $g$ can be expressed as a linear combination of $f$ and $f'$. In any event, clarifications should be edited into the question, not left in comments. –  Jun 08 '15 at 04:53

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