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Consider

$$\log(c_1+c_2 \frac{1}{p(n)-2})$$ where $p(n)$ is decreasing function approaching $2$, and $c_1,c_2$ are constants that do not depend on $n$.

Question What is the asymptotics of this function?

Initially I was thinking of doing Taylor expansion as follows:

$$\log(1+\frac{c_2}{c_1}\frac{1}{p(n)-2})-\log c_1$$ but the term $\frac{c_2}{c_1}\frac{1}{p(n)-2}$ tends to infinity thus doesn't belong to $(-1,1)$, which means Taylor expansion $\log(1+x)=x-x^2/2+O(x^3)$ is not accurate, since it only works for $|x|<1$.

Thus my question is how to deal with this asymptotics?

chloe
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2 Answers2

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Let $q(n) = \frac{1}{p(n)-2}$, which is an increasing function that tends to $+\infty$. Then: $$\frac{\log\bigl(c_1 + c_2 q(n)) - \log(q(n))}{\log(q(n))} = \frac{\log(\frac{c_1}{q(n)} + c_2)}{\log(q(n))}$$ approaches $0$ as $n \to \infty$, that is, $\log(c_1+c_2q(n)) / \log(q(n))$ approaches $1$. It follows that: $$\log(c_1 + c_2 q(n)) \sim \log(q(n)) = -\log(p(n)-2).$$

Najib Idrissi
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Taking an expample.

Suppose $$p(n)=2+\frac{a}{n}+\frac{b}{n^2}+O\left(\frac{1}{n^3}\right)$$ Composing Taylor series from inside to outside $$\frac 1{p(n)-2}=\frac{n}{a}-\frac{b}{a^2}+O\left(\frac{1}{n}\right)$$

$$\log\left(c_1+ \frac{c_2}{p(n)-2}\right)=\log \left(\frac{c_2\, n}{a}\right)+\frac{a^2\, c_1-b\, c_2}{a\, c_2\, n}+O\left(\frac{1}{n^2}\right)$$

Claude Leibovici
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