Taking logarithm preserving asymptotic equivalence

Question

Let for some $c \in (0,1)$, $d \in (0, \infty)$ and $j \in \mathbb{N}$: $$ f(n) \sim \frac{c^n \, d^{j-1}} { (c^n + d)^{j+1}} $$ as $n \to \infty$. "$\sim$" denotes asymptotic equivalence, i.e. the quotient of both sides converges to $1$ as $n \to \infty$. Now, I want check whether $$ \ln (f_n) \sim \log \left( \frac{c^n \, d^{j-1}} { (c^n + d)^{j+1}} \right) $$ holds. I am aware that in general taking the logarithm does not preserve asymptotic equivalence, but nevertheless often does. I suspect that the above holds, but cannot prove it. Any ideas?

Answer 1

Generally, for positive $a_n, b_n$ we have $$a_n \sim b_n \iff \log a_n - \log b_n \to 0\,.$$ The latter implies $\log a_n \sim \log b_n$, except in the case that $(\log a_n)$ has $0$ as an accumulation point. For if $\lvert \log a_n\rvert \geqslant c > 0$ for all $n \geqslant n_0$, then $$\bigl\lvert 1 - \frac{\log b_n}{\log a_n}\biggr\rvert = \frac{\lvert \log a_n - \log b_n\rvert}{\lvert \log a_n\rvert} \leqslant c^{-1} \lvert \log a_n - \log b_n\rvert \to 0\,,$$ i.e. $\frac{\log b_n}{\log a_n} \to 1$ or equivalently $\log a_n \sim \log b_n$. If there is a subsequence with $\log a_{n_k} \to 0$, then $\log a_n \sim \log b_n$ need not hold (but it may hold).

Here we have $$0 < \frac{c^n d^{j-1}}{(c^n+d)^{j+1}} < \frac{d^{j-1}}{d^{j+1}}\cdot c^n = d^{-2}c^n \to 0\,,$$ hence $$\log f_n \sim \log \frac{c^nd^{j-1}}{(c^n+d)^{j+1}}$$ follows.