Why does convolution not maintain asymptotic equality of coefficients? Assume I have four (generating) functions $f$, $f'$, $g$ and $g'$. If that is interesting, we can assume that they all share the same radius of convergence $\rho > 0$. In addition, we know that
$\qquad [z^n]f(z) \sim [z^n]f'(z)$ and
$\qquad [z^n]g(z) \sim [z^n]g'(z)$.
Now we are interested in whether
$\qquad [z^n](f \ast g)(z) \sim [z^n](f' \ast g')(z)$
where $\ast$ denotes (discrete) convolution. That is, we would like to obtain asymptotics for the coefficients of a generating function by replacing factors with "equivalent" functions.
I think this is not always true. Intuitively, coefficients of small $n$ -- which can be arbitrarily bad in $f'$ and $g'$ -- factor into all coefficients of the product; hence the error is blown up and need not vanish in the limit.
What I don't have is


*

*a counter-example and/or

*a rigorous argument for above intuition (if it's accurate).


What are such?
 A: If we take a stupid counterexample, say
$$\begin{gather}
f(z) = \frac{1+z}{1-z} = 1 + 2\sum_{n=1}^\infty z^n,\\
g(z) = \frac{1-z}{1+z} = 1 + 2\sum_{n=1}^\infty (-1)^nz^n,
\end{gather}$$
we see that the pole of each of the two functions on the unit circle is cancelled by the zero of the other when multiplying, $f(z)\cdot g(z) = (f\ast g)(z) \equiv 1$.
But if we add a function $h$ that is holomorphic in $D_r(0)$ for some $r > 1$, then the coefficients of $h$ tend to $0$, so we have asymptotic equality, $[z^n](f+h) \sim [z^n]g$, and also $[z^n](g+h)\sim [z^n]g$, but for the product,
$$(f+h)(z)\cdot (g+h)(z) = f(z)g(z) + \left(f(z)+g(z)\right)h(z) + h(z)^2 = 1 + 2\frac{1+z^2}{1-z^2}h(z) + h(z)^2,$$
the poles are in general not cancelled (only when $h$ has a zero in the appropriate place).
The point is that the radius of convergence of the convolution (Cauchy product) of two power series can be larger than the radius of convergence of the factors, if the singularities of each factor on the boundary of the domain of convergence are cancelled by the behaviour of the other factor, but that cancelling is not preserved under small perturbations, so if cancellation of singularities happens for $f\ast g$, it generally does not happen for $f'\ast g'$, hence the radii of convergence of $f\ast g$ and $f'\ast g'$ are different, but a necessary condition for asymptotic equivalence of the coefficients is that the two power series have the same radius of convergence.
With the condition that the coefficients of the series be positive, an example where the radius of convergence of the convolution of the power series is larger than the radius of the factors may not exist (if one exists, it's at least much harder to see), but we still need not have asymptotic equality of the coefficients preserved by small perturbations. If we consider e.g. for some $C > 1$
$$f(z) = \sum_{n=0}^\infty (n+1)C^n z^n,\quad g(z) = \sum_{n=0}^\infty \frac{C^nz^n}{(n+1)^2},$$
the coefficients of the convolution are
$$\begin{align}
\sum_{k=0}^n (n+1-k)C^{n-k}\frac{C^k}{(k+1)^2} &= C^n\left((n+2)\sum_{k=0}^n \frac{1}{(k+1)^2} - \sum_{k=0}^n \frac{1}{k+1}\right)\\
&= C^n\left(n\frac{\pi^2}{6} + O(1) - O(\log n)\right),
\end{align}$$
and changing the constant coefficient of $g$ already changes the constant factor on the dominant term $\frac{\pi^2}{6} n C^n$.
