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Suppose that two sequences $\{u_n\}$ and $\{v_n\}$ are such that $$ u_n \sim f(n) \qquad \text{and} \qquad v_n \sim g(n) \qquad (n \to \infty),$$ for some smooth functions $f,g$ which tend to $\infty$ as $n \to \infty$. To what extent - and under what conditions - may we give an asymptotic formula for the convolution $\sum_{k=0}^n u_k v_{n-k}$?

  1. It seems likely that the given information is insufficient in general to produce a leading constant (as that constant should be dependent on early terms in our sequence.

  2. It also seems likely that this would be impossible if either one of $u_n$ or $v_n$ grows too rapidly (say, exponentially, with rate $> 1$), as this would cause the dominant terms in the convolution to cluster near $u_0 v_n$ or $u_n v_0$.

If we do not require accuracy in a leading constant, is the growth condition given in (2) sufficient for known methods to give good results? In any case, I'd appreciate a reference to this sort of question (asymptotics of a Cauchy product).

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Note that the smoothness of $f$ and $g$ doesn't really get you anything: any sequence of integers can be interpolated by a smooth function, and if the sequence goes to $+\infty$ then so can the smooth function. – Robert Israel Apr 5 '12 at 3:07
up vote 1 down vote accepted

Consider the generating functions $F(z) = \sum_n u_n z^n$ and $G(z) = \sum_n v_n z^n$. The convolution has generating function $H(z) = F(z) G(z)$. Assuming $u_n$ and $v_n$ grow at most exponentially, these series have nonzero radii of convergence, so $H$ is analytic in a neighbourhood of $0$, and analysis of its closest singularities to 0 may be able to get you the leading terms in the asymptotics of your convolution.

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