The Cauchy product may apply to infinite series[1][2] or power series.[3][4] When people apply it to finite sequences[5] or finite series, that can be seen merely as a particular case of a product of series with a finite number of non-zero coefficients (see discrete convolution).
Not to be confused with Mertens' theorems concerning distribution of prime numbers.
Let (an)n≥0 and (bn)n≥0 be real or complex sequences. It was proved by Franz Mertens that, if the series converges to A and converges to B, and at least one of them converges absolutely, then their Cauchy product converges to AB.[6] The theorem is still valid in a Banach algebra (see first line of the following proof).
It is not sufficient for both series to be convergent; if both sequences are conditionally convergent, the Cauchy product does not have to converge towards the product of the two series, as the following example shows:
which are only conditionally convergent (the divergence of the series of the absolute values follows from the direct comparison test and the divergence of the harmonic series). The terms of their Cauchy product are given by
for every integer n ≥ 0. Since for every k ∈ {0, 1, ..., n} we have the inequalities k + 1 ≤ n + 1 and n – k + 1 ≤ n + 1, it follows for the square root in the denominator that √(k + 1)(n − k + 1) ≤ n +1, hence, because there are n + 1 summands,
for every integer n ≥ 0. Therefore, cn does not converge to zero as n → ∞, hence the series of the (cn)n≥0 diverges by the term test.
For simplicity, we will prove it for complex numbers. However, the proof we are about to give is formally identical for an arbitrary Banach algebra (not even commutativity or associativity is required).
Fix ε > 0. Since by absolute convergence, and since Bn converges to B as n → ∞, there exists an integer N such that, for all integers n ≥ N,
(2)
(this is the only place where the absolute convergence is used). Since the series of the (an)n≥0 converges, the individual an must converge to 0 by the term test. Hence there exists an integer M such that, for all integers n ≥ M,
(3)
Also, since An converges to A as n → ∞, there exists an integer L such that, for all integers n ≥ L,
(4)
Then, for all integers n ≥ max{L, M + N}, use the representation (1) for Cn, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates (2), (3) and (4) to show that
For some , let and . Then by definition and the binomial formula. Since, formally, and , we have shown that . Since the limit of the Cauchy product of two absolutely convergent series is equal to the product of the limits of those series, we have proven the formula for all .
As a second example, let for all . Then for all so the Cauchy product does not converge.
All of the foregoing applies to sequences in (complex numbers). The Cauchy product can be defined for series in the spaces (Euclidean spaces) where multiplication is the inner product. In this case, we have the result that if two series converge absolutely then their Cauchy product converges absolutely to the inner product of the limits.
Let such that (actually the following is also true for but the statement becomes trivial in that case) and let be infinite series with complex coefficients, from which all except the th one converge absolutely, and the th one converges. Then the limit
exists and we have:
Because
the statement can be proven by induction over : The case for is identical to the claim about the Cauchy product. This is our induction base.
The induction step goes as follows: Let the claim be true for an such that , and let be infinite series with complex coefficients, from which all except the th one converge absolutely, and the -th one converges. We first apply the induction hypothesis to the series . We obtain that the series
converges, and hence, by the triangle inequality and the sandwich criterion, the series
converges, and hence the series
converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have:
Therefore, the formula also holds for .
A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms, or in other words as a function with finite support. For any complex-valued functions f, g on with finite support, one can take their convolution:
Then is the same thing as the Cauchy product of and .
More generally, given a monoidS, one can form the semigroup algebra of S, with the multiplication given by convolution. If one takes, for example, , then the multiplication on is a generalization of the Cauchy product to higher dimension.