since converges to as , there exists an integer such that, for all integers , (this is the only place where the absolute convergence is used). Since the series of the converges, the individual must converge to 0 by the term test. Hence there exists an integer such that, for all integers , Also, since converges to as , there exists an integer such that, for all integers , Then, for all integers , use the representation () for , split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates (), () and () to show that By the definition of convergence of a series, as required. Cesàro's theorem In cases where the two sequences are convergent but not absolutely convergent, the Cauchy product is still Cesàro summable. Specifically: If , are real sequences with and then This can be generalised to the case where the two sequences are not convergent but just Cesàro summable: Theorem For and , suppose the sequence is summable with sum A and is summable with sum B. Then their Cauchy product is summable with sum AB. Examples 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. Generalizations 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. Products of finitely many infinite series 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: Proof 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