# Cauchy product

$c_{n}=\sum _{k=0}^{n}a_{k}b_{n-k}.$ ## Series

A particularly important example is to consider the sequences $\textstyle a_{n},b_{n}$ to be terms of two strictly formal (not necessarily convergent) series

$\sum _{n=0}^{\infty }a_{n},\qquad \sum _{n=0}^{\infty }b_{n},$ usually, of real or complex numbers. Then the Cauchy product is defined by a discrete convolution as follows.

$\left(\sum _{n=0}^{\infty }a_{n}\right)\cdot \left(\sum _{m=0}^{\infty }b_{m}\right)=\sum _{j=0}^{\infty }c_{j},\qquad \mathrm {where} \ c_{j}=\sum _{k=0}^{j}a_{k}b_{j-k}$ for n = 0, 1, 2, ...

"Formal" means we are manipulating series in disregard of any questions of convergence. These need not be convergent series. See in particular formal power series.

One hopes, by analogy with finite sums, that in cases in which the two series do actually converge, the sum of the infinite series

$\sum _{j=0}^{\infty }c_{j}$ is equal to the product

$\left(\sum _{n=0}^{\infty }a_{n}\right)\left(\sum _{m=0}^{\infty }b_{m}\right)$ just as would work when each of the two sums being multiplied has only finitely many terms. This is not true in general, but see Mertens' Theorem and Cesàro's theorem below for some special cases.

## Finite summations

The product of two finite series ak and bk with k between 0 and n satisfies the equation:

$\left(\sum _{k=0}^{n}a_{k}\right)\cdot \left(\sum _{k=0}^{n}b_{k}\right)=\sum _{k=0}^{2n}\sum _{i=0}^{k}a_{i}b_{k-i}-\sum _{k=0}^{n-1}\left(a_{k}\sum _{i=n+1}^{2n-k}b_{i}+b_{k}\sum _{i=n+1}^{2n-k}a_{i}\right)$ ## Convergence and Mertens' theorem

Let (an)n≥0 and (bn)n≥0 be real or complex sequences. It was proved by Franz Mertens that, if the series $\textstyle \sum _{n=0}^{\infty }a_{n}$ converges to A and $\textstyle \sum _{n=0}^{\infty }b_{n}$ converges to B, and at least one of them converges absolutely, then their Cauchy product converges to AB.

It is not sufficient for both series to be conditionally convergent, as the following example shows.

### Example

Consider the two alternating series with

$a_{n}=b_{n}={\frac {(-1)^{n}}{\sqrt {n+1}}}\,,$ 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

$c_{n}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{\sqrt {k+1}}}\cdot {\frac {(-1)^{n-k}}{\sqrt {n-k+1}}}=(-1)^{n}\sum _{k=0}^{n}{\frac {1}{\sqrt {(k+1)(n-k+1)}}}$ for every integer n ≥ 0. Since for every we have the inequalities k + 1 ≤ n + 1 and nk + 1 ≤ n + 1, it follows for the square root in the denominator that Template:Sqrtn +1, hence, because there are n + 1 summands,

$|c_{n}|\geq \sum _{k=0}^{n}{\frac {1}{n+1}}\geq 1$ 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.

### Proof of Mertens' theorem

Assume without loss of generality that the series of the $\textstyle \sum _{n=0}^{\infty }a_{n}$ converges absolutely. Define the partial sums

$A_{n}=\sum _{i=0}^{n}a_{i},\quad B_{n}=\sum _{i=0}^{n}b_{i}\quad {\text{and}}\quad C_{n}=\sum _{i=0}^{n}c_{i}$ with

$c_{i}=\sum _{k=0}^{i}a_{k}b_{i-k}\,.$ Then

$C_{n}=\sum _{i=0}^{n}a_{n-i}B_{i}$ by rearrangement, hence

Fix ε > 0. Since $\textstyle \sum _{k\in {\mathbb {N} }}|a_{k}|<\infty$ by absolute convergence, and since Bn converges to B as n → ∞, there exists an integer N such that, for all integers nN,

(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 nM,

Also, since An converges to A as n → ∞, there exists an integer L such that, for all integers nL,

Then, for all integers n ≥ maxTemplate:Mset, use the representation (Template:EquationNote) for Cn, split the sum in two parts, use the triangle inequality for the absolute value, and finally use the three estimates (Template:EquationNote), (Template:EquationNote) and (Template:EquationNote) to show that

{\begin{aligned}|C_{n}-AB|&={\biggl |}\sum _{i=0}^{n}a_{n-i}(B_{i}-B)+(A_{n}-A)B{\biggr |}\\&\leq \sum _{i=0}^{N-1}\underbrace {|a_{\underbrace {n-i} _{\scriptscriptstyle \geq M}}|\,|B_{i}-B|} _{\leq \,\varepsilon /(3N){\text{ by (3)}}}+{}\underbrace {\sum _{i=N}^{n}|a_{n-i}|\,|B_{i}-B|} _{\leq \,\varepsilon /3{\text{ by (2)}}}+{}\underbrace {|A_{n}-A|\,|B|} _{\leq \,\varepsilon /3{\text{ by (4)}}}\leq \varepsilon \,.\end{aligned}} By the definition of convergence of a series, CnAB as required.

## Examples

### Infinite series

$c_{n}=\sum _{i=0}^{n}{\frac {x^{i}}{i!}}{\frac {y^{n-i}}{(n-i)!}}={\frac {1}{n!}}\sum _{i=0}^{n}{\binom {n}{i}}x^{i}y^{n-i}={\frac {(x+y)^{n}}{n!}}$ ## Cesàro's theorem

In cases where the two sequences are convergent but not absolutely convergent, the Cauchy product is still Cesàro summable. Specifically:

${\frac {1}{N}}\left(\sum _{n=1}^{N}\sum _{i=1}^{n}\sum _{k=0}^{i}a_{k}b_{i-k}\right)\to AB.$ This can be generalised to the case where the two sequences are not convergent but just Cesàro summable:

## Generalizations

All of the foregoing applies to sequences in $\textstyle {\mathbb {C} }$ (complex numbers). The Cauchy product can be defined for series in the $\textstyle {\mathbb {R} }^{n}$ 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.

## Relation to convolution of functions

One can also define the Cauchy product of doubly infinite sequences, thought of as functions on $\textstyle \mathbb {Z}$ . In this case the Cauchy product is not always defined: for instance, the Cauchy product of the constant sequence 1 with itself, $\textstyle (\dots ,1,\dots )$ is not defined. This doesn't arise for singly infinite sequences, as these have only finite sums.

One has some pairings, for instance the product of a finite sequence with any sequence, and the product $\textstyle \ell ^{1}\times \ell ^{\infty }$ . This is related to duality of Lp spaces.