Cauchy product of two infinite series

Let ${\textstyle \sum _{i=0}^{\infty }a_{i}}$ and ${\textstyle \sum _{j=0}^{\infty }b_{j}}$ be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows:

${\displaystyle \left(\sum _{i=0}^{\infty }a_{i}\right)\cdot \left(\sum _{j=0}^{\infty }b_{j}\right)=\sum _{k=0}^{\infty }c_{k}}$     where     ${\displaystyle c_{k}=\sum _{l=0}^{k}a_{l}b_{k-l}}$.

Cauchy product of two power series

Consider the following two power series

${\displaystyle \sum _{i=0}^{\infty }a_{i}x^{i}}$     and     ${\displaystyle \sum _{j=0}^{\infty }b_{j}x^{j}}$

with complex coefficients ${\displaystyle \{a_{i}\}}$ and ${\displaystyle \{b_{j}\}}$. The Cauchy product of these two power series is defined by a discrete convolution as follows:

${\displaystyle \left(\sum _{i=0}^{\infty }a_{i}x^{i}\right)\cdot \left(\sum _{j=0}^{\infty }b_{j}x^{j}\right)=\sum _{k=0}^{\infty }c_{k}x^{k}}$     where     ${\displaystyle c_{k}=\sum _{l=0}^{k}a_{l}b_{k-l}}$.

Example

Consider the two alternating series with

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

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

$${\displaystyle c_{n}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{\sqrt {k+1}}}\cdot {\frac {(-1)^{n-k}}{\sqrt {n-k+1}}}=(-1)^{n}\sum _{k=0}^{n}{\frac {1}{\sqrt {(k+1)(n-k+1)}}}}$$

for every integer n ≥ 0. Since for every 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,

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

for every integer n ≥ 0. Therefore, c_n does not converge to zero as n → ∞, hence the series of the (c_n)_n≥0 diverges by the term test.

Proof of Mertens' theorem

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).

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

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

with

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

Then

$${\displaystyle C_{n}=\sum _{i=0}^{n}a_{n-i}B_{i}}$$

by rearrangement, hence

${\displaystyle C_{n}=\sum _{i=0}^{n}a_{n-i}(B_{i}-B)+A_{n}B\,.}$

(1)

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

${\displaystyle |B_{n}-B|\leq {\frac {\varepsilon /3}{\sum _{k\in \mathbb {N} }|a_{k}|+1}}}$

(2)

(this is the only place where the absolute convergence is used). Since the series of the (a_n)_n≥0 converges, the individual a_n must converge to 0 by the term test. Hence there exists an integer M such that, for all integers n ≥ M,

${\displaystyle |a_{n}|\leq {\frac {\varepsilon }{3N(\max _{i\in \{0,\dots ,N-1\}}|B_{i}-B|+1)}}\,.}$

(3)

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

${\displaystyle |A_{n}-A|\leq {\frac {\varepsilon /3}{|B|+1}}\,.}$

(4)

Then, for all integers n ≥ max{L, M + N}, use the representation (1) for C_n, 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

$${\displaystyle {\begin{aligned}|C_{n}-AB|&={\biggl |}\sum _{i=0}^{n}a_{n-i}(B_{i}-B)+(A_{n}-A)B{\biggr |}\\&\leq \sum _{i=0}^{N-1}\underbrace {|a_{\underbrace {\scriptstyle n-i} _{\scriptscriptstyle \geq M}}|\,|B_{i}-B|} _{\leq \,\varepsilon /3{\text{ by (3)}}}+{}\underbrace {\sum _{i=N}^{n}|a_{n-i}|\,|B_{i}-B|} _{\leq \,\varepsilon /3{\text{ by (2)}}}+{}\underbrace {|A_{n}-A|\,|B|} _{\leq \,\varepsilon /3{\text{ by (4)}}}\leq \varepsilon \,.\end{aligned}}}$$

By the definition of convergence of a series, C_n → AB as required.

Theorem

For ${\textstyle r>-1}$ and ${\textstyle s>-1}$, suppose the sequence ${\textstyle (a_{n})_{n\geq 0}}$ is ${\textstyle (C,\;r)}$ summable with sum A and ${\textstyle (b_{n})_{n\geq 0}}$ is ${\textstyle (C,\;s)}$ summable with sum B. Then their Cauchy product is ${\textstyle (C,\;r+s+1)}$ summable with sum AB.

Products of finitely many infinite series

Let ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle n\geq 2}$ (actually the following is also true for ${\displaystyle n=1}$ but the statement becomes trivial in that case) and let ${\textstyle \sum _{k_{1}=0}^{\infty }a_{1,k_{1}},\ldots ,\sum _{k_{n}=0}^{\infty }a_{n,k_{n}}}$ be infinite series with complex coefficients, from which all except the ${\displaystyle n}$th one converge absolutely, and the ${\displaystyle n}$th one converges. Then the limit

$${\displaystyle \lim _{N\to \infty }\sum _{k_{1}+\ldots +k_{n}\leq N}a_{1,k_{1}}\cdots a_{n,k_{n}}}$$

exists and we have:

$${\displaystyle \prod _{j=1}^{n}\left(\sum _{k_{j}=0}^{\infty }a_{j,k_{j}}\right)=\lim _{N\to \infty }\sum _{k_{1}+\ldots +k_{n}\leq N}a_{1,k_{1}}\cdots a_{n,k_{n}}}$$

Proof

Because

$${\displaystyle \forall N\in \mathbb {N}$$ :\sum _{k_{1}+\ldots +k_{n}\leq N}a_{1,k_{1}}\cdots a_{n,k_{n}}=\sum _{k_{1}=0}^{N}\sum _{k_{2}=0}^{k_{1}}\cdots \sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}}

the statement can be proven by induction over ${\displaystyle n}$: The case for ${\displaystyle n=2}$ 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 ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle n\geq 2}$, and let ${\textstyle \sum _{k_{1}=0}^{\infty }a_{1,k_{1}},\ldots ,\sum _{k_{n+1}=0}^{\infty }a_{n+1,k_{n+1}}}$ be infinite series with complex coefficients, from which all except the ${\displaystyle n+1}$th one converge absolutely, and the ${\displaystyle n+1}$-th one converges. We first apply the induction hypothesis to the series ${\textstyle \sum _{k_{1}=0}^{\infty }|a_{1,k_{1}}|,\ldots ,\sum _{k_{n}=0}^{\infty }|a_{n,k_{n}}|}$. We obtain that the series

$${\displaystyle \sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}\cdots \sum _{k_{n}=0}^{k_{n-1}}|a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}|}$$

converges, and hence, by the triangle inequality and the sandwich criterion, the series

$${\displaystyle \sum _{k_{1}=0}^{\infty }\left|\sum _{k_{2}=0}^{k_{1}}\cdots \sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}\right|}$$

converges, and hence the series

$${\displaystyle \sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}\cdots \sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}}$$

converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have:

$${\displaystyle {\begin{aligned}\prod _{j=1}^{n+1}\left(\sum _{k_{j}=0}^{\infty }a_{j,k_{j}}\right)&=\left(\sum _{k_{n+1}=0}^{\infty }\overbrace {a_{n+1,k_{n+1}}} ^{=:a_{k_{n+1}}}\right)\left(\sum _{k_{1}=0}^{\infty }\overbrace {\sum _{k_{2}=0}^{k_{1}}\cdots \sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}} ^{=:b_{k_{1}}}\right)\\&=\left(\sum _{k_{1}=0}^{\infty }\overbrace {\sum _{k_{2}=0}^{k_{1}}\sum _{k_{3}=0}^{k_{2}}\cdots \sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}\cdots a_{n,k_{1}-k_{2}}} ^{=:a_{k_{1}}}\right)\left(\sum _{k_{n+1}=0}^{\infty }\overbrace {a_{n+1,k_{n+1}}} ^{=:b_{k_{n+1}}}\right)\\&=\left(\sum _{k_{1}=0}^{\infty }\overbrace {\sum _{k_{3}=0}^{k_{1}}\sum _{k_{4}=0}^{k_{3}}\cdots \sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}\cdots a_{n,k_{1}-k_{3}}} ^{=:a_{k_{1}}}\right)\left(\sum _{k_{2}=0}^{\infty }\overbrace {a_{n+1,k_{2}}} ^{=:b_{n+1,k_{2}}=:b_{k_{2}}}\right)\\&=\left(\sum _{k_{1}=0}^{\infty }a_{k_{1}}\right)\left(\sum _{k_{2}=0}^{\infty }b_{k_{2}}\right)\\&=\left(\sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}a_{k_{2}}b_{k_{1}-k_{2}}\right)\\&=\left(\sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}\left(\overbrace {\sum _{k_{3}=0}^{k_{2}}\cdots \sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}\cdots a_{n,k_{2}-k_{3}}} ^{=:a_{k_{2}}}\right)\left(\overbrace {a_{n+1,k_{1}-k_{2}}} ^{=:b_{k_{1}-k_{2}}}\right)\right)\\&=\left(\sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}\overbrace {\sum _{k_{3}=0}^{k_{2}}\cdots \sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}\cdots a_{n,k_{2}-k_{3}}} ^{=:a_{k_{2}}}\overbrace {a_{n+1,k_{1}-k_{2}}} ^{=:b_{k_{1}-k_{2}}}\right)\\&=\sum _{k_{1}=0}^{\infty }\sum _{k_{2}=0}^{k_{1}}a_{n+1,k_{1}-k_{2}}\sum _{k_{3}=0}^{k_{2}}\cdots \sum _{k_{n+1}=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}\cdots a_{n,k_{2}-k_{3}}\end{aligned}}}$$

Therefore, the formula also holds for ${\displaystyle n+1}$.