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If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals whose series both converge absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), with sums $A$ and $B$ , and let $(c_{n})$ be their Cauchy product (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ). Then $\sum c_{n}$ converges absolutely, and

$\sum_{n = 0}^{\infty} c_{n} = A B .$

Moreover $\sum_{n = 0}^{\infty} ∣ c_{n} ∣ \leq (\sum_{k = 0}^{\infty} ∣ a_{k} ∣) (\sum_{k = 0}^{\infty} ∣ b_{k} ∣)$ .

Combined with Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum this says that within the absolutely convergent series the product behaves exactly as one would want: it converges, its sum is the product of the sums, and neither factor's order nor the product's order matters.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ with $\sum ∣ a_{k} ∣$ and $\sum ∣ b_{k} ∣$ convergent, sums $L_{a}$ and $L_{b}$ respectively, partial sums $P_{N} = \sum_{k < N} ∣ a_{k} ∣$ and $Q_{m} = \sum_{j < m} ∣ b_{j} ∣$ , and the Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[L1]

The finite identity of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , claim 1: for arbitrary sequences $(x_{k})$ , $(y_{k})$ with partial sums $Y_{m} = \sum_{j < m} y_{j}$ and Cauchy product $(z_{n})$ , one has $\sum_{n < N} z_{n} = \sum_{i < N} x_{i} Y_{N - i}$ for every $N$ .

[L2]

Mertens' theorem, claim 2 of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ : if $\sum x_{k}$ converges absolutely and $\sum y_{k}$ converges, their Cauchy product converges to the product of the sums.

[L3]

$∣ \sum_{k < n} x_{k} ∣ \leq \sum_{k < n} ∣ x_{k} ∣$ (Triangle inequality for finite sums).

[L4]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[L5]

Finite sums are monotone in their terms and scale by a constant factor; the empty sum is $0$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above, and then every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

Direct comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L8]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Proof

technique · direct

1.1givenL4L6

Both $L_{a}$ and $L_{b}$ are nonnegative, and $P_{N} \leq L_{a}$ and $Q_{m} \leq L_{b}$ for all $N$ and $m$ , the terms $∣ a_{k} ∣$ and $∣ b_{j} ∣$ being nonnegative.

1.2givenL4L5

Put $γ_{n} : = \sum_{k = 0}^{n} ∣ a_{k} ∣ ∣ b_{n - k} ∣$ , the Cauchy product of the sequences $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ ; every $γ_{n}$ is nonnegative.

2.1step 1.2L3L4

For every $n$ , $∣ c_{n} ∣ = ∣ \sum_{k = 0}^{n} a_{k} b_{n - k} ∣ \leq \sum_{k = 0}^{n} ∣ a_{k} b_{n - k} ∣ = γ_{n}$ .

2.2step 1.2L1

Applying [L1] to $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ gives $\sum_{n < N} γ_{n} = \sum_{i < N} ∣ a_{i} ∣ Q_{N - i}$ for every $N$ .

3.1step 1.1step 2.2L5

Since $0 \leq Q_{N - i} \leq L_{b}$ and $∣ a_{i} ∣ \geq 0$ , monotonicity and scaling give $\sum_{i < N} ∣ a_{i} ∣ Q_{N - i} \leq \sum_{i < N} ∣ a_{i} ∣ L_{b} = L_{b} P_{N} \leq L_{b} L_{a}$ for every $N$ .

4.1step 1.2step 3.1L6

So $\sum γ_{n}$ is a series of nonnegative terms whose partial sums are bounded above by $L_{a} L_{b}$ ; it therefore converges, with sum at most $L_{a} L_{b}$ .

5.1step 2.1step 4.1L6L7

By step 2.1 and comparison, $\sum ∣ c_{n} ∣$ converges, and its sum is at most that of $\sum γ_{n}$ , hence at most $L_{a} L_{b}$ ; that is, $\sum c_{n}$ converges absolutely and satisfies the displayed bound.

6.1givenL2L8∎

The hypotheses of Mertens' theorem hold, $\sum a_{k}$ converging absolutely and $\sum b_{k}$ converging by step 1.1 and [L8]; so $\sum c_{n}$ converges with sum $A B$ .

Remarks

Only claim 1 of Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ is reused, and it is reused for a different pair of sequences. The identity there is proved for arbitrary real sequences and carries no convergence hypothesis, which is exactly what allows it to be applied here to $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ . Nothing about the absolute values is reproved.
The bound is not an equality. The sum of $\sum ∣ c_{n} ∣$ can be strictly less than $L_{a} L_{b}$ , since cancellation inside each $c_{n}$ is invisible to $γ_{n}$ ; the inequality is all that is claimed and all that is needed.
Absolute convergence of both factors is a genuine strengthening. Mertens' theorem already gives $\sum c_{n} = A B$ with only one factor absolutely convergent; what the second hypothesis buys is that the product series is itself absolutely convergent, hence unconditionally convergent (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum), so that its terms may be reordered in turn.

If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If ∑ak∑ak​ and ∑bk∑bk​ both converge absolutely then their Cauchy product converges absolutely, with sum ABAB

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$