Proof of The Cauchy Product of Two Absolutely Convergent Series

Throughout, $N$ ranges over $\mathbb{N}$, and $D_{n}$ and $c_{n}$ are as in the statement. Elements of a segment $[n]$ are natural numbers by Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers (i), so $a_{i}$ and $b_{j}$ are defined for $(i,j)\in[n]\times[n]$. Natural numbers with zero are read in $\mathbb{R}$ by The Real Line: Standing Notation and Background for Calculus on Intervals (vii), and their equality, sums, differences and order agree with those formed in $\mathbb{R}$ by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vi); the identities and inequalities between natural numbers below are checked in $\mathbb{R}$ in this way. Every $m\in\mathbb{N}$ satisfies $1\le m$ by Arithmetic and Order of the Natural Numbers (x), and every $m\in\mathbb{N}_{0}$ with $1\le m$ lies in $\mathbb{N}$ by Elements of a Segment Are Natural Numbers, and Mixed Transitivity of the Order on the Natural Numbers (i). All index sets below are subsets of finite products of segments (for the sets $U$ and $T'$ of Step 2, $U\subseteq Q_{N}$ by (1a) and $T'\subseteq[N]\times U$), hence finite: segments are finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vii), products of finite sets are finite by Adding an Element to, Subsets of, and Unions and Cartesian Products of Finite Sets Are Finite (iii), and subsets of finite sets are finite by Adding an Element to, Subsets of, and Unions and Cartesian Products of Finite Sets Are Finite (ii).

Step 1: index sets. For $N\in\mathbb{N}$ let $Q_{N}=[N]\times[N]$ and $T_{N}=\{(i,j)\in Q_{N}:i+j\le N+1\}$; so $T_{N}\subseteq Q_{N}$.

(1a) If $n\in[N]$, then $D_{n}\subseteq T_{N}$. Indeed, let $(i,j)\in D_{n}$; then $i\le n\le N$ and $j\le n\le N$, so $(i,j)\in Q_{N}$, and $i+j=n+1\le N+1$.

(1b) For every $(i,j)\in T_{N}$ there is exactly one $n\in[N]$ with $(i,j)\in D_{n}$. Existence: as $1\le i$ and $1\le j$, we have $1\le i+j$, so $n=(i+j)-1\in\mathbb{N}_{0}$, the difference, is defined and $n+1=i+j$. In $\mathbb{R}$, $n=i+(j-1)\ge i\ge1$, so $n\in\mathbb{N}$ and $i\le n$; likewise $j\le n$; and $n=(i+j)-1\le N$ since $i+j\le N+1$. Hence $n\in[N]$, $i,j\in[n]$ and $i+j=n+1$, that is, $(i,j)\in D_{n}$. Uniqueness: if also $(i,j)\in D_{m}$ with $m\in[N]$, then $m+1=i+j=n+1$, so $m=n$.

(1c) $Q_{N}\subseteq T_{N+N}$. Indeed, for $i,j\in[N]$ we have $i\le N\le N+N$ and $j\le N+N$, so $(i,j)\in Q_{N+N}$, and $i+j\le N+N\le(N+N)+1$.

Step 2: sums over $T_{N}$. Let $g:\mathbb{N}\times\mathbb{N}\to\mathbb{R}$ be a map. We show

$$\sum_{n=1}^{N}\sum_{p\in D_{n}}g(p)=\sum_{p\in T_{N}}g(p).\tag{$\ast$}$$

The left side is the sum over $n\in[N]$, by Sums and Products over a Finite Set and over an Interval (iv). Let $U=\bigcup_{n\in[N]}D_{n}$ and $T'=\{(n,p)\in[N]\times U:p\in D_{n}\}$. By Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (vi), with Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i) for the inner sums, whose terms agree with those of $g$ on each $D_{n}$, the left side of $(\ast)$ equals $\sum_{(n,p)\in T'}g(p)$. The map $\pi:T'\to T_{N}$, $(n,p)\mapsto p$, takes its values in $T_{N}$ by (1a). It is injective: if $(n,p),(n',p')\in T'$ and $p=p'$, then $p\in D_{n}\cap D_{n'}$ and $n=n'$ by (1b). It is surjective: for $p\in T_{N}$, (1b) gives $n\in[N]$ with $p\in D_{n}$, so $p\in U$ and $(n,p)\in T'$ with $\pi(n,p)=p$. So $\pi$ is a bijection, and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (iii), applied to $\pi$ and the restriction of $g$ to $T_{N}$, gives $\sum_{(n,p)\in T'}g(\pi(n,p))=\sum_{p\in T_{N}}g(p)$, which is $(\ast)$, as $g(\pi(n,p))=g(p)$ for $(n,p)\in T'$ (Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i)).

Step 3: the sums over $T_{N}$ and $Q_{N}$. For $N\in\mathbb{N}$ let $A_{N}=\sum_{k=1}^{N}a_{k}$, $B_{N}=\sum_{k=1}^{N}b_{k}$, $\bar A_{N}=\sum_{k=1}^{N}|a_{k}|$ and $\bar B_{N}=\sum_{k=1}^{N}|b_{k}|$ be the partial sums, sequences by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (i), applied to $(a_{k})$, $(b_{k})$ and to the sequences $(|a_{k}|)$ and $(|b_{k}|)$ of Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (iv). For $n\in\mathbb{N}$ let $\bar c_{n}=\sum_{(i,j)\in D_{n}}|a_{i}|\,|b_{j}|$; $(\bar c_{n})$ is a sequence in $\mathbb{R}$ by Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (iii), applied to $(|a_{k}|)$ and $(|b_{k}|)$. Let $C_{N}=\sum_{n=1}^{N}c_{n}$ and $\bar C_{N}=\sum_{n=1}^{N}\bar c_{n}$, sequences by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (i). Let $A$, $B$, $\bar A$ and $\bar B$ be the sums of $\sum_{k=1}^{\infty}a_{k}$, $\sum_{k=1}^{\infty}b_{k}$, $\sum_{k=1}^{\infty}|a_{k}|$ and $\sum_{k=1}^{\infty}|b_{k}|$, which exist: $\sum_{k=1}^{\infty}|a_{k}|$ and $\sum_{k=1}^{\infty}|b_{k}|$ converge by hypothesis and Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (iii), and $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=1}^{\infty}b_{k}$ converge by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iii); by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii), $A_{N}\to A$, $B_{N}\to B$, $\bar A_{N}\to\bar A$ and $\bar B_{N}\to\bar B$.

By $(\ast)$ with $g(i,j)=a_{i}b_{j}$ and with $g(i,j)=|a_{i}|\,|b_{j}|$, maps $\mathbb{N}\times\mathbb{N}\to\mathbb{R}$ by Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (ii),

$$C_{N}=\sum_{(i,j)\in T_{N}}a_{i}b_{j},\qquad\bar C_{N}=\sum_{(i,j)\in T_{N}}|a_{i}|\,|b_{j}|.$$

By Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (iii), with both finite sets equal to $[N]$, Sums and Products over a Finite Set and over an Interval (iv) and Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i) (the map $(i,j)\mapsto a_{i}b_{j}$, resp. $|a_{i}|\,|b_{j}|$, on $[N]\times[N]$ agrees with $g$ there),

$$A_{N}B_{N}=\sum_{(i,j)\in Q_{N}}a_{i}b_{j},\qquad\bar A_{N}\bar B_{N}=\sum_{(i,j)\in Q_{N}}|a_{i}|\,|b_{j}|.$$

Step 4: bounds. For all $i,j\in\mathbb{N}$, $0\le|a_{i}|$ and $0\le|b_{j}|$ by Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (iii), so $0\le|a_{i}|\,|b_{j}|$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (vii). Hence $\bar c_{n}\ge0$ for every $n\in\mathbb{N}$, by Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (ix) applied with $S$ and the finite set both equal to $D_{n}$, and, as $T_{N}\subseteq Q_{N}\subseteq T_{N+N}$ by Step 1, Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (ix) and Step 3 give

$$\bar C_{N}\le\bar A_{N}\bar B_{N}\le\bar C_{N+N}\qquad(N\in\mathbb{N}).\tag{$\ast\ast$}$$

By Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i), applied to $(|a_{k}|)$ and $(|b_{k}|)$, $0\le\bar A_{N}\le\bar A$ and $0\le\bar B_{N}\le\bar B$, so $\bar A_{N}\bar B_{N}\le\bar A\bar B$ by Rules of Order in an Ordered Ring: Differences, Sums, Negatives, Products and Squares (vii), and $\bar C_{N}\le\bar A\bar B$ by $(\ast\ast)$. Thus $\bar A\bar B$ is an upper bound of $\{\bar C_{N}:N\in\mathbb{N}\}$, so $(\bar C_{N})$ is bounded above, and, as $\bar c_{n}\ge0$ for every $n$, the series $\sum_{n=1}^{\infty}\bar c_{n}$ converges by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i). Let $\bar C$ be its sum; by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (i) again, $\bar C_{N}\le\bar C$ for every $N$, and $\bar C_{N}\to\bar C$ by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii).

Step 5: absolute convergence. For $n\in\mathbb{N}$, by Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (x) and $|a_{i}b_{j}|=|a_{i}|\,|b_{j}|$ (Rules of Arithmetic and Order in an Ordered Field (x), with Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i)),

$$|c_{n}|\le\sum_{(i,j)\in D_{n}}|a_{i}b_{j}|=\bar c_{n}.$$

Since $\sum_{n=1}^{\infty}\bar c_{n}$ converges, $\sum_{n=1}^{\infty}c_{n}$ converges absolutely by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iv). This is (i); in particular $\sum_{n=1}^{\infty}c_{n}$ converges, by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iii).

Step 6: $\bar C=\bar A\bar B$. Since $\bar C_{N}\le\bar A\bar B$ for every $N$ and $\bar C_{N}\to\bar C$, Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x) gives $\bar C\le\bar A\bar B$. By $(\ast\ast)$ and Step 4, $\bar A_{N}\bar B_{N}\le\bar C_{N+N}\le\bar C$ for every $N$; as $\bar A_{N}\bar B_{N}\to\bar A\bar B$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi) (the sequence $(\bar A_{N}\bar B_{N})$ exists by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v)), Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x) gives $\bar A\bar B\le\bar C$. Hence $\bar C=\bar A\bar B$.

Step 7: the sum. Let $R_{N}=Q_{N}\setminus T_{N}$. Then $Q_{N}=T_{N}\cup R_{N}$ with $T_{N}\cap R_{N}=\emptyset$, so by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (ii) and Step 3,

$$A_{N}B_{N}=C_{N}+\sum_{(i,j)\in R_{N}}a_{i}b_{j},\qquad\bar A_{N}\bar B_{N}=\bar C_{N}+\sum_{(i,j)\in R_{N}}|a_{i}|\,|b_{j}|.$$

Hence, by Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality (x), Rules of Arithmetic and Order in an Ordered Field (x) and Finite Sums of Reals: Congruence, the Product Map (x,y) ↦ f(x)h(y), and the Sequence of Diagonal Sums (i),

$$|A_{N}B_{N}-C_{N}|=\Big|\sum_{(i,j)\in R_{N}}a_{i}b_{j}\Big|\le\sum_{(i,j)\in R_{N}}|a_{i}|\,|b_{j}|=\bar A_{N}\bar B_{N}-\bar C_{N}.$$

By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi) and Step 6, $\bar A_{N}\bar B_{N}-\bar C_{N}\to\bar A\bar B-\bar C=0$, so $(\bar A_{N}\bar B_{N}-\bar C_{N})$, a sequence by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v), is a null sequence, and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xii) with $L=0$ gives $A_{N}B_{N}-C_{N}\to0$. Since $A_{N}B_{N}\to AB$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), the rule for differences in Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), applied to $(A_{N}B_{N})$ and $(A_{N}B_{N}-C_{N})$, sequences by Sequences of Real Numbers Built from Given Ones: Partial Sums, Index Shifts, Predecessor Reindexing, Absolute Values and Termwise Operations (v), gives

$$C_{N}=A_{N}B_{N}-(A_{N}B_{N}-C_{N})\to AB-0=AB.$$

So, by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii) and The Limit of a Convergent Sequence (i), $\sum_{n=1}^{\infty}c_{n}$ converges and its sum is $AB=\big(\sum_{k=1}^{\infty}a_{k}\big)\big(\sum_{k=1}^{\infty}b_{k}\big)$. This is (ii).