Cauchy Products

From Power-Series Products to Series Products

In Multiplication of Power Series, each coefficient of the product was formed by collecting all pairs of terms whose degrees add to the same number. The same finite calculation makes sense for two ordinary series, even when there is no variable or power of $x-a$. The resulting series is called their Cauchy product.

There is an important distinction between defining its coefficients and proving that the resulting series converges to the product of the original sums. Each coefficient is a finite sum, so defining it requires no convergence assumption. The identity between the sums, however, needs hypotheses. We will state the standard theorem for absolutely convergent series and develop some finite identities and examples that clarify what it says.

Definition and Finite Partial Sums

Let $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ be two series. To form a term with total index $n$, pair $a_k$ with $b_{n-k}$ for each $k$ from $0$ to $n$.

Definition (Cauchy Product): The Cauchy product of the series $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ is the series $\sum_{n=0}^{\infty}c_n$, where $$ c_n=\sum_{k=0}^{n}a_kb_{n-k}\qquad(n\geq0). $$ Each coefficient is a finite sum, regardless of whether either original series converges.

For example, the first coefficients are $c_0=a_0b_0$, $c_1=a_0b_1+a_1b_0$, and $c_2=a_0b_2+a_1b_1+a_2b_0$. The formula groups the pairs of indices along diagonals $i+j=n$. Summing the first $N+1$ Cauchy-product terms therefore includes exactly the pairs satisfying $i+j\leq N$.

Proposition (Triangular Partial-Sum Identity): For every integer $N\geq0$, $$ \sum_{n=0}^{N}c_n = \sum_{\substack{i,j\geq0\\i+j\leq N}}a_ib_j. $$

Proof. Substitute the definition of $c_n$ into the partial sum. All sums involved are finite, so their terms can be regrouped:

\sum_{n=0}^{N}c_n =\sum_{n=0}^{N}\sum_{i=0}^{n}a_i b_{n-i} =\sum_{\substack{i,j\geq0\\i+j\leq N}}a_i b_j.

In the last expression, each pair $(i,j)$ occurs exactly once, at $n=i+j$, and the condition $n\leq N$ becomes $i+j\leq N$. This proves the identity. $\square$

The triangular region in this identity is worth distinguishing from the square used in multiplying two finite partial sums. The product $\left(\sum_{i=0}^{N}a_i\right)\left(\sum_{j=0}^{N}b_j\right)$ includes all pairs with $i\leq N$ and $j\leq N$, including pairs for which $i+j>N$. Thus finite multiplication explains the coefficients, but it does not by itself identify the limit of the Cauchy-product partial sums.

The Cauchy Product Theorem

Theorem (Cauchy Product Theorem): Suppose $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ both converge absolutely, with sums $A$ and $B$, respectively. If $c_n=\sum_{k=0}^{n}a_kb_{n-k}$, then $\sum_{n=0}^{\infty}c_n$ converges absolutely and $$ \sum_{n=0}^{\infty}c_n=AB. $$

Absolute convergence is a sufficient hypothesis that allows the infinitely many cross-products to be controlled. The theorem does not say that the Cauchy product must fail when an original series is not absolutely convergent; it says that this particular guarantee is not available from the stated assumptions alone. The proof of the Cauchy Product Theorem will examine how absolute convergence controls the triangular sums and their tails.

Worked Example: Two Geometric Series

Take $a_n=2^{-n}$ and $b_n=3^{-n}$ for $n\geq0$. Both series converge absolutely, so the Cauchy Product Theorem applies. For each $n$,

c_n=\sum_{k=0}^{n}2^{-k}3^{-(n-k)} =3^{-n}\sum_{k=0}^{n}\left(\frac32\right)^k =3^{-n}\frac{(3/2)^{n+1}-1}{(3/2)-1} =3\cdot2^{-n}-2\cdot3^{-n}.

For $n=0$, the formula gives $c_0=3-2=1$, as required by $a_0b_0=1$. For $n=1$, it gives $c_1=3/2-2/3=5/6$; directly, $a_0b_1+a_1b_0=1/3+1/2=5/6$. The original sums are $2$ and $3/2$, so their product is $3$. The coefficients also give

\sum_{n=0}^{\infty}c_n =3\sum_{n=0}^{\infty}2^{-n} -2\sum_{n=0}^{\infty}3^{-n} =3(2)-2\left(\frac32\right)=3.

Here the coefficient computation and the theorem’s convergence guarantee fit together: the finite geometric sum determines each $c_n$, while absolute convergence justifies the conclusion about the sum of all those coefficients.

Two Further Finite Facts

The convolution formula also has useful algebraic structure. In particular, forming a Cauchy product in stages gives the same coefficient as forming it in the other order. This statement concerns coefficients only, so its proof involves finite sums and does not require convergence.

Proposition (Associativity of the Coefficient Operation): Let $c_n=\sum_{i=0}^{n}a_i b_{n-i}$. The coefficient of index $n$ obtained by taking the Cauchy product of $(c_n)$ with $(e_n)$ is $$ \sum_{\substack{i,j,k\geq0\\i+j+k=n}}a_i b_j e_k. $$ The same coefficient results if one first takes the Cauchy product of $(b_n)$ and $(e_n)$, and then takes its Cauchy product with $(a_n)$.

Proof. Write the coefficient obtained by first forming $c_n$ as

\sum_{m=0}^{n}c_m e_{n-m} =\sum_{m=0}^{n}\sum_{i=0}^{m}a_i b_{m-i}e_{n-m}.

In each term, set $j=m-i$ and $k=n-m$. These indices are nonnegative and satisfy $i+j+k=n$. Conversely, any nonnegative $i,j,k$ with $i+j+k=n$ determines exactly one term in the double sum by taking $m=i+j$. Hence the coefficient equals the stated triple sum. If we first combine $b$ and $e$, the resulting coefficient is $\sum_{i=0}^{n}\sum_{j=0}^{n-i}a_i b_j e_{n-i-j}$, which contains exactly the same triples. All rearrangements are of finite sums, proving the claim. $\square$

A finite factor gives another useful case in which the sum identity can be proved without requiring absolute convergence of the other series.

Proposition (Cauchy Product with a Finite Series): Suppose $a_k=0$ for $k>m$, where $m\geq0$, and $\sum_{n=0}^{\infty}b_n=B$ converges. Then the Cauchy product converges and $$ \sum_{n=0}^{\infty}c_n =\left(\sum_{k=0}^{m}a_k\right)B. $$

Proof. For every $n$, the definition gives $c_n=\sum_{k=0}^{\min\{m,n\}}a_kb_{n-k}$. For $N\geq m$, sum this identity from $n=0$ to $n=N$. Since there are only finitely many $k$ with $a_k\ne0$, regrouping gives

\sum_{n=0}^{N}c_n =\sum_{k=0}^{m}a_k\sum_{n=k}^{N}b_{n-k} =\sum_{k=0}^{m}a_k\sum_{r=0}^{N-k}b_r.

For each fixed $k$, $N-k$ tends to infinity with $N$, so the inner partial sum tends to $B$. The outer sum has only $m+1$ terms, and therefore its limit is $\sum_{k=0}^{m}a_kB$. This proves convergence of the Cauchy product and the asserted formula. $\square$

Worked Example: A Finite Factor and a Convergent Series

Let the finite series be $1-2+1$, with coefficients $a_0=1$, $a_1=-2$, $a_2=1$, and $a_k=0$ for $k\geq3$. Let $b_n=(-1)^n/(n+1)$. The alternating harmonic series $\sum_{n=0}^{\infty}b_n$ converges, though it is not absolutely convergent. The finite-factor proposition applies.

The first product coefficients are $c_0=a_0b_0=1$, $c_1=a_0b_1+a_1b_0=-1/2-2=-5/2$, and

c_2=a_0b_2+a_1b_1+a_2b_0 =\frac13+1+1=\frac73.

For $n\geq2$, the three possible contributions are $b_n-2b_{n-1}+b_{n-2}$. The proposition gives the sum without requiring a new infinite rearrangement: the finite coefficient sum is $1-2+1=0$, so the Cauchy product converges to $0\cdot B=0$. This example illustrates why the absolutely convergent theorem is not the only available result, while also showing the special role of a finite factor.

Worked Example: Checking Associativity at One Coefficient

Take three sequences $a_n=2^{-n}$, $b_n=3^{-n}$, and $e_n=4^{-n}$. The coefficient of index $2$ in their combined Cauchy product is the sum over the six triples of nonnegative indices that add to $2$:

a_2b_0e_0+a_0b_2e_0+a_0b_0e_2 +a_1b_1e_0+a_1b_0e_1+a_0b_1e_1 =\frac14+\frac19+\frac1{16}+\frac16+\frac18+\frac1{12} =\frac{115}{144}.

For example, the term $1/6$ comes from the index triple $(1,1,0)$, since $a_1b_1e_0=(1/2)(1/3)(1)=1/6$. Whether the first two sequences are combined first or the last two are combined first, the same six triples occur exactly once. The associativity proposition guarantees this agreement for every coefficient, not just the one calculated here.

What the Hypotheses Do—and Do Not—Say

The Cauchy product is always defined coefficient by coefficient, but an infinite series is more than its list of coefficients: its partial sums must converge. The triangular partial-sum identity makes the issue visible. A finite product of partial sums uses a square of index pairs, while a Cauchy-product partial sum uses a triangle. To pass from finite calculations to an identity of infinite sums, one must control the terms outside the triangle. Absolute convergence supplies the hypothesis in the Cauchy Product Theorem; the theorem’s proof is the next step.

The finite-factor proposition is a useful edge case. Its proof needs only finitely many shifted copies of a convergent series, so it works even when that series is not absolutely convergent. Do not mistake this special case for permission to rearrange arbitrary conditionally convergent series: the finiteness of the first factor is essential to that argument.

Takeaway: The Cauchy product groups cross-products along diagonals: its coefficient of index $n$ is $\sum_{k=0}^{n}a_kb_{n-k}$. Absolute convergence of both original series guarantees that the product series converges to the product of their sums.

Check Your Understanding

Use the definition and the finite identities to answer the following questions.

Write the first three coefficients of the Cauchy product in terms of $a_n$ and $b_n$.
Which pairs of indices are included in the partial sum through coefficient $c_N$?
What convergence hypothesis does the Cauchy Product Theorem require of the two original series?
Why does the associativity proposition require no convergence hypothesis?
In the finite-factor proposition, where is the assumption that only finitely many $a_k$ are nonzero used?

What You'll Learn