Deriving the Cauchy Product Formula

Why the Product Formula Needs Justification

Multiplying two polynomials involves only finitely many terms. To find the coefficient of a particular power, we collect the products that produce that power. For infinite power series, the same pattern suggests a formula, but there is an extra issue: infinitely many terms are involved, so we must justify regrouping them.

Absolute convergence provides that justification. When both series converge absolutely at a point, their product can be represented by collecting together all term-pairs that contribute to each power of the centered variable.

Central idea: The coefficient of the $n$th power in a product is the sum of all products of coefficients whose indices add to $n$. Absolute convergence ensures that these terms may be grouped in this way without changing the sum.

Defining the Cauchy Product

Let two power series have the same center $c$:

\sum_{n=0}^{\infty} a_n(x-c)^n \qquad\text{and}\qquad \sum_{n=0}^{\infty} b_n(x-c)^n.

For each nonnegative integer $n$, define the finite sum

d_n=\sum_{k=0}^{n}a_kb_{n-k} =a_0b_n+a_1b_{n-1}+\cdots+a_nb_0.

The power series $\sum_{n=0}^{\infty}d_n(x-c)^n$ is called the Cauchy product of the original series. The formula accounts for every pair of terms that produces the power $(x-c)^n$: the indices of such a pair must add to $n$.

Definition: The Cauchy product coefficients of the sequences $(a_n)$ and $(b_n)$ are $d_n=\sum_{k=0}^{n}a_kb_{n-k}$, for $n=0,1,2,\ldots$. The formula defines each $d_n$ using only a finite sum.

For example, the constant coefficient is $d_0=a_0b_0$. The coefficient of $(x-c)$ is $d_1=a_0b_1+a_1b_0$, and the coefficient of $(x-c)^2$ is $d_2=a_0b_2+a_1b_1+a_2b_0$. These are exactly the pairs of terms whose powers add to $0$, $1$, or $2$, respectively.

Why Absolute Convergence Allows the Product

The product formula is valid whenever both original series converge absolutely at the point under consideration. More explicitly, fix $x$, put $t=x-c$, and suppose

\sum_{n=0}^{\infty}|a_nt^n|<\infty \qquad\text{and}\qquad \sum_{n=0}^{\infty}|b_nt^n|<\infty.

Write the values of the two series as $A=\sum_{n=0}^{\infty}a_nt^n$ and $B=\sum_{n=0}^{\infty}b_nt^n$. The product $AB$ can first be viewed as a sum over pairs of indices: the term with indices $k$ and $m$ is $a_kb_mt^{k+m}$. Regrouping those terms by the total index $n=k+m$ gives the Cauchy product.

Theorem (Cauchy product): If both power series converge absolutely at $x$, then their product is represented there by $$ \left(\sum_{n=0}^{\infty}a_n(x-c)^n\right) \left(\sum_{n=0}^{\infty}b_n(x-c)^n\right) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_kb_{n-k}\right)(x-c)^n. $$ In particular, this applies at every point inside the common interval of absolute convergence. It may also apply at a boundary point if both series converge absolutely there.

Derivation of the Formula

The essential step is to justify the change from grouping terms by pairs of indices to grouping them by the sum of those indices. The absolute values of all the pair-terms have finite total sum:

\sum_{k=0}^{\infty}\sum_{m=0}^{\infty} |a_kb_mt^{k+m}| = \left(\sum_{k=0}^{\infty}|a_k||t|^k\right) \left(\sum_{m=0}^{\infty}|b_m||t|^m\right) <\infty.

The equality follows by multiplying finite partial sums of nonnegative terms and then taking their limits. The finiteness means that the double series is absolutely convergent. An absolutely convergent series may be rearranged or grouped without changing its sum.

Express the product using pairs of terms.
Finite partial sums multiply by the distributive law. As the partial sums converge to $A$ and $B$, their products converge to $AB$. Absolute convergence ensures that the resulting sum over all pairs has a well-defined value.

Group pairs by their total index.
Each pair $(k,m)$ contributes $a_kb_mt^{k+m}$. For a fixed total $n=k+m$, the possible pairs are $(0,n),(1,n-1),\ldots,(n,0)$.

Collect the coefficient of each power.
The terms with total index $n$ add to $\left(\sum_{k=0}^{n}a_kb_{n-k}\right)t^n=d_nt^n$. Summing these groups over $n$ gives the stated product formula.

In equations, the regrouping is

AB = \sum_{k=0}^{\infty}\sum_{m=0}^{\infty}a_kb_mt^{k+m} = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{n}a_kb_{n-k}\right)t^n.

The last inner sum is finite for each $n$, while the outer series is justified by the absolute convergence of the double series. This is the reason the condition matters: without an applicable convergence justification, rearranging infinitely many terms can change their sum or fail to give a valid representation.

Worked Examples

Worked Example: Multiplying a Geometric Series by a Polynomial

Find the product of $1+x$ and the geometric series $\sum_{n=0}^{\infty}x^n$ for $|x|<1$.

Write $1+x=\sum_{n=0}^{\infty}a_nx^n$, where $a_0=1$, $a_1=1$, and $a_n=0$ for $n\ge2$. For the geometric series, $b_n=1$ for every $n\ge0$. Thus $d_0=a_0b_0=1$. For $n\ge1$, only the terms with $a_0$ and $a_1$ can contribute:

d_n=a_0b_n+a_1b_{n-1}=1\cdot1+1\cdot1=2.

Therefore the product series is $1+2x+2x^2+2x^3+\cdots$. To check it, the geometric series sums to $1/(1-x)$ when $|x|<1$, so the original product is $(1+x)/(1-x)$. The proposed series sums to

1+2\sum_{n=1}^{\infty}x^n = 1+\frac{2x}{1-x} = \frac{1-x+2x}{1-x} = \frac{1+x}{1-x}.

The two expressions agree, and the geometric series is absolutely convergent for $|x|<1$, as required.

Worked Example: Squaring the Geometric Series

Use the Cauchy product to find the series for $\left(\sum_{n=0}^{\infty}x^n\right)^2$, for $|x|<1$.

Both coefficient sequences are $a_n=b_n=1$. For each $n\ge0$, the coefficient is

d_n=\sum_{k=0}^{n}1\cdot1 =\underbrace{1+1+\cdots+1}_{n+1\text{ terms}} =n+1.

The first coefficients are $d_0=1$, $d_1=2$, $d_2=3$, and $d_3=4$. Hence the product series is $\sum_{n=0}^{\infty}(n+1)x^n=1+2x+3x^2+4x^3+\cdots$.

For a check, let $S=\sum_{n=0}^{\infty}(n+1)x^n$. When $|x|<1$, this series converges. Subtracting its shifted multiple gives

S-xS = (1+2x+3x^2+\cdots)-(x+2x^2+3x^3+\cdots) = 1+x+x^2+\cdots = \frac{1}{1-x}.

Thus $(1-x)S=1/(1-x)$, so $S=1/(1-x)^2$. This matches the original product, since each geometric series equals $1/(1-x)$. Absolute convergence holds for both factors throughout $|x|<1$.

Worked Example: Multiplying Two Exponential Series

Apply the formula to $\left(\sum_{n=0}^{\infty}x^n/n!\right)^2$.

Here $a_n=b_n=1/n!$. For a fixed $n$,

d_n = \sum_{k=0}^{n}\frac{1}{k!}\frac{1}{(n-k)!} = \frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{k!(n-k)!} = \frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k} = \frac{2^n}{n!}.

The last equality is the binomial identity $\sum_{k=0}^{n}\binom{n}{k}=2^n$. For example, $d_0=1$, $d_1=1+1=2$, $d_2=1/2+1+1/2=2$, and $d_3=1/6+1/2+1/2+1/6=4/3$. The resulting product series is $\sum_{n=0}^{\infty}2^nx^n/n!$.

By the exponential power series, this is $e^{2x}$. Each factor is $e^x$, so the check is $e^x e^x=e^{2x}$. Both exponential series converge absolutely for every real $x$, so the Cauchy product applies for every real $x$.

A Common Mistake: Pairing Only Matching Indices

A tempting but incorrect shortcut is to multiply only terms with the same index. For example, that shortcut would turn $(1+x+x^2+\cdots)^2$ into $1+x^2+x^4+\cdots$, by multiplying $1$ by $1$, $x$ by $x$, and so on.

This misses cross-products. The coefficient of $x$ in the actual product receives two contributions: $1\cdot x$ and $x\cdot1$. The Cauchy product formula gives

d_1=a_0b_1+a_1b_0=1\cdot1+1\cdot1=2,

not $0$. Indeed, the incorrect series has no $x$-term, whereas the correct product begins $1+2x+3x^2+\cdots$. To find the coefficient of a power, include every index pair whose indices add to its exponent.

When the Formula Matters

The Cauchy product turns multiplication of series into a precise rule for combining their coefficients. It is useful whenever a product of functions is to be represented as a power series: the rule identifies the proposed coefficients, and the absolute-convergence argument establishes that the resulting series really equals the product at points where both original series converge absolutely.

The common center is important because it gives both series the same powers $(x-c)^n$. If their centers differ, their terms are not already expressed in a shared set of powers, so this formula cannot be applied in this direct form. When the centers agree, keep two facts in view: the coefficient rule is finite at each degree, and the equality of the infinite series requires absolute convergence at the point in question.

Check Your Understanding

Use the definitions and convergence condition from this tutorial to answer the following.

Write the Cauchy product coefficient $d_3$ in terms of $a_0,a_1,a_2,a_3$ and $b_0,b_1,b_2,b_3$.
Why does absolute convergence matter when the double sum is grouped by total index?
For $a_n=1$ and $b_n=2$ for all $n\ge0$, find $d_0,d_1,d_2$.
What two contributions make up the coefficient of $x$ in $(1+x+x^2+\cdots)^2$?
State the condition at a point $x$ that allows the Cauchy product theorem to be applied.