Proof of the Cauchy Product

Why the Triangle Needs Controlling

In Cauchy Products, the Cauchy Product Theorem was stated for two absolutely convergent series. Its conclusion has two parts: the series formed from the convolution coefficients converges absolutely, and its sum is the product of the original sums. The finite identities from that tutorial explain how the coefficients are formed, but a proof of the theorem must also control what happens as the number of terms grows.

The key geometry is the difference between a square and a triangle. Multiplying two partial sums through index $M$ uses the square of pairs $0\leq i,j\leq M$. By contrast, the Cauchy-product partial sum through index $N$ uses the triangle $i+j\leq N$. When $N$ is about $2M$, the square fits inside the triangle. Absolute convergence lets us bound the terms in the part of the triangle outside that square. This is the step that turns finite multiplication into a statement about infinite sums.

Throughout, let $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge absolutely, and write their sums as $A$ and $B$. Let $c_n=\sum_{k=0}^{n}a_kb_{n-k}$, as in the definition of the Cauchy product. Define the absolute sums and their tails by

\alpha=\sum_{i=0}^{\infty}|a_i|,\qquad \beta=\sum_{j=0}^{\infty}|b_j|,\qquad \alpha_M=\sum_{i=M+1}^{\infty}|a_i|,\qquad \beta_M=\sum_{j=M+1}^{\infty}|b_j|.

Both total absolute sums are finite, and $\alpha_M$ and $\beta_M$ tend to zero as $M$ tends to infinity.

A Tail Estimate for the Triangle

Lemma (Triangular Tail Estimate): Let $N\geq0$ and $m=\lfloor N/2\rfloor$. For any finite set $F$ of pairs of nonnegative integers satisfying $i+j>N$, $$ \sum_{(i,j)\in F}|a_i b_j| \leq \alpha_m\beta+\alpha\beta_m. $$

Proof. If $i+j>N$, it cannot be the case that both $i\leq m$ and $j\leq m$. In fact, those two inequalities would give $i+j\leq2m\leq N$. Therefore each pair in $F$ has $i\geq m+1$ or $j\geq m+1$. Bound the terms from pairs with $i\geq m+1$ by the sum over all such $i$ and all $j$; bound the remaining terms by the sum over all $i$ and $j\geq m+1$. Counting a pair in both bounds, if necessary, only makes the upper bound larger. Since $F$ is finite and all terms in these bounds are nonnegative,

\sum_{(i,j)\in F}|a_i b_j| \leq \left(\sum_{i=m+1}^{\infty}|a_i|\right) \left(\sum_{j=0}^{\infty}|b_j|\right) + \left(\sum_{i=0}^{\infty}|a_i|\right) \left(\sum_{j=m+1}^{\infty}|b_j|\right) =\alpha_m\beta+\alpha\beta_m.

The right-hand side tends to zero as $N$ tends to infinity, since then $m=\lfloor N/2\rfloor$ tends to infinity. This proves the estimate. $\square$

The finite-set formulation makes the estimate independent of any choice of ordering for the pairs $(i,j)$. It says that any finite collection of terms outside a sufficiently large triangle has small total absolute value. This is precisely the form needed when comparing finite partial sums.

Proof of the Cauchy Product Theorem

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

Proof. First, for each $n$, the triangle inequality gives

|c_n| =\left|\sum_{i=0}^{n}a_i b_{n-i}\right| \leq\sum_{i=0}^{n}|a_i|\,|b_{n-i}|.

For every $N\geq0$, sum these inequalities from $n=0$ to $n=N$. Each pair in the resulting finite sum has nonnegative indices $i,j$ with $i+j\leq N$. These pairs are among those in the square $0\leq i,j\leq N$, so

\sum_{n=0}^{N}|c_n| \leq\sum_{\substack{i,j\geq0\\i+j\leq N}}|a_i|\,|b_j| \leq\left(\sum_{i=0}^{N}|a_i|\right)\left(\sum_{j=0}^{N}|b_j|\right) \leq\alpha\beta.

The partial sums of the nonnegative series $\sum_{n=0}^{\infty}|c_n|$ are increasing and bounded above by $\alpha\beta$. The Tail Criterion for Nonnegative Series, established earlier in the course, therefore gives convergence of $\sum_{n=0}^{\infty}|c_n|$. Thus the Cauchy product converges absolutely.

It remains to identify its sum. Let

P_M=\sum_{i=0}^{M}a_i,\qquad Q_M=\sum_{j=0}^{M}b_j.

Since the original series converge, $P_M\to A$ and $Q_M\to B$, and hence $P_MQ_M\to AB$. For a given $N$, set $M=\lfloor N/2\rfloor$. Every pair in the square $0\leq i,j\leq M$ satisfies $i+j\leq2M\leq N$. The Triangular Partial-Sum Identity from Cauchy Products now gives

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

The square in the second expression is contained in the triangle in the first. Their difference is a finite sum over pairs with $i+j\leq N$ for which $i>M$ or $j>M$. Applying the same tail bounds as in the Triangular Tail Estimate, this time to the absolute values of those extra terms, gives

\left|\sum_{n=0}^{N}c_n-P_MQ_M\right| \leq \alpha_M\beta+\alpha\beta_M.

As $N\to\infty$, $M=\lfloor N/2\rfloor\to\infty$, so the right-hand side tends to zero. Also $P_MQ_M\to AB$. It follows that the partial sums $\sum_{n=0}^{N}c_n$ tend to $AB$. Together with the absolute convergence already proved, this establishes both conclusions of the Cauchy Product Theorem. $\square$

Worked Applications

Worked Example: The Factorial Coefficients

Take $a_n=b_n=1/n!$ for $n\geq0$, where $0!=1$. Both series converge absolutely. The convolution coefficient is

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

The last equality is the finite binomial identity. In particular, the coefficient at index $2$ is

c_2=\frac{1}{0!2!}+\frac{1}{1!1!}+\frac{1}{2!0!} =\frac12+1+\frac12=2, \qquad \frac{2^2}{2!}=2.

This direct check confirms the convolution formula at that index. The Cauchy Product Theorem says that the series with coefficients $2^n/n!$ converges absolutely and has sum equal to the product of the sums of the two series with coefficients $1/n!$. In the notation of the Exponential Series tutorial, this is the identity $E(1)E(1)=E(2)$, also given by the Multiplication Law for the Exponential Series.

Worked Example: Two Geometric Series

Let $a_n=(-1/3)^n$ and $b_n=(1/4)^n$ for $n\geq0$. Both series converge absolutely. Using the finite geometric-sum formula in the convolution,

c_n=\sum_{k=0}^{n}\left(-\frac13\right)^k \left(\frac14\right)^{n-k} =\frac{(1/4)^{n+1}-(-1/3)^{n+1}}{(1/4)-(-1/3)} =\frac{12}{7}\left[\left(\frac14\right)^{n+1} -\left(-\frac13\right)^{n+1}\right].

For $n=0$, this gives $c_0=(12/7)(1/4+1/3)=1$, which agrees with $a_0b_0=1$. For $n=1$, it gives

c_1=\frac{12}{7}\left(\frac{1}{16}-\frac{1}{9}\right) =-\frac{1}{12}.

The convolution formula verifies the same value directly: $c_1=a_0b_1+a_1b_0=1/4-1/3=-1/12$. The original sums are $A=1/(1+1/3)=3/4$ and $B=1/(1-1/4)=4/3$. The theorem therefore guarantees that the Cauchy product converges absolutely and has sum $AB=(3/4)(4/3)=1$.

Worked Example: Conditional Convergence Is Not Enough

For $n\geq0$, take $a_n=b_n=(-1)^n/\sqrt{n+1}$. Each original series converges by the alternating-series test, because $1/\sqrt{n+1}$ decreases to zero. Neither converges absolutely: the absolute-value series is a constant multiple of the divergent $p$-series with exponent $1/2$.

The Cauchy-product coefficient is

c_n=(-1)^n\sum_{k=0}^{n} \frac{1}{\sqrt{(k+1)(n-k+1)}}.

For each $k$, the two positive numbers $k+1$ and $n-k+1$ have sum $n+2$. The arithmetic-geometric mean inequality gives

\sqrt{(k+1)(n-k+1)}\leq\frac{n+2}{2}, \qquad \frac{1}{\sqrt{(k+1)(n-k+1)}}\geq\frac{2}{n+2}.

There are $n+1$ terms in the inner sum, so

|c_n|\geq (n+1)\frac{2}{n+2} =\frac{2(n+1)}{n+2}\geq1.

Thus $c_n$ does not tend to zero. Since the terms of a convergent series must tend to zero, this Cauchy product diverges. This example shows why the absolute-convergence hypothesis in the Cauchy Product Theorem matters: convergence of both original series alone does not guarantee convergence of their Cauchy product.

How to Use the Proof

The argument separates the conclusion into two tasks. First, the coefficientwise triangle inequality bounds every finite partial sum of $\sum |c_n|$ by the product $\alpha\beta$. This proves absolute convergence without yet identifying the sum. Second, the triangular partial sums are compared with rectangular products $P_MQ_M$, whose limits are already known. The tails make the difference between those two finite expressions tend to zero.

A common pitfall is to treat the triangle and square as if they were identical, or to rearrange infinitely many cross-products merely because finite multiplication permits regrouping. They are not identical: for example, when $N\geq1$, the square through $N$ contains pairs with $i+j>N$, such as $(N,N)$. The proof does not discard such terms without justification. Instead, it chooses $N$ so the square through $M=\lfloor N/2\rfloor$ lies inside the triangle through $N$, then bounds the extra terms by absolute tails. That control is what allows the finite identity to pass to the limit.

Takeaway: Absolute convergence controls the cross-products outside a large square or triangle. This makes the triangular Cauchy-product partial sums approach the same limit as the rectangular products of the original partial sums, proving that the product series converges absolutely to $AB$.

Check Your Understanding

Use the tail estimate and the proof to answer the following questions.

Why does $i+j>N$ imply that at least one of $i$ or $j$ exceeds $\lfloor N/2\rfloor$?
Which inequality bounds $\sum_{n=0}^{N}|c_n|$ by a quantity independent of $N$?
Why does the square $0\leq i,j\leq M$ lie inside the triangle $i+j\leq N$ when $M=\lfloor N/2\rfloor$?
What properties of $\alpha_M$ and $\beta_M$ make the difference between the triangular and rectangular sums tend to zero?
In the conditional example, which necessary condition for series convergence fails for the Cauchy product?

What You'll Learn