The Cauchy Product

Prove Mertens' theorem: if $\sum a_n$ converges absolutely to $A$ and $\sum b_n$ converges to $B$ , then the Cauchy product $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ satisfies $\sum c_n = AB$ .

C_N = \sum_{n=0}^{N} c_n = \sum_{k=0}^{N} a_k B_{N-k},\qquad B_M = \sum_{m=0}^{M} b_m

C_N - AB = (A_N - A)B - \sum_{k=0}^{N} a_k R_{N-k},\qquad R_M = B - B_M \to 0

\left|\sum_{k=0}^{N-M-1} a_k R_{N-k}\right| \leq \varepsilon \sum_{k=0}^{\infty} |a_k|

n = 12

,

a_k = b_k = 1/k!

:

C_{12} \approx 7.3890546

and

A_{12}B_{12} \approx 7.3890561

— the two sequences climb to

e^2

together, with

c_{12} = 2^{12}/12! \approx 8.55 \times 10^{-6}

.

Show proof

Write $A_N, B_N, C_N$ for the partial sums. In $C_N = \sum_{n=0}^{N}\sum_{k=0}^{n} a_k b_{n-k}$ , substituting $m = n-k$ turns each term into $a_k b_m$ with $k + m \leq N$ , so reordering the finite sum gives $C_N = \sum_{k=0}^{N} a_k \left(\sum_{m=0}^{N-k} b_m\right) = \sum_{k=0}^{N} a_k B_{N-k}$ . This rearrangement is legal because the sum is finite.
With the remainder $R_M = B - B_M \to 0$ , $B_{N-k} = B - R_{N-k}$ , so $C_N = A_N B - \sum_{k=0}^{N} a_k R_{N-k}$ , hence $C_N - AB = (A_N - A)B - \sum_{k=0}^{N} a_k R_{N-k}$ . The first term tends to $0$ because $A_N \to A$ .
Control the second term. Absolute convergence gives $\alpha = \sum_{k=0}^{\infty} |a_k| < \infty$ . Given $\varepsilon > 0$ , pick $M$ with $|R_m| < \varepsilon$ for $m > M$ . Split the sum at $k = N-M$ : on the head, $|R_{N-k}| < \varepsilon$ , so $\left|\sum_{k=0}^{N-M-1} a_k R_{N-k}\right| \leq \varepsilon \alpha$ .
The tail has only $M+1$ terms: $\left|\sum_{k=N-M}^{N} a_k R_{N-k}\right| \leq \left(\max_{0 \leq m \leq M} |R_m|\right) \cdot \max_{N-M \leq k \leq N} |a_k|$ , and since an absolutely convergent series has $a_k \to 0$ , this maximum tends to $0$ as $N \to \infty$ . Hence $\limsup_{N \to \infty} \left|\sum_{k=0}^{N} a_k R_{N-k}\right| \leq \varepsilon \alpha$ , and $\varepsilon$ was arbitrary, so the sum tends to $0$ .
Both terms tend to $0$ , so $C_N - AB \to 0$ , i.e. $\sum c_n = AB$ . The crux was the $\varepsilon \alpha$ step: the absolutely convergent $\sum |a_k|$ crushes the tail $\sum a_k R_{N-k}$ so that $R_{N-k}$ is uniformly small — the conditionally convergent $b$ side alone cannot do this (see the counterexample in problem 3).

Show that the Cauchy product of the exponential series $\sum_{k=0}^{\infty} 1/k! = e$ with itself is $e^2$ : prove $c_n = \sum_{k=0}^{n} \frac{1}{k!(n-k)!} = \frac{2^n}{n!}$ by the binomial theorem, and justify $\sum c_n = e^2$ via Mertens' theorem.

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!}

\sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} \frac{2^n}{n!} = e^2

n = 10

:

c_{10} = 2^{10}/10! \approx 2.82 \times 10^{-4}

,

C_{10} \approx 7.3889947

,

A_{10}B_{10} \approx 7.3890560

— the Cauchy-product partial sums and the product of partial sums approach

e^2

together.

Show proof

With $a_k = b_k = 1/k!$ , the Cauchy-product term is $c_n = \sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{1}{(n-k)!}$ . Multiplying and dividing by $n!$ gives $\frac{1}{k!(n-k)!} = \frac{1}{n!} \binom{n}{k}$ , so $c_n = \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k}$ .
By the binomial theorem $(1+1)^n = \sum_{k=0}^{n} \binom{n}{k}$ , the inner sum is $2^n$ , so $c_n = \frac{2^n}{n!}$ . The Cauchy-product series is therefore $\sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} \frac{2^n}{n!}$ , which is the exponential series at $x = 2$ , summing to $e^2$ .
This computation is legitimate because the exponential series converges absolutely: the ratio test gives $\frac{1/(k+1)!}{1/k!} = \frac{1}{k+1} \to 0$ , so Mertens' theorem (problem 1) applies and yields $\sum c_n = e \cdot e = e^2$ — the series version of the exponent law $e^1 \cdot e^1 = e^2$ .

Put $a_k = b_k = \frac{(-1)^k}{\sqrt{k+1}}$ . Both series $\sum a_k$ , $\sum b_k$ converge, but show $|c_n| \geq \frac{2(n+1)}{n+2} \geq 1$ to prove the Cauchy product $\sum c_n$ diverges. (Hint: AM-GM.)

|c_n| = \sum_{k=0}^{n} \frac{1}{\sqrt{(k+1)(n-k+1)}} \geq (n+1)\cdot\frac{2}{n+2} \geq 1

(k+1)(n-k+1) \leq \left(\frac{(k+1)+(n-k+1)}{2}\right)^2 = \left(\frac{n+2}{2}\right)^2

c_n \not\to 0 \quad\Longrightarrow\quad \sum c_n \text{ diverges}

n = 20

:

|c_{20}| \approx 2.516872 \geq 1.909091 \geq 1

, so

c_n \not\to 0

.

C_{20} \approx 1.627909

drifts and

A_{20}B_{20} \approx 0.507955

does not settle either.

Show proof

First, both factor series converge: $a_k = \frac{(-1)^k}{\sqrt{k+1}}$ alternates in sign and $\frac{1}{\sqrt{k+1}}$ decreases monotonically to $0$ , so Leibniz's test applies. But $\sum |a_k| = \sum \frac{1}{\sqrt{k+1}}$ is a $p$ -series with $p = \frac12$ , hence diverges — the convergence is conditional.
The Cauchy-product term is $c_n = \sum_{k=0}^{n} \frac{(-1)^k}{\sqrt{k+1}} \cdot \frac{(-1)^{n-k}}{\sqrt{n-k+1}} = (-1)^n \sum_{k=0}^{n} \frac{1}{\sqrt{(k+1)(n-k+1)}}$ , so $|c_n| = \sum_{k=0}^{n} \frac{1}{\sqrt{(k+1)(n-k+1)}}$ .
By AM-GM, $(k+1)(n-k+1) \leq \left(\frac{(k+1)+(n-k+1)}{2}\right)^2 = \left(\frac{n+2}{2}\right)^2$ . Hence each summand $\frac{1}{\sqrt{(k+1)(n-k+1)}} \geq \frac{2}{n+2}$ , and with $n+1$ summands, $|c_n| \geq (n+1)\cdot\frac{2}{n+2} = \frac{2(n+1)}{n+2}$ . Since $\frac{2(n+1)}{n+2} \geq 1$ for $n \geq 0$ , $|c_n| \geq 1$ (in fact $|c_n| \to 2$ ).
Since $|c_n| \geq 1$ , $c_n \not\to 0$ , and the $n$ -th term test gives that $\sum c_n$ diverges. Two convergent series can have a divergent Cauchy product — absolute convergence of one factor in Mertens' theorem (problem 1) is not a removable hypothesis but an essential one.

Examples & Exercises