The Cauchy Product

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 $m > M$.

Only for $N > M$ (all sufficiently large $N$) 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 (for $N > M$).

The correct bound is $\left|\sum_{k=N-M}^{N} a_k R_{N-k}\right| \leq \left(\max_{0 \leq m \leq M} |R_m|\right) \cdot \sum_{k=N-M}^{N} |a_k|$ — a sum of products needs the sum of $|a_k|$, not a single $\max |a_k|$.

For fixed $M$, $\sum_{k=N-M}^{N} |a_k| \le \sum_{k > N-M} |a_k| \to 0$ as $N \to \infty$ (the tail of the absolutely convergent $\sum |a_k|$), so the tail tends to $0$.

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).

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 $e^1 \cdot e^1 = e^2$.

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 $|c_n| \to \pi$: the sum is a Riemann sum for $\int_0^1 \frac{dx}{\sqrt{x(1-x)}} = \pi$; numerically $|c_{100}| \approx 2.85$, $|c_{400}| \approx 3.00$, approaching $\pi$).

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.