It can be proven that if sum from n=1 to infinity a_n=A
It can be proven that if $\sum_{n=1}^{\infty} a_n=A$, $\sum_{n=1}^{\infty} b_n=B$, $\sum_{n=1}^{\infty} c_n=C$ where $c_n=\sum_{v=1}^n a_v b_{n-v+1}$, then $AB=C$ (assuming both original series converge absolutely).
Explanation of Method
The Cauchy product $c_n$ arranges all pairwise products $a_ib_j$ by the diagonal $i+j-1=n$. Comparing the product of partial sums $A_NB_N$ (a sum over the full $N\times N$ grid) to the partial sum $C_N=\sum_{n=1}^Nc_n$ (a sum over only the triangular region $i+j-1\le N$) isolates a remainder $R_N=A_NB_N-C_N$ consisting of the "corner" terms with $i+j-1>N$; absolute convergence of both series forces $R_N\to0$.
- Define $C_N=A_NB_N-R_N$, where $R_N$ collects the terms with $i+j-1>N$.
- Bound $R_N$ using absolute convergence of both series and the tail estimate.
- Conclude $C_N\to AB$, i.e. $C=AB$.
Step 1 · Relate $C_N$ to the product $A_NB_N$
Since $A_NB_N=\sum_{i=1}^N\sum_{j=1}^Na_ib_j$ and $C_N=\sum_{n=1}^Nc_n$ sums exactly the pairs with $i+j-1\le N$:
$$ A_NB_N = C_N + R_N, \qquad R_N = \sum_{\substack{i+j-1>N\\1\le i,j\le N}} a_ib_j. $$
Step 2 · Bound the remainder
Given $\varepsilon>0$, by absolute convergence choose $N_0$ with $\sum_{n>N_0}|a_n|<\varepsilon$ and $\sum_{n>N_0}|b_n|<\varepsilon$. For $N$ large, every term in $R_N$ has $i>N_0$ or $j>N_0$, so:
$$ |R_N| \le \left(\sum_{i>N_0}|a_i|\right)|B| + |A_{N_0}|\left(\sum_{j>N_0}|b_j|\right) < \varepsilon|B| + |A_{N_0}|\varepsilon, $$
which can be made arbitrarily small, so $R_N\to0$.
Step 3 · Conclude
$$ C_N = A_NB_N - R_N \longrightarrow AB - 0 = AB, \qquad\text{so}\qquad C = AB. $$
Summary Table
| Quantity | Behavior |
|---|---|
| $A_NB_N-C_N=R_N$ | "Corner" terms with $i+j-1>N$ |
| $\lim R_N$ | $0$ (by absolute convergence) |
| Conclusion | $C=AB$ |
Additional Notes
- This result (Cauchy's theorem on products of absolutely convergent series) is actually a special case of Mertens' theorem, which only requires one of the two series to converge absolutely and the other merely to converge — a strictly weaker and more delicate hypothesis.
- The example $a_n=b_n=1/n!$ gives $A=B=e$ and, via the binomial-theorem structure of the Cauchy product, $c_n=\sum_{v=1}^n\frac1{v!(n-v+1)!}$, whose sum equals $e^2$ — directly recovering the exponential functional equation $e^x\cdot e^x=e^{2x}$ at $x=1$ from purely series-theoretic grounds.
- If absolute convergence is dropped entirely, the Cauchy product can actually diverge even when both original series converge — the classic counterexample uses $a_n=b_n=(-1)^n/\sqrt n$, both conditionally convergent, whose Cauchy product terms $c_n$ do not even tend to zero.
Result
For absolutely convergent $\sum a_n=A$ and $\sum b_n=B$, the Cauchy product converges to $AB$: