Two strengths of convergence
A series sum a_n converges absolutely when sum |a_n| converges. It is a theorem — not a definition — that absolute convergence implies ordinary convergence: since 0 <= a_n + |a_n| <= 2|a_n|, the comparison test makes sum (a_n + |a_n|) converge, and subtracting sum |a_n| leaves sum a_n convergent. The reverse fails, and the gap has a name.
A series that converges but not absolutely is called conditionally convergent. The headline example is the alternating harmonic series: sum (-1)^{n+1}/n converges to ln 2 (by the alternating series test), yet sum |(-1)^{n+1}/n| = sum 1/n is the divergent harmonic series. Convergence here depends on the order and the cancellation, not on the sizes alone.
Riemann's rearrangement theorem
Here is the astonishing fact. The rearrangement theorem of Riemann says: if sum a_n is conditionally convergent, then for any target value t (finite, +infinity, or -infinity) there is a reordering of the same terms whose new sum is t. Reordering an infinite sum can change its value — addition is not unconditionally commutative once you have infinitely many terms.
Why a conditional series is so malleable.
Split the terms into positives P = {a_n : a_n > 0}
and negatives N = {a_n : a_n < 0}.
Key fact (for a conditionally convergent series):
sum of the positive terms = +infinity,
sum of the negative terms = -infinity,
yet each term a_n -> 0.
(If both piles were finite, the series would converge absolutely;
if only one were infinite, the whole series would diverge.)
Algorithm to hit any target t:
1. Add positive terms in order until the running sum first exceeds t.
2. Add negative terms until the running sum first drops below t.
3. Add positives again until you exceed t; repeat forever.
Because each pile is inexhaustible (sums to +/- infinity) you can
always overshoot, and because a_n -> 0 the overshoots shrink to 0.
So the rearranged partial sums converge exactly to t. QED (sketch)
Concrete: 1 - 1/2 + 1/3 - 1/4 + ... = ln 2,
but the SAME terms reordered can be made to sum to 0, to 100, or to diverge.Cauchy products: multiplying series
To multiply two series we collect terms by their total index, just as when multiplying polynomials. The Cauchy product of sum a_n and sum b_n is sum c_n with c_n = a_0 b_n + a_1 b_{n-1} + … + a_n b_0. Mertens' theorem guarantees that if both series converge and at least one converges absolutely, the Cauchy product converges to the product of the two sums.
Cauchy product of two geometric series (both absolutely convergent for |x| < 1).
Let a_n = x^n and b_n = x^n, so sum a_n = sum b_n = 1/(1 - x).
c_n = sum_{k=0}^n a_k b_(n-k)
= sum_{k=0}^n x^k * x^(n-k)
= sum_{k=0}^n x^n
= (n + 1) x^n.
So the Cauchy product is sum_{n=0}^∞ (n+1) x^n,
and by Mertens it equals the product of the sums:
sum (n+1) x^n = (1/(1-x)) * (1/(1-x)) = 1/(1-x)^2.
Check against calculus: 1/(1-x)^2 = d/dx [1/(1-x)]
= d/dx sum x^n = sum n x^(n-1) = sum (n+1) x^n. Consistent.
Warning: if you Cauchy-multiply the CONDITIONALLY convergent series
sum (-1)^n / sqrt(n+1) by itself, the product series has terms that
do NOT tend to 0 -- the product DIVERGES. Absolute convergence is what
rescues Mertens' theorem.