收敛的两种强度
当 sum |a_n| 收敛时，级数 sum a_n 绝对收敛。这是一条定理——而非定义——即绝对收敛蕴含普通收敛：由于 0 <= a_n + |a_n| <= 2|a_n|，比较判别法使 sum (a_n + |a_n|) 收敛，再减去 sum |a_n| 便得 sum a_n 收敛。反过来不成立，这道缝隙有个名字。
一个收敛但不绝对收敛的级数称为条件收敛。头号例子是交错调和级数：sum (-1)^{n+1}/n 收敛到 ln 2（由交错级数判别法），然而 sum |(-1)^{n+1}/n| = sum 1/n 是发散的调和级数。这里的收敛依赖于次序与抵消，而不只是大小。
黎曼重排定理
这是个惊人的事实。黎曼的重排定理说：若 sum a_n 条件收敛，则对任意目标值 t（有限、+infinity 或 -infinity），都存在对同一批项的某种重新排序，使新的和等于 t。重排一个无穷和可以改变它的值——一旦有了无穷多项，加法就不再无条件可交换。
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.柯西乘积：级数相乘
要把两个级数相乘，我们按总指标收集各项，正如多项式相乘那样。sum a_n 与 sum b_n 的柯西乘积是 sum c_n，其中 c_n = a_0 b_n + a_1 b_{n-1} + … + a_n b_0。Mertens 定理保证：若两个级数都收敛且至少一个绝对收敛，则柯西乘积收敛到这两个和的乘积。
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.