收斂的兩種強度
當 sum |a_n| 收斂時，級數 sum a_n 絕對收斂。這是一條定理——而非定義——即絕對收斂蘊含普通收斂：由於 0 <= a_n + |a_n| <= 2|a_n|，比較判別法使 sum (a_n + |a_n|) 收斂，再減去 sum |a_n| 便得 sum a_n 收斂。反過來不成立，這道縫隙有個名字。
一個收斂但不絕對收斂的級數稱為條件收斂。頭號例子是交錯調和級數：sum (-1)^{n+1}/n 收斂到 ln 2（由交錯級數判別法），然而 sum |(-1)^{n+1}/n| = sum 1/n 是發散的調和級數。這裡的收斂依賴於次序與抵消，而不只是大小。
黎曼重排定理
這是個驚人的事實。黎曼的重排定理說：若 sum a_n 條件收斂，則對任意目標值 t（有限、+infinity 或 -infinity），都存在對同一批項的某種重新排序，使新的和等於 t。重排一個無窮和可以改變它的值——一旦有了無窮多項，加法就不再無條件可交換。
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.柯西乘積：級數相乘
要把兩個級數相乘，我們按總指標收集各項，正如多項式相乘那樣。sum a_n 與 sum b_n 的柯西乘積是 sum c_n，其中 c_n = a_0 b_n + a_1 b_{n-1} + … + a_n b_0。Mertens 定理保證：若兩個級數都收斂且至少一個絕對收斂，則柯西乘積收斂到這兩個和的乘積。
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.