为什么收敛逼迫各项归零
这是关于级数的第一个一般定理，它直接从部分和定义中掉出来。若 sum a_n 收敛，则 a_n -> 0。理由：单独一项 a_n 等于 s_n - s_{n-1}，即相邻部分和之间的差。若 s_n 与 s_{n-1} 都奔向同一极限 s，这个差就被挤到零。
Theorem. If sum a_n converges then a_n -> 0.
Proof. Let s_N be the partial sums and suppose s_N -> s.
For n >= 2 we have a_n = s_n - s_(n-1).
The tail s_(n-1) also tends to s (it's the same sequence shifted).
By the algebra of limits,
a_n = s_n - s_(n-1) -> s - s = 0.
Hence a_n -> 0. QED
Contrapositive (the usable form):
If a_n does NOT tend to 0, then sum a_n diverges.
Example. sum n/(n+1): a_n = n/(n+1) -> 1, not 0.
So sum n/(n+1) diverges. (No partial-sum work needed.)实践中我们用逆否命题：这就是发散的第 n 项判别法。若 a_n 不趋于 0，级数就不可能收敛，到此为止。它是你手头最便宜的检查——永远先跑它。
单行道
全部危险都住在这条单行道里。初学者看到 a_n -> 0 就断定和是有限的。这是本学科里最常见的一个错误，而调和级数存在的意义正是为了反驳它。
调和级数发散
级数 sum 1/n 的各项 1/n -> 0，它却发散。经典证明（约 1350 年 Oresme 所作）把各项分成块，每块从下方用 1/2 控制。于是部分和越过每一个界——慢，却没有上限。
Claim. sum_{n=1}^∞ 1/n diverges, even though 1/n -> 0.
Group the terms in blocks of length 1, 2, 4, 8, ...:
1
+ 1/2
+ (1/3 + 1/4) >= 1/4 + 1/4 = 1/2
+ (1/5 + 1/6 + 1/7 + 1/8) >= 4 * (1/8) = 1/2
+ (1/9 + ... + 1/16) >= 8 * (1/16) = 1/2
+ ...
Each parenthesized block of 2^(k-1) terms is >= 1/2,
because its smallest term is 1/2^k and there are 2^(k-1) of them:
2^(k-1) * (1/2^k) = 1/2.
So s_(2^k) >= 1 + k*(1/2).
The right side -> infinity as k -> infinity,
therefore the partial sums are unbounded and the series diverges. QED