為什麼收斂逼迫各項歸零
這是關於級數的第一個一般定理，它直接從部分和定義中掉出來。若 sum a_n 收斂，則 a_n -> 0。理由：單獨一項 a_n 等於 s_n - s_{n-1}，即相鄰部分和之間的差。若 s_n 與 s_{n-1} 都奔向同一極限 s，這個差就被擠到零。
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.)實踐中我們用逆否命題：這就是發散的第 n 項判別法。若 a_n 不趨於 0，級數就不可能收斂，到此為止。它是你手頭最便宜的檢查——永遠先跑它。
單行道
全部危險都住在這條單行道裡。初學者看到 a_n -> 0 就斷定和是有限的。這是本學科裡最常見的一個錯誤，而調和級數存在的意義正是為了反駁它。
調和級數發散
級數 sum 1/n 的各項 1/n -> 0，它卻發散。經典證明（約 1350 年 Oresme 所作）把各項分成塊，每塊從下方用 1/2 控制。於是部分和越過每一個界——慢，卻沒有上限。
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