比值判别法：与几何级数作比较
比值判别法通过观察相邻项的比来度量级数收缩有多快。若 |a_{n+1}/a_n| -> L，则当 L < 1 时 sum a_n 绝对收敛，当 L > 1 时发散；在 L = 1 时判别法失效。其想法是：最终每一项大约是上一项的 L 倍，于是级数表现得像一个比为 L 的几何级数。
Why L < 1 forces convergence (the geometric domination).
Suppose |a_(n+1)/a_n| -> L < 1. Pick r with L < r < 1.
Then there is an N so that for all n >= N:
|a_(n+1)| <= r * |a_n|.
Iterating from N:
|a_(N+k)| <= r^k * |a_N|.
So the tail sum_{k>=0} |a_(N+k)| <= |a_N| * sum_{k>=0} r^k
= |a_N| / (1 - r) < infinity.
The tail is dominated by a convergent geometric series,
hence sum |a_n| converges => sum a_n converges absolutely. QED
Worked example: sum 1/n! (the series for e minus... well, e).
a_(n+1)/a_n = n!/(n+1)! = 1/(n+1) -> 0 = L < 1.
Ratio test: CONVERGES (absolutely).根值判别法及其更锐利的触角
根值判别法考察大小的 n 次方根：若 |a_n| 的 n 次方根的上极限为 L，则当 L < 1 时 sum a_n 绝对收敛，当 L > 1 时发散。用上极限使根值判别法比比值判别法略强——只要比值判别法能给答案，根值判别法给出相同的答案，而根值判别法有时在比值判别法卡住的地方仍能奏效。
Root test on sum n / 2^n.
nth root of |a_n| = (n / 2^n)^(1/n)
= n^(1/n) / 2.
Since n^(1/n) -> 1 as n -> infinity,
lim sup = 1/2 < 1.
Therefore the series CONVERGES absolutely.
Where the root test beats the ratio test:
a_n = 2^(-n) if n even, a_n = 3^(-n) if n odd.
The ratio a_(n+1)/a_n oscillates wildly (no limit) -> ratio test fails.
But (|a_n|)^(1/n) is either 1/2 or 1/3, so lim sup = 1/2 < 1
-> root test gives CONVERGES.交错符号：莱布尼茨判别法
当符号翻转时，抵消能挽救一个在绝对值上本会发散的级数。交错级数判别法（莱布尼茨）说：若 b_n 为正、递减、且 b_n -> 0，则交错级数 sum (-1)^{n+1} b_n 收敛。部分和以越来越短的步长来回跳动，把极限困在它们之间。
Alternating harmonic series sum_{n=1}^∞ (-1)^(n+1) / n = 1 - 1/2 + 1/3 - 1/4 + ...
Check the three hypotheses with b_n = 1/n:
(1) b_n = 1/n > 0. yes
(2) b_n decreasing: 1/(n+1) < 1/n. yes
(3) b_n -> 0. yes
=> the alternating series test gives CONVERGENCE.
Why it works (the nested-interval picture):
Even partial sums increase: s_2 < s_4 < s_6 < ...
Odd partial sums decrease: s_1 > s_3 > s_5 > ...
and s_(2k) < s_(2k+1), with s_(2k+1) - s_(2k) = b_(2k+1) -> 0.
So both subsequences squeeze to one common limit = the sum.
Bonus error bound: |s - s_N| <= b_(N+1).
(the tail is no bigger than the first omitted term)
For this series the sum is ln 2 = 0.6931...