Ration Test
Given a series s n = ∑ k ∈ [ n ] a n s_n = \sum_{k \in [n]} a_n s n = ∑ k ∈ [ n ] a n a n ≠ 0 a_n \neq 0 a n = 0 ( a n ) (a_n) ( a n )
lim ∣ a n + 1 a n ∣ = r ≤ 1
\lim \left| \frac{a_{n + 1} }{ a_n } \right| = r \le 1
lim a n a n + 1 = r ≤ 1 then ( s n ) (s_n) ( s n )
Proof. The condition implies that
∃ N > 0 , s . t . , ∃ r ≤ r ′ < 1 , ∀ n ≥ N , a n + 1 ≤ r ′ a n
\exists N > 0, s.t., \exists r \le r' < 1, \forall n \ge N, a_{n + 1} \le r' a_n
∃ N > 0 , s . t . , ∃ r ≤ r ′ < 1 , ∀ n ≥ N , a n + 1 ≤ r ′ a n It follows that for n > m ≥ N n > m \ge N n > m ≥ N
∣ s n − s m ∣ = ∣ a m + 1 + a m + 1 + . . . + a n ∣ ≤ ∣ a N ∣ ∑ k = 0 n − m − 1 ( r ′ ) m − N + k ≤ ∣ a N ∣ ∑ k = 0 ∞ ( r ′ ) m − N + k ≤ ∣ a N ∣ 1 − ( r ′ ) ( r ′ ) m − N
\begin{aligned}
|s_n - s_m|
&= | a_{m + 1} + a_{m + 1} + ... + a_n| \\
&\le |a_N| \sum_{k = 0}^{n - m - 1} (r')^{m - N + k} \\
&\le |a_N| \sum_{k = 0}^{ \infty } (r')^{m - N + k} \\
&\le \frac{|a_N| }{1 - (r') } (r')^{m - N}
\end{aligned}
∣ s n − s m ∣ = ∣ a m + 1 + a m + 1 + ... + a n ∣ ≤ ∣ a N ∣ k = 0 ∑ n − m − 1 ( r ′ ) m − N + k ≤ ∣ a N ∣ k = 0 ∑ ∞ ( r ′ ) m − N + k ≤ 1 − ( r ′ ) ∣ a N ∣ ( r ′ ) m − N If m m m ∣ s n − s m ∣ |s_n - s_m| ∣ s n − s m ∣ ( s n ) (s_n) ( s n )
□ \square □
Abel's Test
If the series s n = ∑ k ∈ [ n ] a n s_n = \sum_{k \in [n]} a_n s n = ∑ k ∈ [ n ] a n ( b n ) (b_n) ( b n )
b 1 ≥ b 2 ≥ b 3 ≥ . . . ≥ 0
b_1 \ge b_2 \ge b_3 \ge ... \ge 0
b 1 ≥ b 2 ≥ b 3 ≥ ... ≥ 0 then the series ∑ k ≥ 1 a k b k \sum_{k \ge 1} a_k b_k ∑ k ≥ 1 a k b k
Lemma: Partial Sum
Define s 0 = 0 s_0 = 0 s 0 = 0
∑ k ∈ [ n ] x k y k = ∑ k ∈ [ n ] ( s k − s k − 1 ) b k = ∑ k ∈ [ n ] s k b k − ∑ k ∈ [ n − 1 ] s k b k + 1 = s n b n + ∑ k ∈ [ n − 1 ] s k ( b k − b k + 1 ) .
\begin{aligned}
\sum_{k \in [n]} x_k y_k
&= \sum_{k \in [n]} (s_k - s_{k - 1}) b_k \\
&= \sum_{k \in [n]} s_k b_k - \sum_{k \in [n-1]} s_k b_{k + 1} \\
&= s_n b_n + \sum_{k \in [n-1]} s_k (b_k - b_{k + 1}).
\end{aligned}
k ∈ [ n ] ∑ x k y k = k ∈ [ n ] ∑ ( s k − s k − 1 ) b k = k ∈ [ n ] ∑ s k b k − k ∈ [ n − 1 ] ∑ s k b k + 1 = s n b n + k ∈ [ n − 1 ] ∑ s k ( b k − b k + 1 ) . Proof.
( b n ) (b_n) ( b n ) ( b n ) (b_n) ( b n ) ( s n ) (s_n) ( s n ) ( b n ) (b_n) ( b n ) ( s n b n ) (s_n b_n) ( s n b n )
It remains to prove ∑ k ∈ [ n − 1 ] s k ( b k − b k + 1 ) \sum_{k \in [n-1]} s_k (b_k - b_{k + 1}) ∑ k ∈ [ n − 1 ] s k ( b k − b k + 1 ) ( s n ) (s_n) ( s n ) ∣ s n ∣ ≤ M | s_n | \le M ∣ s n ∣ ≤ M n > m n > m n > m
∣ ∑ k = m n − 1 s k ( b k − b k + 1 ) ∣ ≤ M ∑ k = m n − 1 ( b k − b k + 1 ) = M ( b m − b n )
\left| \sum_{k = m}^{n - 1} s_k (b_k - b_{k + 1}) \right| \le M \sum_{k = m}^{n - 1} (b_k - b_{k + 1} ) = M (b_m - b_n)
k = m ∑ n − 1 s k ( b k − b k + 1 ) ≤ M k = m ∑ n − 1 ( b k − b k + 1 ) = M ( b m − b n ) The fact ( b n ) (b_n) ( b n ) b m − b n → 0 b_m - b_n \rightarrow 0 b m − b n → 0 m m m ∣ ∑ k = m n − 1 s k ( b k − b k + 1 ) ∣ → 0 |\sum_{k = m}^{n - 1} s_k (b_k - b_{k + 1})| \rightarrow 0 ∣ ∑ k = m n − 1 s k ( b k − b k + 1 ) ∣ → 0
□ \square □
Dirichlet's Test
If the series s n = ∑ k ∈ [ n ] a n s_n = \sum_{k \in [n]} a_n s n = ∑ k ∈ [ n ] a n ( b k ) (b_k) ( b k )
{ b 1 ≥ b 2 ≥ b 3 ≥ . . . ≥ 0 a n d lim b k = 0
\begin{cases}
b_1 \ge b_2 \ge b_3 \ge ... \ge 0 \quad and \\
\lim b_k = 0
\end{cases}
{ b 1 ≥ b 2 ≥ b 3 ≥ ... ≥ 0 an d lim b k = 0 then the series ∑ k ≥ 1 a k b k \sum_{k \ge 1} a_k b_k ∑ k ≥ 1 a k b k
Proof.
Let ∣ s n ∣ ≤ M |s_n| \le M ∣ s n ∣ ≤ M ( b n ) → 0 (b_n) \rightarrow 0 ( b n ) → 0 ( s n b n ) → 0 (s_n b_n) \rightarrow 0 ( s n b n ) → 0 ∑ k ∈ [ n − 1 ] s k ( b k − b k + 1 ) \sum_{k \in [n-1]} s_k (b_k - b_{k + 1}) ∑ k ∈ [ n − 1 ] s k ( b k − b k + 1 )
□ \square □
Reference
[1]. Stephen Abbott, Understanding Analysis, Second Edition.
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