# SequencesA sequence is a function whose domain is N N N . (N N N is the set of all positive integers.) In notation, we write f ( n ) = a n f(n) = a_n f ( n ) = a n . # Theoreml i m x → ∞ f ( x ) = L , x ∈ R ⇒ l i m n → ∞ f ( n ) = L , n ∈ N lim_{x \to \infty}f(x) = L, x \in R \Rightarrow lim_{n \to \infty}f(n) = L,n \in N l i m x → ∞ f ( x ) = L , x ∈ R ⇒ l i m n → ∞ f ( n ) = L , n ∈ N .l i m n → ∞ ∣ a n ∣ = 0 ⇒ l i m n → ∞ a n = 0 lim_{n \to \infty}| a_n | = 0 \Rightarrow lim_{n \to \infty} a_n = 0 l i m n → ∞ ∣ a n ∣ = 0 ⇒ l i m n → ∞ a n = 0 .{ r n } n = 1 ∞ \{ r^n \}^\infty_{n=1} { r n } n = 1 ∞ converges iff − 1 < r ≤ 1 -1 < r \le 1 − 1 < r ≤ 1 .Let { a n } \{ a_n \} { a n } be monotonic
(單調) and bounded
(有界) ⇒ { a n } \Rightarrow \{ a_n \} ⇒ { a n } converges. # SeriesSeries: ∑ i = 1 ∞ a i = a 1 + a 2 + . . . + a n + . . . \sum^{\infty}_{i=1} a_i = a_1 + a_2 + ... +a_n + ... ∑ i = 1 ∞ a i = a 1 + a 2 + . . . + a n + . . . The convergence/divergence of a series:(n-th) partial sum = S n = ∑ i = 1 n a i S_n = \sum^{n}_{i=1} a_i S n = ∑ i = 1 n a i ,{ S n } \{ S_n \} { S n } converges/diverges ⇔ ∑ i = 1 ∞ a i \Leftrightarrow \sum^{\infty}_{i=1} a_i ⇔ ∑ i = 1 ∞ a i converges/diverges. # Theorem (Geometric Series.)∑ i = 1 ∞ r i \sum^{\infty}_{i=1}r^i ∑ i = 1 ∞ r i converges ⇔ ∣ r ∣ < 1 \Leftrightarrow |r| < 1 ⇔ ∣ r ∣ < 1 and ∑ i = 1 ∞ r i = r 1 − r \sum^{\infty}_{i=1}r^i = \frac{r}{1-r} ∑ i = 1 ∞ r i = 1 − r r .
# Theorem∑ i = 1 ∞ a i \sum^{\infty}_{i=1} a_i ∑ i = 1 ∞ a i converges ⇒ l i m n → ∞ a n = 0 \Rightarrow lim_{n \to \infty} a_n = 0 ⇒ l i m n → ∞ a n = 0
remark lim n → ∞ 1 n = 0 \lim_{n \to \infty}\frac{1}{n} = 0 lim n → ∞ n 1 = 0 ,but ∑ n = 1 ∞ 1 n \sum^{\infty}_{n=1} \frac{1}{n} ∑ n = 1 ∞ n 1 is divergent.
# The Integral Test and Estimates of SumsWhat kind of series can apply the integral test to check its convergence?a i a_i a i satisfy: (i) essentially positive (ii) decreasing (iii) f ( x ) = a x f(x) = a_x f ( x ) = a x is continuous
# The Integral Test∑ n = 1 ∞ 1 n p \sum^{\infty}_{n=1} \frac{1}{n^p} ∑ n = 1 ∞ n p 1 convergences, p > 1 p>1 p > 1 ∑ n = 1 ∞ 1 n p \sum^{\infty}_{n=1} \frac{1}{n^p} ∑ n = 1 ∞ n p 1 divergences, p ≤ 1 p \le 1 p ≤ 1 # Estimates of SumsLet { a i } \{a_i\} { a i } can apply the integral test, and let R n = S − S n R_n = S - S_n R n = S − S n , Here S = ∑ i = 1 ∞ a i S = \sum^{\infty}_{i=1} a_i S = ∑ i = 1 ∞ a i and S n = ∑ i = 1 n a i S_n = \sum^{n}_{i=1} a_i S n = ∑ i = 1 n a i Then ∫ n + 1 ∞ a x d x ≤ R n ≤ ∫ n ∞ a x d x \int^{\infty}_{n+1} a_x dx \le R_n \le \int^{\infty}_n a_x dx ∫ n + 1 ∞ a x d x ≤ R n ≤ ∫ n ∞ a x d x
# The Comparison Tests減法比較:a n ≥ b n a_n \ge b_n a n ≥ b n and ∑ a n \sum a_n ∑ a n converges ⇒ ∑ b n \Rightarrow \sum b_n ⇒ ∑ b n converges.a n ≥ b n a_n \ge b_n a n ≥ b n and ∑ b n \sum b_n ∑ b n diverges ⇒ ∑ b n \Rightarrow \sum b_n ⇒ ∑ b n diverges. 除法比較:l i m n → ∞ a n b n = c > 0 lim_{n \to \infty} \frac{a_n}{b_n} = c > 0 l i m n → ∞ b n a n = c > 0 ⇒ \Rightarrow ⇒ Both ∑ a n \sum a_n ∑ a n and ∑ b n \sum b_n ∑ b n converge and diverge.l i m n → ∞ a n b n = 0 lim_{n \to \infty} \frac{a_n}{b_n} = 0 l i m n → ∞ b n a n = 0 If ∑ b n \sum b_n ∑ b n converges ⇒ ∑ a n \Rightarrow \sum a_n ⇒ ∑ a n converges. If ∑ a n \sum a_n ∑ a n diverges ⇒ ∑ b n \Rightarrow \sum b_n ⇒ ∑ b n diverges. # Alternating Series# Theorema n a_n a n is decreasing.a n → 0 a_n \to 0 a n → 0 as n → ∞ n \to \infty n → ∞ .⇒ ∑ n = 1 ∞ ( − 1 ) n + 1 a n \Rightarrow \sum^{\infty}_{n=1}(-1)^{n+1}a_n ⇒ ∑ n = 1 ∞ ( − 1 ) n + 1 a n converges.# Error Estimate∑ n = 1 ∞ ( − 1 ) n + 1 a n \sum^{\infty}_{n=1}(-1)^{n+1}a_n ∑ n = 1 ∞ ( − 1 ) n + 1 a n satisfies 1. and 2.∣ R n ∣ = ∣ S − S n ∣ ≤ ∣ S n + 1 − S n ∣ ≤ a n + 1 |R_n| = |S - S_n| \le |S_{n+1} - S_n| \le a_{n+1} ∣ R n ∣ = ∣ S − S n ∣ ≤ ∣ S n + 1 − S n ∣ ≤ a n + 1 .
# Absolute Convergence and The Ratio and Root test# Define∑ n = 1 ∞ a n \sum_{n=1}^{\infty} a_n ∑ n = 1 ∞ a n is said to be absolute convergence
(AC) if ∑ n = 1 ∞ ∣ a n ∣ \sum_{n=1}^{\infty} |a_n| ∑ n = 1 ∞ ∣ a n ∣ is convergent.If ∑ n = 1 ∞ a n \sum_{n=1}^{\infty} a_n ∑ n = 1 ∞ a n is convergent but not AC,then ∑ n = 1 ∞ a n \sum_{n=1}^{\infty} a_n ∑ n = 1 ∞ a n is called conditional convergence
(CC). # TheoremAC ⇒ \Rightarrow ⇒ convergence.
# Ratio Test比前後像縮小的比例
l i m n → ∞ ∣ a n + 1 a n ∣ = L lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = L l i m n → ∞ ∣ a n a n + 1 ∣ = L 適用對象:有 "!" 項。 # Root Test開n n n 次根號也為檢查尾巴項的行為
l i m n → ∞ ∣ a n ∣ n = L lim_{n \to \infty} \sqrt[n]{|a_n|} = L l i m n → ∞ n ∣ a n ∣ = L 適用對象:有\sqrt[n] 或( ) n ()^n ( ) n 項。 # ConclusionL < 1 ⇒ L < 1 \Rightarrow L < 1 ⇒ AC.L > 1 ⇒ L > 1 \Rightarrow L > 1 ⇒ Divergence.L = 1 ⇒ L = 1 \Rightarrow L = 1 ⇒ The test fails.# Power Series# Define∑ n = 0 ∞ c n ( x − a ) n = c 0 + c 1 ( x − a ) + c 2 ( x − a ) 2 + . . . \sum_{n=0}^{\infty}c_n(x-a)^n = c_0+c_1(x-a)+c_2(x-a)^2 + ... ∑ n = 0 ∞ c n ( x − a ) n = c 0 + c 1 ( x − a ) + c 2 ( x − a ) 2 + . . . is called a power series about a a a (or in ( x − a ) (x-a) ( x − a ) ).
# TheoremFor series ∑ n = 1 ∞ c n ( x − a ) n \sum^{\infty}_{n=1}c_n(x-a)^n ∑ n = 1 ∞ c n ( x − a ) n ,either one of the following 3 possibilities holds:
r = 0 r = 0 r = 0 : 只有一點收斂r < ∞ r < \infty r < ∞ : 兩端點須被檢驗,才可知其收斂與否r = ∞ r = \infty r = ∞ : 每一點皆收斂# Representations of Functions as Power Series# Using Geometric Series1 1 − x = 1 + x + x 2 + . . . = ∑ n = 0 ∞ x n , ∣ x ∣ < 1 \frac{1}{1-x} = 1+x+x^2+... = \sum_{n=0}^{\infty}x^n, |x| <1 1 − x 1 = 1 + x + x 2 + . . . = ∑ n = 0 ∞ x n , ∣ x ∣ < 1 .
note Power series 在絕對收斂的範圍可作逐項微分或積分的動作 有時須微分或積分 不只一次
才能觀察出來 # Taylor and Maclaurin Series# DefineTaylor series of f f f at x = a x = a x = a is defined to be ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k \sum_{k=0}^{\infty}\frac{f^(k)(a)}{k!}(x-a)^k ∑ k = 0 ∞ k ! f ( k ) ( a ) ( x − a ) k . If a = 0 a=0 a = 0 ,the corresponding Taylor series is called Maclaurin series. ∑ k = 0 ∞ f ( k ) ( a ) k ! ( x − a ) k \sum_{k=0}^{\infty}\frac{f^(k)(a)}{k!}(x-a)^k ∑ k = 0 ∞ k ! f ( k ) ( a ) ( x − a ) k is called the n n n th-degree Taylor polynomial of f f f at a a a .Remark f ( x ) f(x) f ( x ) is not necessarily equal to its Taylor series.
# Taylor InequalityLet f ( x ) − T n ( x ) = R n ( x ) f(x)-T_n(x) = R_n(x) f ( x ) − T n ( x ) = R n ( x ) ,where T n = ∑ k = 0 n f ( k ) ( a ) k ! ( x − a ) k T_n = \sum_{k=0}^{n}\frac{f^(k)(a)}{k!}(x-a)^k T n = ∑ k = 0 n k ! f ( k ) ( a ) ( x − a ) k is called the n-th degree Taylor polynomial of f f f at a a a .
If ∣ f ( n + 1 ) ( x ) ∣ ≤ M |f^{(n+1)}(x)| \le M ∣ f ( n + 1 ) ( x ) ∣ ≤ M for ∣ x − a ∣ ≤ d |x-a| \le d ∣ x − a ∣ ≤ d , then ∣ R n ( x ) ∣ ≤ M ( n + 1 ) ! ∣ x − a ∣ n + 1 |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1} ∣ R n ( x ) ∣ ≤ ( n + 1 ) ! M ∣ x − a ∣ n + 1 for ∣ x − a ∣ ≤ d |x-a| \le d ∣ x − a ∣ ≤ d .
# Reference莊重 - 微積分 (二) Calculus II - 103 學年度