# Sequences

  • A sequence is a function whose domain is NN. (NN is the set of all positive integers.)
  • In notation, we write f(n)=anf(n) = a_n.

# Theorem

  1. limx→∞f(x)=L,x∈R⇒limn→∞f(n)=L,n∈Nlim_{x \to \infty}f(x) = L, x \in R \Rightarrow lim_{n \to \infty}f(n) = L,n \in N.
  2. limn→∞∣an∣=0⇒limn→∞an=0lim_{n \to \infty}| a_n | = 0 \Rightarrow lim_{n \to \infty} a_n = 0.
  3. {rn}n=1∞\{ r^n \}^\infty_{n=1} converges iff −1<r≤1-1 < r \le 1.
  4. Let {an}\{ a_n \} be monotonic (單調) and bounded (有界) ⇒{an}\Rightarrow \{ a_n \} converges.

# Series

  • Series: ∑i=1∞ai=a1+a2+...+an+...\sum^{\infty}_{i=1} a_i = a_1 + a_2 + ... +a_n + ...
  • The convergence/divergence of a series:(n-th) partial sum = Sn=∑i=1naiS_n = \sum^{n}_{i=1} a_i,
    {Sn}\{ S_n \} converges/diverges ⇔∑i=1∞ai\Leftrightarrow \sum^{\infty}_{i=1} a_i converges/diverges.

# Theorem (Geometric Series.)

∑i=1∞ri\sum^{\infty}_{i=1}r^i converges ⇔∣r∣<1\Leftrightarrow |r| < 1 and ∑i=1∞ri=r1−r\sum^{\infty}_{i=1}r^i = \frac{r}{1-r}.

# Theorem

∑i=1∞ai\sum^{\infty}_{i=1} a_i converges ⇒limn→∞an=0\Rightarrow lim_{n \to \infty} a_n = 0

remark

lim⁡n→∞1n=0\lim_{n \to \infty}\frac{1}{n} = 0,but ∑n=1∞1n\sum^{\infty}_{n=1} \frac{1}{n} is divergent.

# The Integral Test and Estimates of Sums

What kind of series can apply the integral test to check its convergence?
aia_i satisfy:
(i) essentially positive
(ii) decreasing
(iii) f(x)=axf(x) = a_x is continuous

# The Integral Test

  • ∑n=1∞1np\sum^{\infty}_{n=1} \frac{1}{n^p} convergences, p>1p>1
  • ∑n=1∞1np\sum^{\infty}_{n=1} \frac{1}{n^p} divergences, p≤1p \le 1

# Estimates of Sums

Let {ai}\{a_i\} can apply the integral test, and let Rn=S−SnR_n = S - S_n,
Here S=∑i=1∞aiS = \sum^{\infty}_{i=1} a_i and Sn=∑i=1naiS_n = \sum^{n}_{i=1} a_i
Then ∫n+1∞axdx≤Rn≤∫n∞axdx\int^{\infty}_{n+1} a_x dx \le R_n \le \int^{\infty}_n a_x dx

# The Comparison Tests

  • 減法比較:
    • an≥bna_n \ge b_n and ∑an\sum a_n converges ⇒∑bn\Rightarrow \sum b_n converges.
    • an≥bna_n \ge b_n and ∑bn\sum b_n diverges ⇒∑bn\Rightarrow \sum b_n diverges.
  • 除法比較:
    • limn→∞anbn=c>0lim_{n \to \infty} \frac{a_n}{b_n} = c > 0
      ⇒\Rightarrow Both ∑an\sum a_n and ∑bn\sum b_n converge and diverge.
    • limn→∞anbn=0lim_{n \to \infty} \frac{a_n}{b_n} = 0
      • If ∑bn\sum b_n converges ⇒∑an\Rightarrow \sum a_n converges.
      • If ∑an\sum a_n diverges ⇒∑bn\Rightarrow \sum b_n diverges.

# Alternating Series

# Theorem

  1. ana_n is decreasing.
  2. an→0a_n \to 0 as n→∞n \to \infty.
    ⇒∑n=1∞(−1)n+1an\Rightarrow \sum^{\infty}_{n=1}(-1)^{n+1}a_n converges.

# Error Estimate

∑n=1∞(−1)n+1an\sum^{\infty}_{n=1}(-1)^{n+1}a_n satisfies 1. and 2.
∣Rn∣=∣S−Sn∣≤∣Sn+1−Sn∣≤an+1|R_n| = |S - S_n| \le |S_{n+1} - S_n| \le a_{n+1}.

# Absolute Convergence and The Ratio and Root test

# Define

  1. ∑n=1∞an\sum_{n=1}^{\infty} a_n is said to be absolute convergence (AC) if ∑n=1∞∣an∣\sum_{n=1}^{\infty} |a_n| is convergent.
  2. If ∑n=1∞an\sum_{n=1}^{\infty} a_n is convergent but not AC,then ∑n=1∞an\sum_{n=1}^{\infty} a_n is called conditional convergence (CC).

# Theorem

AC ⇒\Rightarrow convergence.

# Ratio Test

比前後像縮小的比例

  • limn→∞∣an+1an∣=Llim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = L
  • 適用對象:有 "!" 項。

# Root Test

開nn 次根號也為檢查尾巴項的行為

  • limn→∞∣an∣n=Llim_{n \to \infty} \sqrt[n]{|a_n|} = L
  • 適用對象:有\sqrt[n] 或()n()^n 項。

# Conclusion

  1. L<1⇒L < 1 \Rightarrow AC.
  2. L>1⇒L > 1 \Rightarrow Divergence.
  3. L=1⇒L = 1 \Rightarrow The test fails.

# Power Series

# Define

∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+...\sum_{n=0}^{\infty}c_n(x-a)^n = c_0+c_1(x-a)+c_2(x-a)^2 + ... is called a power series about aa(or in (x−a)(x-a)).

# Theorem

For series ∑n=1∞cn(x−a)n\sum^{\infty}_{n=1}c_n(x-a)^n,either one of the following 3 possibilities holds:

  1. r=0r = 0: 只有一點收斂
  2. r<∞r < \infty: 兩端點須被檢驗,才可知其收斂與否
  3. r=∞r = \infty: 每一點皆收斂

# Representations of Functions as Power Series

# Using Geometric Series

11−x=1+x+x2+...=∑n=0∞xn,∣x∣<1\frac{1}{1-x} = 1+x+x^2+... = \sum_{n=0}^{\infty}x^n, |x| <1.

note
  1. Power series 在絕對收斂的範圍可作逐項微分或積分的動作
  2. 有時須微分或積分 不只一次 才能觀察出來

# Taylor and Maclaurin Series

# Define

  1. Taylor series of ff at x=ax = 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.
  2. If a=0a=0,the corresponding Taylor series is called Maclaurin series.
  3. ∑k=0∞f(k)(a)k!(x−a)k\sum_{k=0}^{\infty}\frac{f^(k)(a)}{k!}(x-a)^k is called the nn th-degree Taylor polynomial of ff at aa.
Remark

f(x)f(x) is not necessarily equal to its Taylor series.

# Taylor Inequality

Let f(x)−Tn(x)=Rn(x)f(x)-T_n(x) = R_n(x),where Tn=∑k=0nf(k)(a)k!(x−a)kT_n = \sum_{k=0}^{n}\frac{f^(k)(a)}{k!}(x-a)^k is called the n-th degree Taylor polynomial of ff at aa.

If ∣f(n+1)(x)∣≤M|f^{(n+1)}(x)| \le M for ∣x−a∣≤d|x-a| \le d, then ∣Rn(x)∣≤M(n+1)!∣x−a∣n+1|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1} for ∣x−a∣≤d|x-a| \le d.


# Reference

  • 莊重 - 微積分 (二) Calculus II - 103 學年度
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