# Chapter 7 Infinite Series (無窮級數)

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## (無窮級數)

### Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

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## Sequences (序列)

An infinite sequence of real numbers is denoted by {an} = {an}n=1={a1, a2,· · · , an,· · · },

where anis the nth term (第 n 項) of the sequence for n∈ N.

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### Def. (序列的收斂性)

A sequence{an} converges to a limit L, denoted by

nlim→∞an= L, if∀ ε > 0, ∃ M > 0 s.t.

n > M =⇒ |an− L| < ε.

Otherwise, we say that{an} diverges if the limit does not exist.

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## 示意圖 (承上頁)

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### Thm 7.1 (函數在∞ 處的極限 ⇒ 序列的收斂性)

Let f be a real-valued function having the limit

xlim→∞f(x) = L.

Ifan= f(n) ∀ n ∈ N, then

nlim→∞an= L.

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### pf: Let ε > 0 be given arbitrarily. Since lim

x→∞f(x) = L,∃ M > 0 s.t.

x > M =⇒ |f(x) − L| < ε.

Withan= f(n) for all n∈ N, ifn > M, then

|an− L| = |f(n) − L| < ε.

Thus, it follows from the Def. that

nlim→∞an= lim

n→∞f(n) = L.

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### Example (Thm 7.1 的反敘述不成立)

Consider f(x) = sin(πx) for all x > 0 and let an= f(n) for n∈ N.

Then

nlim→∞an= lim

n→∞f(n) = lim

n→∞sin(nπ) = lim

n→∞0 = 0, but we know that

xlim→∞f(x) = lim

x→∞sin(πx) does not exist!

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### Thm 7.2 (Limit Laws for Sequences)

Suppose that lim

n→∞an= L and lim

n→∞bn= K. Then

1 lim

n→∞(an± bn) = L± K.

2 lim

n→∞(c· an) = c· L for all c ∈ R.

3 lim

n→∞(an· bn) = L· K.

4 lim

n→∞

an bn = L

K ifbn̸= 0 ∀ n ∈ N andK̸= 0.

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### Thm 7.3 (Squeeze Theorem for Sequences)

If∃ M > 0 s.t.an≤ cn≤ bn ∀ n > M, and

nlim→∞an= L = lim

n→∞bn,

then the sequence{cn} converges to the same limit L, i.e.,

nlim→∞cn= L.

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n

### }?

Sincen!≥ 2n for all n≥ 4, it follows that

|cn| = (−1)n n!

= 1 n! 1

2n, and hence we immediately obtain

−1

2n ≤ cn 1 2n for alln≥ 4.

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### Thm 7.4 (Absolute Value Thoerem)

Let{an} be a sequence of real numbers. Then

nlim→∞|an| = 0 ⇐⇒ limn

→∞an= 0.

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## Monotonic and Bounded Sequences (1/2)

### Def. (單調序列的定義)

A sequence{an} is said to be monotonic (單調的) if its terms are nondecreasing (遞增的)

an≤ an+1 ∀ n ∈ N, or its terms arenonincreasing (遞減的)

an≥ an+1 ∀ n ∈ N.

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## Monotonic and Bounded Sequences (2/2)

### Def. (有界序列的定義)

(1) The sequence {an} isbounded above (於上有界) if∃ M ∈ R s.t. an≤ M ∀ n ∈ N.

(2) The sequence {an} isbounded below (於下有界) if∃ N ∈ R s.t. an≥ N ∀ n ∈ N.

(3) The sequence{an} is bounded (有界的)if it isbounded above and bounded below, i.e. ∃ M > 0 s.t. |an| ≤ M ∀ n ∈ N.

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### Thm 7.5 (Bounded Monotonic Sequences)

If the sequence{an} isbounded and monotonic, then it converges, i.e.,∃! L ∈ R s.t. limn

→∞an= L.

### Example 9 (Thm 7.5 的例子)

(a) The sequence {an} = {1/n} is bounded and nonincreasing, since

|an| ≤ 1 and an= 1 n 1

n + 1 =an+1 ∀ n ∈ N. So, it must converge by Thm 7.5 with lim

n→∞an= 0.

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### Thm 7.5 (Bounded Monotonic Sequences)

If the sequence{an} isbounded and monotonic, then it converges, i.e.,∃! L ∈ R s.t. limn

→∞an= L.

### Example 9 (Thm 7.5 的例子)

(a) The sequence {an} = {1/n} is bounded and nonincreasing, since

|an| ≤ 1 and an= 1 n 1

n + 1 =an+1 ∀ n ∈ N.

So, it must converge by Thm 7.5 with lim

n→∞an= 0.

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## (級數與收斂)

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### Def. (Partial Sums of a Series)

(a) An infinite series (無窮級數) of real numbers is of the form Xan=

X n=1

an= a1+ a2+ a3+· · · .

(b) For each n∈ N, the nth partial sum (第 n 個部分和) ofP an is defined by

Sn= Xn

i=1

ai= a1+ a2+· · · + an.

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### Def. (Convergence of a Series)

(a) We say thatP

an converges if the sequence{Sn} converges with lim

n→∞Sn= S. In this case, S is called the sum of the series and write S =P

an. (b) We say thatP

an diverges if the sequence{Sn} diverges.

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## Two Questions for Series

### Two Questions

1 Does a given series converge or diverge?

2 What is the sum of a convergent series?

### Note: these questions are not always easy to answer,

especially the second one. (通常需要藉助數值方法求得近似和!)

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## Geometric Series (幾何級數)

For a̸= 0 and ratio (公比) r ̸= 0, the geometric series is given by X

n=0

arn= a + ar + ar2+ ar3+· · · .

### Thm 7.6 (幾何級數的收斂性)

X n=0

arn=



 a

1− r if |r| < 1, diverges if|r| ≥ 1.

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### Example 3 (Thm 7.6 的例子)

(a)

X n=0

3 2n =

X n=0

3

1 2

n

= 3

1− (1/2) = 6 by Thm 7.6.

(b) X n=0

3 2

n

diverges because the ratio|r| = |3/2| = 3/2 ≥ 1.

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### Thm 7.7 (Properties of Infinite Series)

Suppose thatP

an= A and P

bn= B are convergent series.

1 P

(c· an) = c· P an

= c· A for all c ∈ R.

2 P

(an± bn) = P an

± P bn

= A± B.

3 In general, we know that X(anbn)̸= X

an X bn

, X an bn

̸=

Pan Pbn.

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### Thm 7.8

IfP

an converges, then lim

n→∞an= 0.

(收斂級數第 n 項所形成的序列必定趨近 0!)

### pf: If the series

P

an= S converges, then we know that

nlim→∞Sn= S = lim

n→∞Sn−1, where Sn=

Pn i=1

ai is the nth partial sum ofP

an. So, we immediately obtain

nlim→∞an= lim

n→∞(Sn− Sn−1) = S− S= 0.

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### Thm 7.9 (The nth Term Test; 第 n 項測試法)

If lim

n→∞an̸= 0, the series P

an diverges.

(若第 n 項不趨近到 0, 則級數必定發散!)

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## (積分與比較測試法)

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### Thm 7.10 (The Integral Test; 積分測試法)

If f ispositive, conti. and↘ on [1, ∞), andan= f(n) ∀ n ∈ N, then

X n=1

an and Z

1

f(x) dxboth converge or both diverge.

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### Def. (p-級數的定義)

(a)

X n=1

1

np is called a p-series (p-級數) with p > 0.

(b) If p = 1, then X n=1

1

n is called a harmonic series (調和級數).

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### Thm 7.11 (p-級數的收斂與發散)

The p-series

X n=1

1

np converges forp > 1, and diverges for 0 < p≤ 1.

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### 計算 p-級數的和 (檔名: p_series.m)

p = 2.7; data = [];

for k = 0:8 N = 10^k;

n = 1:N;

S_N = sum(1./(n.^p));

data = [data;N S_N];

end

semilogx(data(:,1),data(:,2),'bo-');

title('Partial sums of a p-series with p = 2.7');

xlabel('\bf n');

ylabel('\bf S_n');

P

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## 程式執行結果 (承上例)

n=1

n1p 的收斂與發散，其中 p = 2.7 和 p = 0.7:

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### Thm 7.12 (Direct Comparison Test; 直接比較法)

If0 < an≤ bn for all n∈ N, then

1 P

bnconverges =P

anconverges, and

2 P

an diverges =P

bn diverges.

### 口訣:

(1) 大的級數收斂保證小的級數也收斂!

(2) 小的級數發散保證大的級數也發散!

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### Thm 7.13 (Limit Comparison Test; 極限比較法)

Letan, bn> 0 ∀ n ∈ Nwith lim

n→∞

an bn = L.

1 If0 < L <∞, then P

anand P

bn both converge or both diverge.

2 IfL = 0, then P

bn converges =P

an converges, and Pan diverges =P

bn diverges.

3 IfL =∞, then P

anconverges =P

bn converges, and Pbndiverges =P

an diverges.

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## (其他收斂測試法)

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### Alternating Series (交錯級數)

An alternating series is of the form

S = X n=1

(−1)n+1an= X n=1

(−1)n−1an= a1− a2+ a3− a4+· · · ,

where the nth terman> 0 for all n∈ N.

### Questions

Does the alternating series always converge?

How to estimate the sum S of a convergent alternating series?

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### Thm 7.14 (Alternating Series Test; 交錯級數測試法)

If the sequence ofpositive terms{an} satisfies

1 ∃ M > 0 s.t. an≥ an+1 for all n > M, and

2 lim

n→∞an= 0,

then the alternating series S =P

(−1)n+1an converges with the property

|Sn− S| ≤ an+1 ∀ n ∈ N.

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## Absolute and Conditional Convergence

### Def. (Types of Convergence for a Series)

(1) P

an is absolutely convergent (絕對收斂) if P

|an| converges.

(2) P

an is conditionally convergent (條件收斂) if P an converges, but P

|an| diverges.

### Thm 7.16 (絕對收斂保證原級數收斂)

P|an| converges =⇒P

anconverges.

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## Absolute and Conditional Convergence

### Def. (Types of Convergence for a Series)

(1) P

an is absolutely convergent (絕對收斂) if P

|an| converges.

(2) P

an is conditionally convergent (條件收斂) if P an converges, but P

|an| diverges.

### Thm 7.16 (絕對收斂保證原級數收斂)

P|an| converges =⇒P

anconverges.

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### Example (Thm 7.16 的反例)

The series P

n=1 (−1)n

n converges by the Alternating Series Test, but P

n=1|an| = P

n=1 1

n is a divergent p-series with p = 1!

In fact, any conditionally convergent series gives a counterexample of Thm 7.16.

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### Thm 7.17 (The Ratio Test; 比值法)

Letan̸= 0 ∀ n ∈ Nwith ρ = lim

n→∞

an+1 an

.

1 ρ < 1 =P

an converges absolutely.

2 ρ > 1 or ρ =∞ =P

an diverges.

3 ρ = 1 =⇒ the test is inconclusive.

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### Thm 7.18 (The Root Test; 根式法)

Letan̸= 0 ∀ n ∈ Nwith ρ = lim

n→∞

pn

|an| = limn

→∞

|an|1n .

1 ρ < 1 =P

an converges absolutely.

2 ρ > 1 or ρ =∞ =P

an diverges.

3 ρ = 1 =⇒ the test is inconclusive.

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## (泰勒多項式與近似)

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### Def. (Taylor and Maclaurin Polynomials)

Suppose that f has n derivatives at c∈ dom(f).

(1) A polynomial of the form

Pn(x) = Xn k=0

f(k)(c)

k! (x− c)k

= f(c) + f(c)(x− c) + · · · + f(n)(c)

n! (x− c)n is called the nth Taylor poly. (n 階泰勒多項式) for f at c.

(2) Ifc = 0, then Pn(x) = Xn k=0

f(k)(0)

k! xk is called the nth Maclaurin poly. (n 階馬克勞林多項式) for f.

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## (x) 的示意圖 (承上例)

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### Def. (The Remainder of P

n(x))

Let f have (n + 1) derivatives on an interval I containing c.

(1) Rn(x)≡ f(x) − Pn(x) is called the remainder (剩餘項) associated with Pn(x).

(2) |Rn(x)| = |f(x) − Pn(x)| is the error associated with Pn(x).

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### Thm 7.19 (Taylor’s Theorem; 泰勒定理)

If f has (n + 1) derivatives on an interval I containing c, then

∀ x ∈ I,∃ z between x and c s.t.

f(x) = Xn k=0

f(k)(c)

k! (x− c)k+ Rn(x) = Pn(x) + Rn(x), where the Lagrange form of the remainder is given by

Rn(x) = f(n+1)(z)

(n + 1)!(x− c)n+1.

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## (冪級數)

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### Def. (以 c 點為中心的冪級數)

An infinite series of the form

X n=0

an(x− c)n= a0+ a1(x− c) + a2(x− c)2+· · ·

is called a power series (冪級數) centered at c∈ R.

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## 冪級數的收斂行為

### Three Types of Convergence for Power Series

For a power series P

n=0

an(x− c)n, you will see that

1 it converges only at x = c or

2 it converges only for|x − c| < R with R > 0 or

3 it converges for all x∈ R.

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## Radius and Interval of Convergence

Denote

R = Radius of Convergence (收斂半徑), I = Interval of Convergence (收斂區間).

### Type I of Convergence

X n=0

an(x− c)n converges only at x = c.

=⇒R = 0 and I = {c}.

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### Type II of Convergence

X n=0

an(x− c)nconverges absolutely for |x − c| < R, and diverges for |x − c| > R.

=⇒R > 0 and I = (c − R, c + R).

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### Type III of Convergence

X n=0

an(x− c)nconverges for all x∈ R.

=⇒R = ∞ and I = (−∞, ∞).

### Example 4 (Type III 的例子)

Find the radius of convergence for the power series X

n=0

(−1)n

(2n + 1)!x2n+1.

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### Note

In Section 7.8, we will further show that sin x =

X n=0

(−1)n

(2n + 1)!x2n+1 = x−x3 3! +x5

5! x7

7! +· · · ∀ x ∈ R.

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### Endpoint Convergence (端點收斂性)

Besides the interval of convergence I = (c− R, c + R) in Type II, we also have

I = [c− R, c + R), (c − R, c + R] or [c − R, c + R]

for different endpoint convergence of a power series.

### Example 5 (端點收斂的例子)

Find the interval of convergence of the power series X xn

= x +x2 +x3

+ x4

+· · · .

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## Differentiation and Integration of Power Series

For the cases of Type II or Type III, we consider a real-valued function f defined by

f(x) = X n=0

an(x− c)n ∀ x ∈ I,

where I = (c− R, c + R) with R > 0, or I = (−∞, ∞).

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### Two Questions

Is f differentiable and integrable on the open interval I?

If yes, what are f (x) and Z

f(x) dx?

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### Thm 7.21 (冪級數的微分與積分公式)

Let f(x) =

X n=0

an(x− c)n be well-defied on I = (c− R, c + R).

1 Term-by-term differentiation (逐項微分):

f(x) = X n=0

d

dx[an(x− c)n] = X n=1

nan(x− c)n−1 ∀ x ∈ I.

2 Term-by-term integration (逐項積分):

Z

f(x) dx = X n=0

h Z an(x− c)ndxi

=C+ X n=0

an

n + 1(x− c)n+1 ∀ x ∈ I,

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### Remarks

(1) The radii of convergence of f (x) and Z

f(x) dx are the same as that of f(x) =P

an(x− c)n.

(2) But, their intervals of convergence may differ from f.

(微分和積分後的收斂半徑與 f 相同，但收斂區間略有不同!)

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## (以冪級數作為函數的表示式)

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### Geometric Power Series (幾何冪級數)

The function f(x) = 1

1− x is well-defined for x̸= 1.

f has a geometric power series centered at x = 0, i.e.,

f(x) = 1 1− x =

X n=0

xn= 1 + x + x2+ x3+· · · for |x| < 1.

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## 示意圖 (承上頁)

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### Operations with Power Series

Let f(x) =

X n=0

anxn and g(x) = X n=0

bnxn be well-defined.

1 For any k∈ R, f(kx) =X

n=0

an(kx)n= X n=0

(ankn)xn.

2 For any N∈ N, f(xN) = X n=0

an(xN)n= X n=0

anxnN.

3 f(x)± g(x) =X

n=0

anxn

±X

n=0

bnxn

= X n=0

(an± bn)xn.

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## (泰勒級數與馬克勞林級數)

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### Thm 7.22 (The Form of a Convergent Power Series)

If f is a real-valued function defined on I = (c− R, c + R) by

f(x) = X n=0

an(x− c)n ∀ x ∈ I,

then f has derivatives of all orders on I, and moreover, we have an= f(n)(c)

n! ∀ n ∈ N, with 0! = 1 and f(0) = f.

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### Def. (泰勒級數的定義)

Suppose that f has derivatives of all orders at c.

(1) A power series of the form X

n=0

f(n)(c)

n! (x− c)n= f(c) + f(c)(x− c) +f ′′(c)

2! (x− c)2+· · · is called the Taylor series (泰勒級數) for f at c.

(2) Ifc = 0, a power series of the form X n=0

f(n)(0)

n! xnis called the Maclaurin series (馬克勞林級數) for f.

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## Convergence of Taylor Series

Suppose that f has derivatives of all orders on an open interval I containing c. It follows from Taylor’s Thm (Thm 7.19) that

∀ x ∈ I,∃ z between x and c s.t.

f(x) = Xn k=0

f(k)(c)

k! (x− c)k+ Rn(x) = Pn(x) + Rn(x), where the Lagrange form of the remainder is given by

Rn(x) = f(n+1)(z)

(n + 1)!(x− c)n+1.

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### Question

When does the Taylor series for f at c always converge to f on the open interval I?

### Thm 7.23 (泰勒級數的收斂性)

f(x) =

X n=0

f(n)(c)

n! (x− c)n conv. on I⇐⇒ lim

n→∞Rn(x) = 0 ∀ x ∈ I.

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### Question

When does the Taylor series for f at c always converge to f on the open interval I?

### Thm 7.23 (泰勒級數的收斂性)

f(x) =

X n=0

f(n)(c)

n! (x− c)n conv. on I⇐⇒ lim

n→∞Rn(x) = 0 ∀ x ∈ I.

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### Remark

In order to prove that

nlim→∞Rn(x) = lim

n→∞

f(n+1)(z)

(n + 1)!(x− c)n+1= 0 ∀ x ∈ I, we often use the Squeeze Theorem and the following fact

nlim→∞

|x − c|n+1

(n + 1)! = 0 ∀ x ∈ R.

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### Useful Taylor or Maclaurin Series (1/2)

(1) 1

1− x = X n=0

xn for −1 < x < 1.

(2) ln x = X n=1

(−1)n−1

n (x− 1)nfor 0 < x≤ 2.

(3) ex= X n=0

xn

n! for −∞ < x < ∞.

(4) sin x =

X (−1)n x2n+1 for −∞ < x < ∞.

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### Useful Taylor or Maclaurin Series (2/2)

(5) cos x =

X n=0

(−1)n

(2n)!x2n for −∞ < x < ∞.

(6) sin−1x = X n=0

(2n)!

(2nn!)2(2n + 1)x2n+1 for −1 ≤ x ≤ 1.

(7) tan−1x = X n=0

(−1)n

2n + 1x2n+1 for−1 ≤ x ≤ 1. (8) Binomial Series (二項級數) withk∈ R:

(1+x)k = 1+

X n=1

k(k− 1) · · · (k− n + 1)

n! xn for −1 < x < 1.

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## References

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