A number s ∈ R is called the sum of the series ∞ X k=1 ak if s = lim n→∞sn= lim n→∞ n X k=1 ak

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Sequences, Series, and Power Series

Definition Let {a_k}^∞_k=1 be a sequence of real numbers. Then

∞

X

k=1

a_k = a₁ + a₂+ a₃+ · · · is called aninfinite series (or just aseries) and

s_n =

n

X

k=1

a_k is called then^th partial sum of

∞

X

k=1

a_k.

The series

∞

X

k=1

a_k is called convergent if the sequence {s_n} is convergent, or equivalently

∞

X

k=1

a_k is convergent if lim

n→∞R_n = lim

n→∞

∞

X

k=n+1

a_k = lim

n→∞

∞

X

k=1

a_k−

n

X

k=1

a_k

!

= 0 .

A number s ∈ R is called the sum of the series

∞

X

k=1

ak if

s = lim

n→∞s_n= lim

n→∞

n

X

k=1

a_k =

∞

X

k=1

a_k i.e.

∞

X

k=1

a_k is convergent and it conveges to s.

The series is called divergent if the sequence {s_n} is divergent.

Theorems 1. If

∞

X

k=1

a_k and

∞

X

k=1

b_k are convergent series, and if c ∈ R, then so are the series

∞

X

k=1

c a_k,

∞

X

k=1

(a_k+ b_k) and

∞

X

k=1

(a_k− b_k), with respectively

∞

X

k=1

c a_k = c

∞

X

k=1

a_k,

∞

X

k=1

(a_k+ b_k) =

∞

X

k=1

a_k+

∞

X

k=1

b_k and

∞

X

k=1

(a_k− b_k) =

∞

X

k=1

a_k−

∞

X

k=1

b_k.

2. (A Test for Divergence) If lim

k→∞a_k does not exist or if lim

k→∞a_k ̸= 0, then the series

∞

X

k=1

a_k is

divergent. Equivalently, if the series

∞

X

k=1

ak is convergent, then lim

k→∞ak = 0.

3. (Geometric Series) If r ̸= 1 is a real number, then the geometric series

∞

X

k=1

r^k = lim

n→∞

n

X

k=1

r^k= lim

n→∞

r − rⁿ⁺¹ 1 − r

( converges to r

1 − r if |r| < 1, diverges if |r| > 1.

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Examples

1. Suppose that

s_n =

n

X

k=1

a_k = 2n

3n + 5 for each n = 1, 2, . . . .

Then ∞

X

k=1

a_k = lim

n→∞s_n = lim

n→∞

2n

3n + 5 = 2 3. 2. The sum of the geometric series 5 − 10

3 +20 9 −40

27 + · · · is 5 − 10

3 +20 9 −40

27 + · · · = 5 · 1

1 − (−²₃) = 3.

3. The harmonic series

∞

X

k=1

1

k = 1 + 1 2+ 1

3+ 1

4+ · · · is divergent since

s2ⁿ =

2ⁿ

X

k=1

1

k = 1+ 1 2

+

1

2 + 1 + 1 2²

+· · ·+

1

2ⁿ⁻¹+ 1 + · · · 1 2ⁿ

≥ 1+1 2+2

2²+· · ·+2ⁿ⁻¹

2ⁿ = 1+n 2.

4. The sum of the series

∞

X

k=1

3

k(k + 1) + 1 2^k

is

∞

X

k=1

3

k(k + 1)+ 1 2^k

= lim

n→∞

" _n X

k=1

3 k −

n

X

k=1

3 k + 1

# +

1 2

1 −¹₂ = 3 · 1 + 1 = 4.

The Integral TestSuppose that f is a continuous,positive, decreasingfunction on [1, ∞), and let a_k= f (k). Then

• the series

∞

X

k=1

ak is convergent if and only if the improper integral Z ∞

1

f (x) dx is convergent.

In other words, (i) if

Z ∞ 1

f (x) dx is convergent, then

∞

X

k=1

a_k is convergent.

(ii) if Z ∞

1

f (x) dx is divergent, then

∞

X

k=1

a_k is divergent.

Proof For each k = 1, 2, . . . , let ak = f (k) and for each n = 1, 2, . . . let R_n =

∞

X

k=n+1

a_k =

∞

X

k=1

a_k−

n

X

k=1

a_k = s − s_n be the n^th remainder term.

Since f is decreasing on [1, ∞), we have

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ak = f (k) ≥ f (x) ≥ f (k + 1) = a_k+1 for each x ∈ [k, k + 1] and for each k = 1, 2, . . . ,

Z ∞ n

f (x) dx

(∗)

≥ R_n=

∞

X

k=n+1

a_k = a_n+1+ a_n+2+ · · ·

(†)

≥ Z ∞

n+1

f (x) dx for each n ≥ 1.

Thus Z ∞

1

f (x)dx is convergent ⇐⇒ lim^def

n→∞

Z ∞ n

f (x)dx = 0,

⇐⇒(∗) (†) lim

n→∞R_n = lim

n→∞

∞

X

k=n+1

a_k = 0 ⇐⇒^def

∞

X

k=1

a_k is convergent.

Examples 1. The p-series

∞

X

k=1

1

k^p is convergent if p > 1 and divergent if p ≤ 1.

2. Test the series

∞

X

n=1

1

n²+ 1 for convergence or divergence.

3. Determine whether the series

∞

X

n=1

lnn

n converges or diverges.

[Note that if f (x) = lnx

x , x ≥ 1, then f^′(x) = 1 − lnx

x² < 0 for x > e, and f (x) is positive, decreasing on [e, ∞). ]

4.

(a) Approximate the sum of the series

∞

X

k=1

1

k³ by using the sum of the first 10 terms.

Estimate the error involved in this approximation.

[Solution: s₁₀ ≈ 1.1975 and since R₁₀ = s − s₁₀

(∗)

≤ Z ∞

10

1

x³dx = 1

200 = 0.005, the size of the error is at most 0.005. ]

(b) How many terms are required to ensure that the sum is accurate to within 0.0005?

[Solution: Accuracy to within 0.0005 means that we have to find a value of n such that R_n ≤ 0.0005. Since R_n ^(∗)≤

Z ∞ n

1

x³ dx = 1

2n², we want 1

2n² < 0.0005 =⇒ n² > 1000 or n >√

1000 ≈ 31.6. So we need 32 terms to ensure accuracy to within 0.0005. ]

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5. Note that if we add s_n to each side of estimates (∗), (†) for the remainder R_n= s − s_n, we get a lower bound and an upper bound for s.

s_n+ Z ∞

n

f (x) dx ≥ s_n+R_n = s ≥ s_n+ Z ∞

n+1

f (x) dx =⇒

Z ∞ n

f (x) dx ≥ s−s_n ≥ Z ∞

n+1

f (x) dx.

The Comparison Tests

(The Direct Comparison Test) Suppose that

∞

X

k=1

a_kand

∞

X

k=1

b_k are series withpositive terms.

(a) If

∞

X

k=1

bk = lim

n→∞

n

X

k=1

bkis convergent and ak ≤ bkfor all k, then

∞

X

k=1

akis also convergent.

(b) If

∞

X

k=1

b_k is divergent and a_k≥ b_k for all k, then

∞

X

k=1

a_k = lim

n→∞

n

X

k=1

a_k is also divergent.

(The Limit Comparison Test) Suppose that

∞

X

k=1

a_k and

∞

X

k=1

b_k are series withpositive terms.

(a) If lim

k→∞

a_k bk

= r ∈ (0, ∞), then either both series converge or both diverge.

(b) If lim

k→∞

a_k

b_k = 0 and if

∞

X

k=1

b_k is convergent, then

∞

X

k=1

a_k is convergent.

(c) If lim

k→∞

a_k

b_k = ∞ and if

∞

X

k=1

a_k is convergent, then

∞

X

k=1

b_k is convergent.

Examples

∞

X

k=1

5

2k²+ 4k + 3 converges or diverges.

2. Test the series

∞

X

k=1

1

2^k− 1 for convergence or divergence.

∞

X

k=1

2k²+ 3k

√5 + k⁵ converges or diverges.

4. Use the sum of the first 100 terms to approximate the sum of the series

∞

X

k=1

1

k³+ 1. Estimate the error involved in this approximation.

[Solution: Let

R_n=

∞

X

k=n+1

1

k³+ 1, T_n =

∞

X

k=n+1

1 k³ ≤

Z ∞ n

1

x³ dx = 1 2n².

Then R₁₀₀

(∗)

≤ Z ∞

100

1

x³ dx = 1

2(100)² = 0.00005. ]

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Alternating Series and Absolute Convergence

Analternating series is a series whose terms are alternately positive and negative. The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.

Alternating Series Test

If bk> 0, b_k ≥ b_k+1 for all k ≥ 1 and

if lim

k→∞b_k= 0,

then the alternating series

∞

X

k=1

(−1)^k−1bk = b1− b2+ b3− b4+ b5− b6+ · · · is convergent.

Furthermore,

if

∞

X

k=1

(−1)^k−1b_k = s ∈ R, i.e. the alternating series converges to s ∈ R, and

if Rn = s − s_n=

∞

X

k=n+1

(−1)^k−1b_k, then for each n = 1, 2, . . . we have

|R_n| = |s − s_n|

= (b_n+1− b_n+2) + (b_n+3− b_n+4) + (b_n+5− b_n+6) + · · ·

= b_n+1− (b_n+2− b_n+3) − (b_n+4− b_n+5) − · · ·

≤ b_n+1 since b_n+k− b_n+k+1≥ 0 for all k ≥ 2.

Examples

1. The alternating harmonic series

∞

X

k=1

(−1)^k−1

k is convergent by the Alternating Series Test.

∞

X

k=1

(−1)^k3k

4k − 1 is divergent since lim

k→∞

3k

4k − 1 = 3 4 ̸= 0.

3. Determine the convergence of the series

∞

X

k=1

(−1)^k+1k² k³+ 1 . 4. Find the sum of the series

∞

X

k=0

(−1)^k

k! correct to three decimal places.

[Solution: First observe that the series

∞

X

k=0

(−1)^k

k! is convergent by the Alternating Series Test. Since b7 = 1

7! = 1

5040 < 1

5000 = 0.0002 and s6 =

6

X

k=0

(−1)^k

k! ≈ 0.368056, so we have s ≈ 0.368 correct to three decimal places.]

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Definitions

A series X

a_k is called absolutely convergent if the series of absolute values X

|a_k| is convergent.

A series X

ak is called conditionally convergent if it is convergent but not absolutely convergent; that is, if X

a_k converges butX

|a_k| diverges.

Theorem If a series X

a_k is absolutely convergent, then it is convergent.

Proof Since X

a_k is absolutely convergent and |R_n| ≤

∞

X

k=n+1

|a_k| for all n, we have 0 ≤

n→∞lim |R_n| ≤ lim

n→∞

∞

X

k=n+1

|a_k| = 0 =⇒ lim

n→∞|R_n| = 0. Hence, the series X

a_k is convergent.

Examples

1. The alternating series

∞

X

k=1

(−1)^k−1

k² is absolutely convergent since

∞

X

k=1

(−1)^k−1 k²

=

∞

X

k=1

1 k² is a convergent p-series (p = 2 > 1).

∞

X

k=1

(−1)^k−1

k is conditionally convergent since

∞

X

k=1

(−1)^k−1 k is convergent by the Alternating Series Test and

∞

X

k=1

(−1)^k−1 k

=

∞

X

k=1

1

k is a divergent p-series.

∞

X

n=1

cos n

n² is convergent or divergent.

[Solution: Since

cos n n²

≤ 1

n² for all n and since the p-series

∞

X

n=1

1

n² is convergent, the series

∞

X

n=1

cos n

n² is absolutely convergent by the Comparison Test, and hence it is convergent.]

4. Determine whether the series is absolutely convergent, conditionally convergent, or divergent. (a)

∞

X

k=1

(−1)^k k³ , (b)

∞

X

k=1

(−1)^k

√3

k , (c)

∞

X

k=1

(−1)^kk 2k + 1 Definition By a rearrangement of an infinite seriesX

a_k we mean a series obtained by simply changing the order of the terms. For instance, a rearrangement of X

a_k could start as follows:

a₁+ a₂+a₅+ a₃+ a₄+a₁₀+ a₆+ a₇+a₁₅+ · · · It turns out that ifX

ak is absolutely convergent series with sum s, then any rearrangement of Xa_k has the same sum s.

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However, any conditionally convergent series can be rearranged to give a different sum. To illustrate this fact let’s consider the alternating harmonic series

(∗) 1 −1 2 +1

3−1 4+ 1

5− 1 6 +1

7−1

8 + · · · = s = ln2, where we assume that ln(1 + x) =

∞

X

k=1

(−1)^k−1x^k

k for −1 < x ≤ 1. If we multiply by 1

2, we get 1

2 − 1 4+ 1

6− 1

8+ · · · = 1 2s = 1

2ln2 Inserting zeros between the terms of this series, we have

(†) 0 + 1

2 + 0−1

4 + 0 + 1

6+ 0−1

8 + · · · = 1 2s = 1

2ln2 Now we add the series in (∗) and (†):

(∗∗) 1 + 1 3−1

2+ 1 5+ 1

7−1 4 +1

9 + 1

11− · · · = 3 2s = 3

2ln2

Notice that the series in (∗∗) contains the same terms as in (∗) but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are different. In fact, Riemann proved that

• if X

ak is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of X

a_k that has a sum equal to r.

The Ratio and Root Tests

The Ratio TestSuppose that a_k̸= 0 for all k = 1, 2, . . . . (i) If lim

k→∞

|a_k+1|

|a_k| = L < 1, then the series

∞

X

k=1

ak is absolutely convergent (and therefore convergent).

(ii) If lim

k→∞

|a_k+1|

|ak| = L > 1 or lim

k→∞

|a_k+1|

|ak| = ∞, then the series

∞

X

k=1

a_k is divergent.

(iii) If lim

k→∞

|ak+1|

|a_k| = 1, the Ratio Test is inconclusive; that is, no conclusion can be drawn about the convergence or divergence of

∞

X

k=1

a_k.

Examples

1. Test the series

∞

X

n=1

(−1)ⁿn³

3ⁿ for absolute convergence.

2. Determine the convergence of

∞

X

n=1

1 n!.

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3. Determine the convergence of

∞

X

n=1

(−1)ⁿ⁻¹ (2n − 1)!. 4. Determine the convergence of

∞

X

n=1

nⁿ n!. The Root Test

(i) If lim

k→∞

p|ak _k| = L < 1, then the series

∞

X

k=1

ak is absolutely convergent (and therefore convergent).

(ii) If lim

k→∞

p|ak _k| = L > 1 or lim

k→∞

p|ak _k| = ∞, then the series

∞

X

k=1

a_k is divergent.

(iii) If lim

k→∞

p|ak _k| = 1, the Root Test is inconclusive.

Examples

1. Test the convergence of the series

∞

X

n=1

2n + 3 3n + 2

n

.

∞

X

n=1

1 + 1

n

n²

converges or diverges.

Strategy for Testing Series

Examples In the following examples, don’t work out all the details but simply indicate which tests should be used.

1.

∞

X

n=1

n − 1

2n + 1 [Solution: Use the Test for Divergence.]

2.

∞

X

n=1

√n³+ 1

3n³+ 4n²+ 2 [Solution: Use the Limit Comparison Test.]

3.

∞

X

n=1

(−1)ⁿ n²

n⁴+ 1 [Solution: Use the Alternating Series Test. We can also observe that the series converges absolutely and hence converges.]

4.

∞

X

k=1

2^k

k! [Solution: Use the Ratio Test.]

5.

∞

X

n=1

1

2 + 3ⁿ [Solution: Use the Comparison or the Limit Comparison Test.]

Power Series

Definition A power series in x is a series of the form

∞

X

k=0

a_kx^k = a₀+ a₁x + a₂x²+ a₃x³+ · · · = lim

n→∞

n

X

k=0

a_kx^k,

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where x is a variable and the a_k’s are constants called the coefficients of the series.

For each number that we substitute for x, the series is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x.

The sum of the power series is a function

s(x) =

∞

X

k=0

a_kx^k = a₀+ a₁x + a₂x²+ · · · + a_kx^k+ · · · = lim

n→∞

n

X

k=0

a_kx^k

whose domain is the set of all x for which the series converges. Notice that s(x) resembles a polynomial. The only difference is that s(x) has infinitely many terms.

More generally, a series of the form

∞

X

k=0

ak(x − a)^k = a0+ a1(x − a) + a2(x − a)²+ · · · = lim

n→∞

n

X

k=0

ak(x − a)^k

is called a power series in (x − a) or a power series centered at a or a power series about a.

Proposition Suppose that a_k ̸= 0 for all k = 1, 2, . . . , and for a fixed point x ̸= a, suppose that lim

k→∞

p|ak _k(x − a)^k| = lim

k→∞

p|ak _k|

|x − a| = L < 1 or

lim

k→∞

|a_k+1|

|a_k|

|x − a| = L < 1.

Then

∞

X

k=1

a_k(y − a)^k is absolutely convergent for all |y − a| ≤ |x − a|.

Theorem For a power series

∞

X

k=0

a_k(x − a)^k, there are only three possibilities:

(1) The series converges only when x = a.

(2) The series converges for all x.

(3) There is a positive number R,calledthe radius of convergenceof the power series, such that –

∞

X

k=0

a_k(x − a)^k converges if |x − a| < Rand

–

∞

X

k=0

a_k(x − a)^k diverges if |x − a| > R.

By convention,

the radius of convergence is R = 0 in case (1) and

R = ∞ in case (2).

The interval of convergence of a power series is the interval that consists of all values of x for which the series converges.

In case (1), the interval consists of just a single point a.

In case (2), the interval is (−∞, ∞).

In case (3), the interval is one of (a−R, a+R), [a−R, a+R), (a−R, a+R] or [a−R, a+R].

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Proposition (radius of convergence) Suppose that a_k ̸= 0 for all k = 1, 2, . . . , and suppose that

lim

k→∞

p|ak _k| = L or lim

k→∞

|a_k+1|

|a_k| = L for some 0 ≤ L ≤ ∞.

Then the radius of convergence R of the power series

∞

X

k=1

ak(x − a)^k is given by (i) R = 1/L if 0 < L < ∞.

Proof Since

k→∞lim

p|ak _k(x − a)^k| = lim

k→∞

p|ak _k||x − a|

(<L · 1/L =1 for each |x − a| < R = 1/L,

>L · 1/L =1 for each |x − a| > R = 1/L, or

k→∞lim

ak+1(x − a)^k+1

|a_k(x − a)^k| = lim

k→∞

|a_k+1|

|a_k| |x−a|

(<L · 1/L =1 for each 0 < |x − a| < R = 1/L,

>L · 1/L =1 for each |x − a| > R = 1/L,

so

∞

X

k=1

a_k(x − a)^k

– converges for each |x − a| < R = 1/L, – diverges for each |x − a| > R = 1/L,

and R = 1/L is the radius of convergence of the power series

∞

X

k=1

a_k(x − a)^k. (ii) R = ∞ if L = 0.

(iii) R = 0 if L = ∞.

Examples

1. For what values of x is the series

∞

X

n=0

xⁿ convergent?

[Solution: By the Ratio Test and the Divergence Test, the series converges (absolutely) only when |x| < 1.]

∞

X

n=1

(x − 3)ⁿ

n convergent?

[Solution: By the Ratio Test, the Alternating Series Test and the p-series Test, the series converges only when 2 ≤ x < 4.]

∞

X

n=0

n!xⁿ convergent?

[Solution: By the Ratio Test, the series converges only when x = 0.]

∞

X

n=0

xⁿ

(2n)! convergent?

[Solution: By the Ratio Test, the series converges (absolutely) when x ∈ (−∞, ∞).]

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Representations of Functions as Power Series Examples

1. Since the power series

∞

X

k=0

x^k converges absolutely for |x| < 1, and since

∞

X

k=0

x^k= lim

n→∞

n

X

k=0

x^k = lim

n→∞

1 − xⁿ⁺¹

1 − x = 1

1 − x for |x| < 1,

we say that

∞

X

k=0

x^k, is a power series representation of 1

1 − x for x ∈ (−1, 1).

2. Express 1

1 + x² as the sum of a power series and find the interval of convergence.

[Solution: 1

1 + x² = 1

1 − (−x²) =

∞

X

k=0

(−x²)^k converges for | − x²| < 1 ⇐⇒ |x| < 1.]

Differentiation and Integration of Power Series Theorem If the power series

∞

X

k=0

a_k(x − a)^k has radius of convergence R > 0, then the function f defined by

f (x) =

∞

X

k=0

a_k(x − a)^k = a₀+ a₁(x − a) + a₂(x − a)² + · · · is differentiable (and therefore continuous) and

(i) f^′(x) = d dx

∞

X

k=0

a_k(x − a)^k =

∞

X

k=0

d

dx a_k(x − a)^k =

∞

X

k=1

k a_k(x − a)^k−1 for each x ∈ (a − R, a + R),

(ii) Z

f (x) dx = Z ^∞

X

k=0

a_k(x − a)^kdx =

∞

X

k=0

Z

a_k(x − a)^kdx = C +

∞

X

k=0

a_k

k + 1(x − a)^k+1 on the interval (a − R, a + R).

(iii) the radii of convergence of

∞

X

k=1

k a_k(x − a)^k−1 and

∞

X

k=0

ak

k + 1(x − a)^k+1 are both R,

(iv) f has derivatives of all order n = 0, 1, 2 . . . on (a − R, a + R) and for each x ∈ (a − R, a + R),

f⁽ⁿ⁾(x) = dⁿ dxⁿ

∞

X

k=0

a_k(x−a)^k =

∞

X

k=0

dⁿ

dxⁿ ak(x − a)^k =

∞

X

k=n

k(k−1) · · · (k−n+1)a_k(x−a)^k−n.

Examples

1. Express 1

(1 − x)² as a power series by differentiating the Equation 1 1 − x =

∞

X

n=0

xⁿ. What is the radius of convergence?

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[Solution: Since 1 1 − x =

∞

X

n=0

xⁿ for |x| < 1, and by differentiating both sides, we get

1 (1 − x)² =

∞

X

n=1

nxⁿ⁻¹ =

∞

X

n=0

(n + 1)xⁿ for |x| < 1.

Since lim

n→∞(n + 1)^1/n = 1, the radius of convergence R = 1. ] 2. Express tan⁻¹x as a power series by integrating the equation 1

1 + x² =

∞

X

n=0

(−1)ⁿx²ⁿ. What is the radius of convergence?

[Solution: For |x| < 1, since tan⁻¹x = tan⁻¹z|^x₀ =

Z x 0

1 1 + z² dz

= Z x

0

∞

X

n=0

(−1)ⁿz²ⁿdz =

∞

X

n=0

Z x 0

(−1)ⁿz²ⁿdz =

∞

X

n=0

(−1)ⁿ 2n + 1x²ⁿ⁺¹

and since

∞

X

n=0

(−1)ⁿ

2n + 1x²ⁿ⁺¹ converges when x = ±1, we have

tan⁻¹x =

∞

X

n=0

(−1)ⁿ

2n + 1x²ⁿ⁺¹ for all |x| ≤ 1. ] Examples (from Section 17.4)

1. Use power series y =

∞

X

n=0

c_nxⁿ to solve the equation y^′ = ry, where r is a constant.

0 = y^′− ry =

∞

X

n=1

nc_nxⁿ⁻¹−

∞

X

n=0

rc_nxⁿ=

∞

X

n=0

[(n + 1)c_n+1− rc_n]xⁿ

=⇒ c_n+1 = rc_n

n + 1 for n = 0, 1, 2, . . . (called a recursive relation)

=⇒ c_n+1 = rc_n

n + 1 = r²c_n−1

(n + 1)n = · · · = rⁿ⁺¹c₀ (n + 1)!

=⇒ y = c₀

∞

X

n=0

rⁿ

n!xⁿ= c₀e^rx

2. Use power series y =

∞

X

n=0

c_nxⁿ to solve the equation y^′′+ ry = 0, where r > 0 is a constant.

0 = y^′′+ ry =

∞

X

n=2

n(n − 1)c_nxⁿ⁻²+

∞

X

n=0

rc_nxⁿ =

∞

X

n=0

[(n + 2)(n + 1)c_n+2+ rc_n]xⁿ

=⇒ c_n+2 = −rcn

(n + 2)(n + 1) for n = 0, 1, 2, . . . (called a recursive relation)

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=⇒ c_2n= (−r)c_2n−2

(2n)(2n − 1) = (−r)²c_2n−4

(2n)(2n − 1)(2n − 2)(2n − 3) = · · · = (−r)ⁿc₀

(2n)! , n = 0, 1, 2, . . . c_2n+1 = (−r)c_2n−1

(2n + 1)(2n) = (−r)²c_2n−3

(2n + 1)(2n)(2n − 1)(2n − 2) = · · · = (−r)ⁿc₁

(2n + 1)!, n = 0, 1, 2, . . .

=⇒ y = c₀

∞

X

n=0

(−1)ⁿrⁿ

(2n)! x²ⁿ+ c₁

∞

X

n=0

(−1)ⁿrⁿ

(2n + 1)!x²ⁿ⁺¹ = c₀cos√

rx + c₁sin√ rx

3. Show that J₀(x) =

∞

X

n=0

(−1)ⁿx²ⁿ

2²ⁿ(n!)² , the Bessel function of order 0, is a solution of the Bessel equation x²y^′′+ xy^′+ x²y = 0.

If y =

∞

X

n=0

c_nxⁿ, then x²y =

∞

X

n=0

c_nxⁿ⁺², and

y^′ =

∞

X

n=1

nc_nxⁿ⁻¹, y^′′=

∞

X

n=2

n(n − 1)c_nxⁿ⁻²

=⇒ xy^′ =

∞

X

n=1

nc_nxⁿ, x²y^′′=

∞

X

n=2

n(n − 1)c_nxⁿ

=⇒ xy^′ = c₁x +

∞

X

n=2

nc_nxⁿ, x²y^′′=

∞

X

n=2

n(n − 1)c_nxⁿ

set k=n−2

n=k+2=⇒ xy^′ = c1x +

∞

X

k=0

(k + 2)ck+2x^k+2, x²y^′′ =

∞

X

k=0

(k + 2)(k + 1)ck+2x^k+2

set k=n

=⇒ xy^′ = c₁x +

∞

X

n=0

(n + 2)c_n+2xⁿ⁺², x²y^′′ =

∞

X

n=0

(n + 2)(n + 1)c_n+2xⁿ⁺²

=⇒ 0 = x²y^′′+ xy^′+ x²y = c1x +

∞

X

n=0

[(n + 2)(n + 1)cn+2+ (n + 2)cn+2+ cn]xⁿ⁺²

=⇒ 0 = c₁x +

∞

X

n=0

[(n + 2)²c_n+2+ c_n]xⁿ⁺²

=⇒ c₁ = 0, c_n+2 = (−1)c_n

(n + 2)² for n = 0, 1, 2, . . . (called a recursive relation)

=⇒ c_2n+1 = 0, c_2n =(−1)c_2n−2

(2n)² = (−1)²c_2n−4

[2²n²][2²(n − 1)²] = · · · = (−1)ⁿc₀

2²ⁿ(n!)² for n = 0, 1, 2, . . .

=⇒ y = c0

∞

X

n=0

(−1)ⁿx²ⁿ 2²ⁿ(n!)²

set c0=1

=

∞

X

n=0

(−1)ⁿx²ⁿ

2²ⁿ(n!)² = J0(x).

(a) Find the domain of J₀(x). [Solution: By the Ratio Test, the series converges for all values of x. In other words, the domain of the Bessel function J₀ is (−∞, ∞).]

(b) Find the derivative of J0(x). [Solution: J₀^′(x) =

∞

X

n=0

d dx

(−1)ⁿx²ⁿ 2²ⁿ(n!)² =

∞

X

n=1

(−1)ⁿ(2n)x²ⁿ⁻¹ 2²ⁿ(n!)² .]

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Taylor and Maclaurin Series

Taylor’s Theorem If f (x) has derivatives of all orders in an open interval I = (a − R, a + R) containing a, then for each positive integer n and for each x ∈ I,

f (x) = P_n(x) + R_n(x) =

n

X

k=0

f^(k)(a)

k! (x − a)^k+ 1 n!

Z x a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt,

where f^(k)(a) = d^kf

dx^k(a) is the k^th derivative of f at a for k ≥ 1, f⁽⁰⁾(a) = f (a), and

Pn(x) =

n

X

k=0

f^(k)(a)

k! (x − a)^k = f (a) +f^′(a)

1! (x − a) +f^′′(a)

2! (x − a)²+ · · · + 1 n!

dⁿf

dxⁿ(a)(x − a)ⁿ is called then^th-degree Taylor polynomial of f at a

Rn(x) = f (x) − P_n(x) = 1 n!

Z x a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt is called the remainder of order n for the approximation of f (x) by P_n(x) over I.

Proof Using integration by parts formula Z x

a

u dv = uv|^x_a− Z x

a

v du repeatedly, we get f (x) − f (a) =

Z x a

f^′(t) dt = − Z x

a

f^′(t) d(x − t), u = f^′(t), dv = −d(x − t)

= −f^′(t)(x − t)|^x_a+ Z x

a

f^′′(t) (x − t) dt

= f^′(a)(x − a) − 1 2!

Z x a

f^′′(t) d(x − t)², u = f^′′(t), dv = −d(x − t)² 2!

= f^′(a)(x − a) − 1

2!f^′′(t)(x − t)²|^x_a+ 1 2!

Z x a

f^′′′(t)(x − t)²dt

= f^′(a)(x − a) + 1

2!f^′′(a)(x − a)² − 1 3!

Z x a

f^′′′(t) d(x − t)³

· · · ·

(∗)= f^′(a)(x − a) + · · · + 1

n!f⁽ⁿ⁾(a)(x − a)ⁿ+ 1 n!

Z x a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt

The Remainder Estimation TheoremIf there exists a positive constant such that |f⁽ⁿ⁺¹⁾(x)| ≤ M for all |x − a| ≤ R, then the remainder term R_n(x) in Taylor’s Theorem satisfies the inequality

|Rn(x)| ≤ M

(n + 1)!|x − a|ⁿ⁺¹ for |x − a| ≤ R.

Proof Taylor’s Theorem implies that

|Rn(x)|^(∗)= |f (x)−Pn(x)| ≤ 1 n!

Z x a

|f⁽ⁿ⁺¹⁾(t)| (x − t)ⁿdt

≤ M

(n + 1)!

(x − t)ⁿ⁺¹ |^x_a

= M

(n + 1)!|x−a|ⁿ⁺¹. Lagrange Remainder In fact, since f⁽ⁿ⁺¹⁾(x) is continuous for each |x − a| < R and by the Mean Value Theorem for an integral, there exists a c between a and x such that

R_n(x) ^(∗)= f (x) − P_n(x) = 1 n!

Z x a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt = −1 (n + 1)!

Z x a

f⁽ⁿ⁺¹⁾(t) d(x − t)ⁿ⁺¹

= −f⁽ⁿ⁺¹⁾(c) (n + 1)!

Z x a

d(x − t)ⁿ⁺¹dt = f⁽ⁿ⁺¹⁾(c)

(n + 1)! (x − a)ⁿ⁺¹.

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TheoremIf lim

n→∞R_n(x) = 0 for each |x − a| < R, then f has a power series expansion at a, that is

f (x) = lim

n→∞

n

X

k=0

f^(k)(a)

k! (x − a)^k =

∞

X

k=0

f^(k)(a)

k! (x − a)^k for each |x − a| < R, where the series

∞

X

k=0

f^(k)(a)

k! (x − a)^k is called the Taylor series of the function f at a(or about a orcentered at a).

In case a = 0, the Taylor series becomes f (x) =

∞

X

k=0

f^(k)(0)

k! x^k = f (0) + f^′(0)

1! x +f^′′(0)

2! x² +f^′′′(0)

3! x³+ · · · for |x| < R and it is called the Maclaurin series of f.

Examples

1. Find the Maclaurin series of the function f (x) = 1

1 ∓ x and its radius of convergence.

[Solution: 1 1 ∓ x =

∞

X

n=0

(±x)ⁿ and R = 1 by the Ratio Test.]

2. Find the Maclaurin series of the function f (x) = e^x and its radius of convergence.

[Solution: e^x =

∞

X

n=0

1

n!xⁿ and R = ∞ by the Ratio Test.]

3. Find the Maclaurin series of the function f (x) = sin x and its radius of convergence.

[Solution: sin x =

∞

X

n=0

(−1)ⁿ x²ⁿ⁺¹

(2n + 1)! and R = ∞ by the Ratio Test.]

4. Find the Maclaurin series of the function f (x) = cos x and its radius of convergence.

[Solution: cos x =

∞

X

n=0

(−1)ⁿ x²ⁿ

(2n)! and R = ∞ by the Ratio Test.]

5. Find the Maclaurin series of the function f (x) = tan⁻¹x and its radius of convergence.

[Solution: tan⁻¹x =

∞

X

n=0

(−1)ⁿ x²ⁿ⁺¹

2n + 1 and R = 1 by the Ratio Test.]

6. Find the Maclaurin series of the function f (x) = ln(1 + x) and its radius of convergence.

[Solution: ln(1 + x) =

∞

X

n=1

(−1)ⁿ⁻¹xⁿ n =

∞

X

k=0

(−1)^kx^k+1

k + 1 and R = 1 by the Ratio Test.]

7. Find the Maclaurin series of the function f (x) = (1 + x)^k and its radius of convergence.

[Solution: (1 + x)^k=

k

X

n=0

k n

xⁿ and R = 1 by the Ratio Test.]

8. Find the Maclaurin series for (a) f (x) = x cos x and (b) f (x) = ln(1 + 3x²).

[Solution: x cos x =

∞

X

n=0

(−1)ⁿx²ⁿ⁺¹

(2n)! and ln(1+3x²) =

∞

X

n=1

(−1)ⁿ⁻¹3ⁿx²ⁿ

n =

∞

X

k=0

(−1)^k3^k+1x^2k+2 k + 1 ]

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9. Find the first three nonzero terms in the Maclaurin series for (a) e^xsin x and (b) tan x.

[Solution: (a) x + x²+1

3x³+ · · · ; (b) x + 1

3x³+ 2

15x⁵+ · · · ] Examples

(a) Approximate the function f (x) =√³

x by a Taylor polynomial of degree 2 at a = 8.

[Solution: T₂(x) = 2 + 1

12(x − 8) − 1

288(x − 8)².]

(b) How accurate is this approximation when 7 ≤ x ≤ 9?

[Solution: Because x ≥ 7, we have x^8/3 ≥ 7^8/3 and so f^′′′(x) = 10

27· 1

x^8/3 ≤ 10 27 · 1

7^8/3 < 0.0021 =⇒ |R₂(x)| ≤ 0.0021

3! |x − 8|³ < 0.0004. ] Some Proofs

(a) (The Direct Comparison Test) Suppose that

∞

X

k=1

ak and

∞

X

k=1

bk are series with positive

terms. If

∞

X

k=1

b_k = lim

n→∞

n

X

k=1

b_k is convergent and a_k ≤ b_k for all k, then

∞

X

k=1

a_k is also convergent.

Proof Since

∞

X

k=1

bk = lim

n→∞

n

X

k=1

bk is convergent,

0 = lim

n→∞

∞

X

k=n+1

b_k ≥ lim

n→∞

∞

X

k=n+1

a_k≥ 0 =⇒ lim

n→∞

∞

X

k=n+1

a_k = 0 by the squeeze theorem,

and that

∞

X

k=1

a_k = lim

n→∞

n

X

k=1

a_k is convergent.

(b) (The Limit Comparison Test) Suppose that

∞

X

k=1

a_k and

∞

X

k=1

b_k are series with positive terms.

If lim

k→∞

a_k

b_k = r ∈ (0, ∞), then either both series converge or both diverge.

Proof Let ε = r

2 > 0. Since lim

k→∞

a_k

b_k = r ∈ (0, ∞), there is an M ∈ N such that

ak

b_k − r

< ε = r

2 for all k ≥ M

⇐⇒ −r 2 < a_k

b_k − r < r

2 ⇐⇒ r 2 < a_k

b_k < 3r

2 for all k ≥ M

⇐⇒ r

2b_k < a_k< 3r

2 b_k for all k ≥ M.

=⇒ 0 < r 2

∞

X

k=n

bk<

∞

X

k=n

ak < 3r 2

∞

X

k=n

bk for all n ≥ M.

If lim

k→∞

a_k

b_k = 0 and if

∞

X

k=1

b_k is convergent, then

∞

X

k=1

a_k is convergent.

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Proof Since lim

k→∞

a_k

bk = 0, there is an M ∈ N such that 0 < a_k

b_k =

a_k b_k − 0

< 1 for all k ≥ M ⇐⇒ 0 < a_k

b_k < 1 for all k ≥ M

⇐⇒ 0 < a_k< b_k for all k ≥ M. =⇒ 0 <

∞

X

k=n

a_k <

∞

X

k=n

b_k for all n ≥ M.

If lim

k→∞

a_k

b_k = ∞ and if

∞

X

k=1

a_k is convergent, then

∞

X

k=1

b_k is convergent.

Proof Since lim

k→∞

ak

b_k = ∞, there is an M ∈ N such that ak

b_k > 1 for all k ≥ M. Thus we have

a_k > b_k for all k ≥ M =⇒

∞

X

k=n

a_k >

∞

X

k=n

b_k> 0 for all n ≥ M.

(c) (The Ratio Test)Suppose that a_k̸= 0 for all k = 1, 2, . . . , and suppose that lim

k→∞

|a_k+1|

|a_k| = L < 1, then the series

∞

X

k=1

a_k is absolutely convergent (and therefore convergent).

Proof Given 1 − L

2 > ε > 0, since lim

k→∞

|ak+1|

|a_k| = L < 1, there is an M ∈ N such that

|ak+1|

|a_k| − L

< ε < 1 − L

2 for all k ≥ M

=⇒ |a_k+1|

|ak| < L + 1 − L

2 = 1 + L

2 < 1 for all k ≥ M

=⇒ |a_k+1| < 1 + L 2

|a_k| for all k ≥ M

=⇒ |a_k| < 1 + L 2

|a_k−1| < 1 + L 2

2

|a_k−2| < · · · < 1 + L 2

k−M

|a_M| for all k ≥ M

=⇒

∞

X

k=n

|a_k| ≤

∞

X

k=n

1 + L 2

k−M

|a_M| = 1 + L 2

n−M

· |a_M|

1 − (1 + L)/2 for all n ≥ M

(d) (The Root Test) If lim

k→∞

p|ak _k| = L < 1, then the series

∞

X

k=1

a_k is absolutely convergent (and therefore convergent).

Proof Given 1 − L

2 > ε > 0, since lim

k→∞

p|ak _k| = L < 1, there is an M ∈ N such that

p|ak _k| − L

< ε for all k ≥ M

=⇒ p|a^k _k| − L < ε < 1 − L

2 for all k ≥ M

=⇒ 0 ≤ p|a^k _k| ≤ 1 + L 2

< 1 for all k ≥ M

=⇒ |a_k| ≤ 1 + L 2

k

for all k ≥ M

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=⇒

∞

X

k=n

|a_k| ≤

∞

X

k=n

1 + L 2

k

= 1 + L 2

n

· 1

1 − (1 + L)/2 for all n ≥ M (e) (Lagrange Remainder) If f⁽ⁿ⁺¹⁾(t) is continuous on [a, x], then there exists a c ∈ [a, x]

such that Z x

a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt = f⁽ⁿ⁺¹⁾(c) Z x

a

(x − t)ⁿdt = f⁽ⁿ⁺¹⁾(c)

n + 1 (x − a)ⁿ⁺¹.

Proof Since f⁽ⁿ⁺¹⁾(t) is continuous and (x − t)ⁿ ≥ 0 on [a, x], there exist m and M such that m ≤ f⁽ⁿ⁺¹⁾(t) ≤ M for each t ∈ [a, x] and

m Z x

a

(x − t)ⁿdt ≤ Z x

a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt ≤ M Z x

a

(x − t)ⁿdt

=⇒ m ≤

Rx

a f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt Rx

a(x − t)ⁿdt ≤ M.

By the Intermediate Value Theorem, there is a point c ∈ [a, x] such that

f⁽ⁿ⁺¹⁾(c) = Rx

a f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt Rx

a(x − t)ⁿdt =⇒

Z x a

f⁽ⁿ⁺¹⁾(t) (x − t)ⁿdt = f⁽ⁿ⁺¹⁾(c)

n + 1 (x − a)ⁿ⁺¹.