Academ!Sequences & Series

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Formulas

Limit of a sequenceThe limit of a sequence {an} is L

limn→∞

an = L or an → L as n → ∞

if we can make the terms of an closer and closer to L as we we make n larger and larger. If

limn→∞

an

exists, the sequence converges (is convergent). Otherwise it diverges (is divergent).

Precise definition of the limit of a sequenceThe limit of a sequence {an} is L

limn→∞

an = L or an → L as n → ∞

if for every ϵ > 0 there is a corresponding integer N such that

if n > N then |an − L | < ϵ

Limit laws for sequencesIf {an} and {bn} are convergent sequences and c is a constant, then

limn→∞

(an + bn) = limn→∞

an + limn→∞

bn limn→∞

(an − bn) = limn→∞

an − limn→∞

bn

limn→∞

can = c limn→∞

an limn→∞

c = c

limn→∞

(anbn) = limn→∞

an ⋅ limn→∞

bn limn→∞

an

bn=

limn→∞ an

limn→∞ bn if lim

n→∞bn ≠ 0

limn→∞

apn = [ lim

n→∞an]

p

if p > 0 and an > 0

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Squeeze theorem for sequencesIf an ≤ bn ≤ cn for n ≥ n0 and lim

n→∞an = lim

n→∞cn = L then lim

n→∞bn = L.

Absolute value of a sequenceIf lim

n→∞|an | = 0, then lim

n→∞an = 0.

Convergence of a sequence rn

The sequence {rn} is convergent if −1 < r ≤ 1 and divergent for all other values of r.

limn→∞

rn = 0 if −1 < r < 1

= 1 if r = 1

Increasing, decreasing, and monotonic sequencesA sequence {an} is

increasing if an < an+1 for all n ≥ 1, (a1 < a2 < a3 < . . . )

decreasing if an > an+1 for all n ≥ 1, (a1 > a2 > a3 > . . . )

monotonic if it’s either increasing or decreasing

Bounded sequencesA sequence {an} is

bounded above if there’s a number M such that an ≤ M for all n ≥ 1

bounded below if there’s a number m such that m ≤ an for all n ≥ 1

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a bounded sequence if it’s bounded above and below

Monotonic sequence theoremEvery bounded, monotonic sequence is convergent.

Partial sum of the series

Given a series ∞

∑n=1

an = a1 + a2 + a3 + . . . , let sn denote its nth partial sum.

sn =n

∑i=1

ai = a1 + a2 + … + an

If the sequence {sn} converges and limn→∞

sn = s exists as a real number, then the series

∑ an converges and we can say

a1 + a2 + … + an + … = s or ∞

∑n=1

an = s

The number s is the sum of the series. If the sequence {sn} diverges, then the series diverges.

Convergence and sum of a geometric seriesThe geometric series

∞

∑n=1

arn−1 = a + ar + ar2 + . . .

converges if |r | < 1, otherwise it diverges (if |r | ≥ 1). The sum of the convergent series is

∞

∑n=1

arn−1 =a

1 − r|r | < 1

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Convergence of an

If the series ∞

∑n=1

an converges, then limn→∞

an = 0.

Test for divergence

If limn→∞

an doesn’t exist or if limn→∞

an ≠ 0, then the series ∞

∑n=1

an diverges.

Laws of convergent seriesIf the series an and bn both converge, then so do these (where c is a constant):

∞

∑n=1

can = c∞

∑n=1

an

∞

∑n=1

(an + bn) =∞

∑n=1

an +∞

∑n=1

bn

∞

∑n=1

(an − bn) =∞

∑n=1

an −∞

∑n=1

bn

Integral test for convergenceSuppose f is a continuous, positive, decreasing function on [1,∞), and let an = f (n).

If ∫∞

1f (x) dx converges, then

∞

∑n=1

an converges.

If ∫∞

1f (x) dx diverges, then

∞

∑n=1

an diverges.

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Remainder estimate for the integral testSuppose f (k) = ak, where f is a continuous, positive, decreasing function for x ≥ n and

∑ an converges. If Rn = s − sn, then

∫∞

n+1f (x) dx ≤ Rn ≤ ∫

∞

nf (x) dx

p-Series test for convergence

The p-series ∞

∑n=1

1np

converges if p > 1

diverges if p ≤ 1

Comparison test for convergenceSuppose that ∑ an and ∑ bn are series with positive terms.

If ∑ bn converges and an ≤ bn for all n, then ∑ an converges.

If ∑ bn diverges and an ≥ bn for all n, then ∑ an diverges.

Limit comparison test for convergenceSuppose that ∑ an and ∑ bn are series with positive terms.

If limn→∞

an

bn= c

where c is a finite number and

where c > 0

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then either both series converge or both diverge.

Alternating series test for convergenceIf the alternating series

∞

∑n=1

(−1)n−1bn = b1 − b2 + b3 − b4 + b5 − b6 + … bn > 0

satisfies

bn+1 ≤ bn for all n

limn→∞

bn = 0

then the series converges.

Alternating series estimation theoremIf s = ∑ (−1)n−1bn is the sum of an alternating series that satisfies

bn+1 ≤ bn

limn→∞

bn = 0

then |Rn | = |s − sn | ≤ bn+1.

Absolute convergenceA series ∑ an is absolutely convergent if the series of absolute values ∑ |an | is

convergent.

If a series ∑ an is absolutely convergent, then it’s convergent.

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Conditional convergenceA series ∑ an is conditionally convergent if it’s convergent but not absolutely convergent.

Ratio test for convergence

If limn→∞

an+1

an= L < 1, then the series

∞

∑n=1

an is absolutely convergent.

If limn→∞

an+1

an= L > 1 or lim

n→∞

an+1

an= ∞, then the series

∞

∑n=1

an is divergent.

If limn→∞

an+1

an= 1, the ratio test is inconclusive about the convergence of

∞

∑n=1

an, which

means we’ll have to use a different convergence test to determine convergence.

Root test for convergence

If limn→∞

n |an | = L < 1, then the series ∞

∑n=1

an is absolutely convergent.

If limn→∞

n |an | = L > 1 or limn→∞

n |an | = ∞, then the series ∞

∑n=1

an is divergent.

If limn→∞

n |an | = 1, the root test is inconclusive about the convergence of ∞

∑n=1

an, which

means we’ll have to use a different convergence test to determine convergence.

Convergence of power series

Given a power series ∞

∑n=0

cn(x − a)n,

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the series converges only when x = a or

the series converges for all x or

there is a positive number R such that the series converges if |x − a | < R and diverges if |x − a | > R

Differentiation and integration of power seriesIf the power series ∑ cn(x − a)n has a radius of convergence R > 0, then the function f

defined by

f (x) = c0 + c1(x − a) + c2(x − a)2 + … =∞

∑n=0

cn(x − a)n

is differentiable (and therefore continuous) on the interval (a − R, a + R) and

f′�(x) = c1 + 2c2(x − a) + 3c3(x − a)2 + … =∞

∑n=1

ncn(x − a)n−1

∫ f (x) dx = C + c0(x − a) + c1(x − a)2

2+ c2

(x − a)3

3+ … = C +

∞

∑n=0

cn(x − a)n+1

n + 1

Note: The radii of convergence of these two power series is R.

Power series representation (expansion)If f has a power series representation (expansion) at a, that is, if

f (x) =∞

∑n=0

cn(x − a)n |x − a | < R

then its coefficients are given by

cn =f (n)(a)

n!

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and the power series has the form

f (x) =∞

∑n=0

f (n)(a)n!

(x − a)n

= f (a) +f′�(a)1!

(x − a) +f′ �′�(a)

2!(x − a)2 +

f′�′�′�(a)3!

(x − a)3 + . . .

which is the taylor series of the function f at a (or about a or centered at a).

Taylor seriesThe taylor series of a function f at a (or about a or centered at a) is

f (x) =∞

∑n=0

f (n)(a)n!

(x − a)n

= f (a) +f′�(a)1!

(x − a) +f′ �′�(a)

2!(x − a)2 +

f′�′�′�(a)3!

(x − a)3 + . . .

Remainder of the taylor seriesIf f (x) = Tn(x) + Rn(x) where Tn is the nth-degree taylor polynomial of f at a and

limn→∞

Rn(x) = 0

for |x − a | < R, then f is equal to the sum of its taylor series on the interval |x − a | < R.

Taylor’s inequalityIf

| f (n+1)(x) | ≤ M for |x − a | ≤ d

then the remainder Rn(x) of the taylor series satisfies the inequality

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|Rn(x) | ≤M

(n + 1)!|x − a |n+1 for |x − a | ≤ d

Maclaurin seriesThe maclaurin series is a specific instance of the taylor series where a = 0. In other words, it’s just the taylor series of a function f at 0 (or about 0 or centered at 0).

f (x) =∞

∑n=0

f (n)(0)n!

(x − 0)n

=∞

∑n=0

f (n)(0)n!

xn

= f (0) +f′�(0)1!

(x − 0) +f′�′�(0)

2!(x − 0)2 +

f′�′�′�(0)3!

(x − 0)3 + . . .

= f (0) +f′�(0)1!

x +f′�′�(0)

2!x2 +

f′�′�′�(0)3!

x3 + . . .

Common maclaurin series and their radii of convergence1

1 − x=

∞

∑n=0

xn = 1 + x + x2 + x3 + . . . R = 1

ex =∞

∑n=0

xn

n!= 1 +

x1!

+x2

2!+

x3

3!+ … R = ∞

sin x =∞

∑n=0

(−1)n x2n+1

(2n + 1)!= x −

x3

3!+

x5

5!−

x7

7!+ … R = ∞

cos x =∞

∑n=0

(−1)n x2n

(2n)!= 1 −

x2

2!+

x4

4!−

x6

6!+ … R = ∞

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tan−1 x =∞

∑n=0

(−1)n x2n+1

2n + 1= x −

x3

3!+

x5

5!−

x7

7!+ … R = 1

ln(1 + x) =∞

∑n=1

(−1)n−1 xn

n= x −

x2

2+

x3

3−

x4

4+ … R = 1

(1 + x)k =∞

∑n=0

(kn)xn = 1 + kx +

k(k − 1)2!

x2 +k(k − 1)(k − 2)

3!x3 + . . . R = 1

Exponential series

ex =∞

∑n=0

xn

n! for all x

e =∞

∑n=0

1n!

= 1 +11!

+12!

+13!

+ . . .

Binomial seriesIf k is any real number and |x | < 1, then

(1 + x)k =∞

∑n=0

(kn)xn = 1 + kx +

k(k − 1)2!

x2 +k(k − 1)(k − 2)

3!x3 + . . .

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