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Lecture 3 & 4

Infinite sequences and series

Sequences

Infinite sequences are functions whose domain is the set of natural numbers

ℕ = {1, 2, 3, 4, . . . , 𝑛, . . . }.

Definition. Infinite sequence is a function whose domain is the set of natural

numbers ℕ, which assigns to each number 𝑛 ∈ ℕ exactly one real number 𝑎𝑛 ∈ ℝ,

such that [𝑛, 𝑎𝑛 ] ∈ 𝑝:

𝑛 ∈ ℕ ∃! (𝑎𝑛 ∈ ℝ ) ([𝑛, 𝑎𝑛 ] ∈ 𝑝)

Notation of sequences:

{𝑎𝑛 }𝑛=1∞ = 𝑎𝑛 = {𝑎1, 𝑎2, 𝑎3, … }

𝑎𝑛 is called nth member of the sequence, 𝑛 - corresponding independent variable

Graph of a sequence: isolated points, Fig. 2.1.

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Fig. 2.1. Graph of sequence 𝑎𝑛 = 1

𝑛 . Contains points [𝑛, 𝑎𝑛 ], 𝑛 ∈ ℕ, 𝑎𝑛 ∈ ℝ.

Sequence can be defined:

listing of several initial members

defining the rule for calculation of 𝑎𝑛 ,

recurrently - 𝑎𝑛+1 is defined from 𝑎𝑛

e.g.

𝑛2 = 1, 4, 9, 16, …

𝑛

𝑛+1 =

1

2,

2

3,

3

4,

4

5, …

2𝑛+(−1)𝑛

𝑛 = 1,

5

2,

5

3,

9

4,

9

5,

13

6, …

𝑛

𝑎𝑛

2

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Example. Put down first 5 members of a sequence whose nth member is given as:

𝑎𝑛 =𝑛+2

2𝑛+1.

𝑛 = 1, … ,5 𝑎𝑛 = 1,4

5,

5

7,

6

9,

7

11, …

Example. Find the formula for nth member of sequence:

𝑏𝑛 = 0, −3

2, −

8

3, −

15

4, −

24

5, … .

Written in the form of compound fractions:

𝒏 1 2 3 4 5

𝒃𝒏 0 −11

2 −2

2

3 −3

3

4 −4

4

5

𝑎𝑛 decreases with increasing n, but is by 1

𝑛 larger than −𝑛: 𝑎𝑛 = −𝑛 +

1

𝑛=

1

𝑛− 𝑛. The

formula holds also for 𝑛 = 1.

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Example. Put down first 6 members of a sequence given by recursive formula:

𝑎1 = 𝑎2 = 1, 𝑎𝑛+1 = 𝑎𝑛−1 + 𝑎𝑛

Solve by substituting numbers:

𝑎1 = 1

𝑎2 = 1

𝑎3 = 𝑎1 + 𝑎2 = 1 + 1 = 2

𝑎4 = 𝑎2 + 𝑎3 = 1 + 2 = 3

𝑎5 = 𝑎3 + 𝑎4 = 2 + 3 = 5

𝑎6 = 𝑎4 + 𝑎5 = 3 + 5 = 8

Then: 𝑎𝑛 = {1, 1, 2, 3, 5, 8, … }

It is called sequence of Fibonacci1

________________________

1 Leonardo Pisano (1170-1250), known as Fibonacci, was Italian mathematician

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Arithmetic sequence:

𝑎𝑛+1 = 𝑎𝑛 + 𝑑

Number d is difference. Calculation of 𝑎𝑛 from 𝑑:

𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑 𝑎𝑛 = 𝑎𝑚 + (𝑛 − 𝑚)𝑑

Sum 𝑆𝑛 of first 𝑛 members of arithmetic sequence:

𝑆𝑛 =𝑛

2(𝑎1 + 𝑎𝑛)

Example. Calculate the sum of first 15 members of arithmetic sequence, if:

𝑎4 = 9 a 𝑎7 = 15

Calculate 𝑑 from 𝑎4 and 𝑎7:

𝑎4: 𝑎1 + 3𝑑 = 9 𝑎7: 𝑎1 + 6𝑑 = 15

We obtain: 𝑎1 = 3 and 𝑑 = 2. Then 𝑎15 = 𝑎1 + 14𝑑 = 31

Sum:

𝑆15 =𝑛

2 𝑎1 + 𝑎𝑛 =

15

2 3 + 31 = 255

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Geometric sequence is defined by recurrent formula:

𝑎𝑛+1 = 𝑎𝑛𝑞

number q - quotient.

Calculation of 𝑎𝑛 :

𝑎𝑛 = 𝑎1𝑞𝑛−1 𝑎𝑛 = 𝑎𝑚𝑞𝑛−𝑚

Sum 𝑆𝑛 of first 𝑛 members of geometric sequence with q 1 and 𝑎 ≡ 𝑎1:

𝑆𝑛 = 𝑎 + 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + … + 𝑎𝑞𝑛−1

𝑞𝑆𝑛 = 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + 𝑎𝑞4 + … + 𝑎𝑞𝑛

𝑆𝑛 − 𝑞𝑆𝑛 = 𝑎 − 𝑎𝑞𝑛

𝑆𝑛 =𝑎(1−𝑞𝑛 )

1−𝑞

Graphs of geometric sequences depending on quotient q, Fig. 2.2.

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A B

C D

Fig. 2.2. Graph of geometric sequence 𝑎𝑛 = 𝑎1𝑞𝑛−1 for 𝑎1 = 1 and various

values of 𝑞. A. 𝑞 = −1,2 B. 𝑞 = −0,8 C. 𝑞 = 0,8 D. 𝑞 = 1,2.

𝑞 = 1,2

𝑛 𝑛

𝑎𝑛 𝑎𝑛 𝑞 = 0,8

𝑛 𝑛

𝑎𝑛

𝑎𝑛 𝑞 = −1,2 𝑞 = −0,8

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Example. Calculate the sum of first 10 members of geometric sequence, if:

𝑏5 =2

81 and 𝑏7 =

2

729

First we calculate constants 𝑏1 a 𝑞 from 𝑏5 a 𝑏7 :

𝑏5: 𝑏1𝑞4 =2

81 𝑏7: 𝑏1𝑞6 =

2

729

Solution: 𝑏1 = 2 and 𝑞 =1

3, then the sum 𝑆10:

𝑆10 = 𝑏11−𝑞10

1−𝑞= 2

1−1

310

1−1

3

= 2,99995

With increasing value of independent variable n the values of corresponding

members 𝑎𝑛 can increase, decrease or converge to certain number.

Sequence is called increasing [decreasing, non-decreasing or non-increasing] if for

each 𝑛 ∈ ℕ holds:

𝑎𝑛 < 𝑎𝑛+1 [𝑎𝑛 > 𝑎𝑛+1, 𝑎𝑛 ≤ 𝑎𝑛+1, 𝑎𝑛 ≥ 𝑎𝑛+1].

Sequence {𝑎𝑛}𝑛=1∞ is called bounded below [above] if for each 𝑛 ∈ ℕ exists 𝐻 ∈ ℝ,

such that: 𝑎𝑛 > 𝐻 (𝑎𝑛 < 𝐻).

Sequence is called bounded if it is bounded below and above.

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Example. Prove that sequence 𝑛+1

𝑛+2 𝑛=1

∞ is growing and bounded above.

For increasing sequence: 𝑎𝑛 < 𝑎𝑛+1

𝑛+1

𝑛+2<

𝑛+1 +1

𝑛+1 +2

𝑛+1

𝑛+2<

𝑛+2

𝑛+3

𝑛 + 1 𝑛 + 3 < (𝑛 + 2)2

𝑛2 + 4𝑛 + 3 < 𝑛2 + 4𝑛 + 4

3 < 4

therefore: sequence is increasing

For 𝑛 = 1 we get the lower bound 𝐻𝑙 = 𝑎1 =2

3 , therefore: 𝑎𝑛 ≥ 𝐻𝑠 =

2

3.

We estimate the upper bound as: 𝐻𝑢 = 1.

𝑛+1

𝑛+2< 1

𝑛 + 1 < 𝑛 + 2

1 < 2

Therefore 𝑛+1

𝑛+2 𝑛=1

∞ is increasing and bounded: 𝐻𝑙 =

2

3 and 𝐻𝑢 = 1.

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Limit of sequence

Graphs of sequences on Fig. 2.3. show sequence {𝑛2} which with increasing value

of 𝑛 increases beyond any bounds (sequence is divergent). On the other hand,

sequence 1

𝑛2 is with increasing 𝑛 approaching zero (sequence is converging

to zero).

A B

Fig. 2.3. Graphs of infinite sequences. A. 𝑎𝑛 = 𝑛2 . B. 𝑏𝑛 = 1

𝑛2 .

𝑛 𝑛

𝑎𝑛 𝑏𝑛

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Closeness to a point is related to unlimited approaching.

𝜀-environment of a point 𝑏, 𝑂𝜀(𝑏), is defined as the interval (𝑏 − 𝜀, 𝑏 + 𝜀) with a

center in 𝑏 and width 𝜀 > 0, or as a set 𝑂𝜀 𝑏 = {𝑥 ∈ ℝ; |𝑏 − 𝑥| < 𝜀}. An infinite

sequence is approaching its limit 𝑏 if for any small 𝜀 points 𝑎𝑛 lie within the

environment 𝑂𝜀(𝑏), Fig. 2.4:

∀𝜀 ∈ ℝ+, 𝜀 << 1 ∃𝑛𝜀 ; 𝑛 > 𝑛𝜀 ⇒ |𝑎𝑛– 𝑏| < 𝜀

Fig. 2.4. Graph of sequence 𝑎𝑛 = 3𝑛+2

𝑛+1 . Pre 𝑛 ≥ 5 its points 𝑎𝑛 lie within a

narrow band around number 3 (3 − 𝜀, 3 + 𝜀), 0 < 𝜀 ≪ 1.

_________________________

1 Symbol „≪“ in the inequality 𝑐 ≪ 𝑑 means 𝑐 is much smaller than 𝑑.

𝑛

𝑎𝑛

𝑂𝜀(𝑏)

𝑛𝜀

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Definition. We say that number 𝑏 is the limit of sequence {𝑎𝑛}𝑛=1∞ , if for all é 𝜀 >

0, 𝜀 ∈ ℝ exists 𝑛𝜀 ∈ ℕ, that for 𝑛 > 𝑛𝜀 :

𝑏 − 𝜀 < 𝑎𝑛 < 𝑏 + 𝜀 or |𝑏 − 𝑎𝑛 | < 𝜀

we write: lim𝑛→∞ 𝑎𝑛 = 𝑏

and say „limit of 𝑎𝑛 for 𝑛 increasing to infinity converges to the finite number 𝑏“.

Such sequence is called convergent. Sequence which does not have a limit is

divergent.

Example. Prove that the limit of sequence 3𝑛+2

𝑛+1 from Fig. 1.4., is equal to:

lim𝑛→∞3𝑛+2

𝑛+1= 3.

By definition for all 𝑛 > 𝑛𝜀 and 𝜀 > 0:

3 − 𝜀 <3𝑛+2

𝑛+1 < 3 + 𝜀

Right hand side: 3𝑛+2

𝑛+1 < 3 + 𝜀 after rearrangement: 0 < 𝜀𝑛 + 𝜀 + 1, which is true

for all 𝑛 ∈ ℕ.

Left hand side: 3 − 𝜀 <3𝑛+2

𝑛+1 can be converted to:

1

𝑛+1< 𝜀 shows for what 𝑛𝜀 the

inequality hold if we choose parameter 𝜀:

𝜀 𝟎, 𝟑𝟑 𝟎, 𝟏𝟓 𝟎, 𝟎𝟗 𝟎, 𝟎𝟓 𝟎, 𝟎𝟏

𝑛𝜀 2 5 10 20 100

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Example. Prove that the limit of sequence 1

𝑛 is equal to: lim𝑛→∞

1

𝑛= 0.

If the limit is equal to 0, then:

0 − 𝜀 <1

𝑛 < 0 + 𝜀

- left hand side is always true: 𝑛 > −1

𝜀 for 𝑛 ∈ ℕ.

- right hand side: 𝑛 >1

𝜀 for any 𝜀 we can always find: 𝑛 > 𝑛𝜀 >

1

𝜀 so that the this

inequality is fulfilled. Therefore: lim𝑛→∞1

𝑛= 0.

Theorem. Limit of sequence 1

𝑛𝑘 is equal to: lim𝑛→∞1

𝑛𝑘= 0 where 𝑘 ∈ ℕ

Existence of a limit:

Theorem. Each bounded monotonic sequence1 has a limit. Each sequence has at

most one limit.

_________________________

1 The term monotonic function or sequence means that a curve in its domain only increases or

only decreases and does not combine increase with a decrease.

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Definition. The sequence {𝑎𝑛}𝑛=1∞ has a infinite limit of +∞ (−∞), if for each

𝐺 ∈ ℝ exists 𝑛 such, that for all 𝑛 > 𝑛 holds: 𝑎𝑛 > 𝐺 (𝑎𝑛 < 𝐺), which can be

written as:

lim𝑛→∞ 𝑎𝑛 = ∞ (lim𝑛→∞ 𝑎𝑛 = −∞)

Properties of limits:

Theorem. Let us have two convergent sequences 𝑎𝑛 and 𝑏𝑛 , which have limits

equal to lim𝑛→∞ 𝑎𝑛 = 𝑎, lim𝑛→∞ 𝑏𝑛 = 𝑏

lim𝑛→∞(𝑎𝑛 + 𝑏𝑛 ) = 𝑎 + 𝑏

lim𝑛→∞(𝑎𝑛 − 𝑏𝑛 ) = 𝑎 − 𝑏

lim𝑛→∞ 𝑘 ⋅ 𝑎𝑛 = 𝑘 ⋅ 𝑎 kde 𝑘 ∈ ℝ

lim𝑛→∞ |𝑎𝑛 | = |𝑎|

lim𝑛→∞( 𝑎𝑛 ⋅ 𝑏𝑛 ) = 𝑎 ⋅ 𝑏

lim𝑛→∞𝑎𝑛

𝑏𝑛=

𝑎

𝑏, ak 𝑏𝑛 , 𝑏 ≠ 0

lim𝑛→∞ 𝑏𝑎𝑛 = 𝑏𝑎 ak 𝑏 ∈ ℝ, 𝑏 > 0

lim𝑛→∞ 𝑎𝑛𝑏 = 𝑎𝑏 ak 𝑏 ∈ ℝ, 𝑎 > 0

lim𝑛→∞ 1 +1

𝑛

𝑛= 𝑒

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Example. Determine the limit of sequence: lim𝑛→∞2𝑛3+3𝑛2−𝑛+1

5𝑛3−2𝑛2+4𝑛−7.

We convert this limit of the rational function to a collection of limits of the type

lim𝑛→∞1

𝑛𝑘, by multiplying it with a factor:

1/𝑛𝑚

1/𝑛𝑚, where 𝑚 is the highest power 𝑛𝑚

occurring in the fraction.

lim𝑛→∞2𝑛3+3𝑛2−𝑛+1

5𝑛3−2𝑛2+4𝑛−7= lim𝑛→∞

2𝑛3+3𝑛2−𝑛+1

5𝑛3−2𝑛2+4𝑛−7⋅

1

𝑛3

1

𝑛3

= lim𝑛→∞

2+3

𝑛−

1

𝑛2+1

𝑛3

5−2

𝑛+

4

𝑛2−7

𝑛3

=

=lim 𝑛→∞ 2+lim 𝑛→∞

3

𝑛−lim 𝑛→∞

1

𝑛2+lim 𝑛→∞1

𝑛3

lim 𝑛→∞ 5−lim 𝑛→∞2

𝑛+lim 𝑛→∞

4

𝑛2−lim 𝑛→∞7

𝑛3

=2+0−0+0

5−0+0−0=

2

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Example. Determine the limit of sequence: lim𝑛→∞ 3𝑛+2

2𝑛−1

2+1

𝑛.

lim𝑛→∞ 3𝑛+2

2𝑛−1

2+1

𝑛= lim𝑛→∞

3𝑛+2

2𝑛−1

lim 𝑛→∞ 2+1

𝑛

= 3

2

2=

9

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Example. Determine the limit of sequence: lim𝑛→∞ 𝑛+1

𝑛

𝑛2

3𝑛3+1.

lim𝑛→∞ 𝑛+1

𝑛

𝑛2

3𝑛3+1= lim𝑛→∞ 1 +

1

𝑛

𝑛⋅𝑛

3𝑛3+1= 𝑒

lim 𝑛→∞𝑛

3𝑛3+1 = 𝑒0

3+0 = 𝑒0 = 1

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The theorem characterizing the properties of limits can be cautiously applied also to

infinite limits. For:

lim𝑛→∞ 𝑎𝑛 = ∞ and lim𝑛→∞ 𝑏𝑛 = ∞

then: lim𝑛→∞(𝑎𝑛 + 𝑏𝑛) = ∞ which can be written as: ∞ + ∞ = ∞.

In a simplified notation:

∞ + ∞ = ∞, −∞ −∞ = −∞

∞ ± 𝑐 = ∞, −∞ ± 𝑐 = −∞ where 𝑐 ∈ ℝ

𝑐 ⋅∞ = ∞, 𝑐 ⋅ −∞ = −∞ where 𝑐 ∈ ℝ, 𝑐 > 0

𝑑 ⋅∞ = −∞, 𝑑 ⋅ −∞ = ∞ where 𝑑 ∈ ℝ, 𝑑 < 0

𝑐∞ = ∞, 𝑐−∞ = 0 where 𝑐 ∈ ℝ, 𝑐 > 1

𝑑∞ = 0, 𝑑−∞ = ∞ where 𝑑 ∈ ℝ, 0 < 𝑑 < 1

𝑐

∞= 0,

𝑐

−∞= 0 where 𝑐 ∈ ℝ

When evaluating limits we can find undefined expressions:

0 ⋅∞, 0 ⋅ −∞ , ∞ −∞,0

0,

±∞

±∞, 1∞, 1−∞ these must be solved individually.

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Infinite series

Let 𝑎𝑛 𝑛=1∞ = 1,

1

3,

1

9,

1

27, … ,

1

3𝑛−1, … be an infinite sequence with 𝑎𝑛 =

1

3𝑛−1

(𝑛 ∈ ℕ). With help of 𝑎𝑛 we will form a new sequence {𝑆𝑛} with members equal

to the sum of first 𝑛 members of 𝑎𝑛 :

𝑆1 = 1

𝑆2 = 1 +1

3= 1

1

3

𝑆3 = 1 +1

3+

1

9= 1

4

9

𝑆4 = 1 +1

3+

1

9+

1

27= 1

13

27

𝑆𝑛 = 1 +1

3+

1

9+

1

27+ ⋯ +

1

3𝑛−1=

1

3𝑛−1𝑛1

It can be shown that limit of this sequence {𝑆𝑛} obtained as partial sums of the

sequence 𝑎𝑛 will have for 𝑛 → ∞ the shape:

𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎𝑛 + … = 𝑎𝑛∞𝑛=1

The limit in the example shown above of the sequence 𝑎𝑛 =1

3𝑛−1 will be equal to:

lim𝑛→∞ 𝑆𝑛 =3

2.

However sum of an infinite series may not exist (can be infinite). It exists only if the

sequence of partial sums converges i.e. has a finite limit.

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Theorem. We say that infinite series 𝑎𝑛∞𝑛=1 is convergent (has a finite sum), if

the sequence of partial sums deriving from it {𝑆𝑛} is convergent and has a limit:

lim𝑛→∞ 𝑆𝑛 = 𝑆. Then also the sum of the infinite series equal to 𝑆:

𝑎𝑛∞𝑛=1 = lim𝑛→∞ 𝑆𝑛 = 𝑆

Example . Calculate the sum of infinite series: 1 +1

2+

1

4+

1

8+

1

16+ … =

1

2𝑘∞𝑘=0 .

First we form the series of partial sums:

𝑆𝑘 = 1

2𝑘0𝑘=0 ,

1

2𝑘1𝑘=0 ,

1

2𝑘2𝑘=0 ,

1

2𝑘3𝑘=0 , … =

= 1, 1 +1

2, 1 +

1

2+

1

4, 1 +

1

2+

1

4+

1

8, … = 1,

3

2,

7

4,

15

8, … = 2 −

1

2𝑘−1

The sum exist if 𝑆𝑘 = {2 −1

2𝑘−1} converges:

lim𝑘→∞ 2 −1

2𝑘−1 = 2 − lim𝑘→∞1

2𝑘−1= 2 − 0 = 2

Then:

𝑆 = 1

2𝑘∞𝑘=0 = lim

𝑘→∞ 𝑆𝑘 = lim𝑘→∞ 2 −

1

2𝑘−1 = 2

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Theorem. Let 𝑎𝑛∞𝑛=1 be an infinite series and suppose the limit: lim𝑛→∞

𝑎𝑛+1

𝑎𝑛 =

𝐿 exists (D’Alembert criterion of convergence)1

If 𝐿 < 1 then the series converges, if 𝐿 > 1 then the series diverges, if 𝐿 = 1

then it is not possible to decide based on this criterion.

Example. Decide whether the infinite series: 𝑛

𝑒𝑛∞𝑛=1 converges or not.

D’Alembert criterion:

𝐿 = lim𝑛→∞ 𝑎𝑛+1

𝑎𝑛 = lim𝑛→∞

𝑛+1

𝑒𝑛+1𝑛

𝑒𝑛 = lim𝑛→∞

𝑛+1 𝑒𝑛

𝑛𝑒𝑛+1 =

1

e⋅ lim𝑛→∞

𝑛+1

𝑛=

=1

e⋅ lim𝑛→∞

1+1

𝑛

1= =

1

e⋅

1+0

1=

1

e< 1

Since 𝐿 < 1 the series converges. The sum is, however, unknown.

_________________________

1 Jean le Rond D'Alembert (1717-1783) French mathematician

20

Example. Decide whether the infinite series: 1

𝑛 !∞𝑛=1 converges or not.

D’Alembert criterion:

𝐿 = lim𝑛→∞ 𝑎𝑛+1

𝑎𝑛 = lim𝑛→∞

1

𝑛+1 !1

𝑛 !

= lim𝑛→∞ 𝑛 !

𝑛+1 ! = lim𝑛→∞

𝑛 !

𝑛+1 ⋅𝑛 ! =

= lim𝑛→∞1

𝑛+1= lim𝑛→∞

1

𝑛

1+1

𝑛

=0

1+0= 0 < 1

Since 𝐿 < 1 the series converges.

The sum of an infinite series can be obtained for geometric infinite series.

The sum of first n members of a geometric sequence {𝑎𝑞𝑛} is: 𝑆𝑛 =𝑎(1−𝑞𝑛 )

1−𝑞.

Infinite geometric series is the sum of infinite number of the members of geometric

sequence:

𝑎 + 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + … + 𝑎𝑞𝑛 + … = 𝑎𝑞𝑛∞𝑛=0

its sum is defined as the limit 𝑆𝑛 :

𝑆 = 𝑎𝑞𝑛∞𝑛=0 = lim𝑛→∞ 𝑆𝑛 = lim𝑛→∞

𝑎(1−𝑞𝑛 )

1−𝑞=

𝑎

1−𝑞⋅ lim𝑛→∞ 1 − 𝑞𝑛

If the quotient q:

|𝑞| < 1 then lim𝑛→∞ 1 − 𝑞𝑛 = 1 exists: 𝑆 = 𝑎𝑞𝑛∞𝑛=0 =

𝑎

1−𝑞

|𝑞| > 1 then lim𝑛→∞ 1 − 𝑞𝑛 does not exist and the sum of the infinite

geometric series diverges (is infinite)

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21

Example. Determine whether infinite series: 2 −1

4

𝑛∞𝑛=0 converges and if yes,

then find its sum:

Geometric series, coefficient: 𝑎 = 2 and quotient: 𝑞 = −1

4, −

1

4 =

1

4< 1,

Therefore series converges and its sum is:

2(−1

4)𝑛∞

𝑛=0 =𝑎

1−𝑞=

2

1−(−1

4)

=8

5

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Power series

So far we have considered infinite series whose members were constants (numbers)

𝑎𝑛 ∈ ℝ:

𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎𝑛 + … = 𝑎𝑛∞𝑛=1

Power series are composed of terms which contain coefficients an as well as the

independent variable 𝑥 and its powers 𝑥𝑛 :

𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + … + 𝑎𝑛𝑥𝑛 + … = 𝑎𝑛∞𝑛=0 𝑥𝑛

where: 𝑛 ∈ ℕ

Power series - polynomial of infinite degree (𝑛 → ∞)

By inserting a value of independent variable: 𝑥 = 𝑟 power series turns to infinite

numerical series: 𝑎𝑛∞𝑛=0 𝑟𝑛 (which may converge for the given 𝑟).

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23

Example. Find the sum of power series: 𝑥𝑛∞𝑛=0 .

For this series: 𝑎0 = 𝑎1 = ⋯ = 𝑎𝑛 = … = 1

It represents a after insertion 𝑥 = 𝑟 a geometric series with 𝑎 = 1 and quotient r.

If |𝑟| < 1 then this series converges to the sum 𝑆:

𝑆 = 𝑟𝑛∞𝑛=0 =

1

1−𝑟.

if for example 𝑟 =1

2 then 𝑆 =

1

2

𝑛∞𝑛=0 =

1

1−1

2

= 2.

if for example 𝑟 = −2

3 then 𝑆 = −

2

3

𝑛∞𝑛=0 =

1

1+2

3

=3

5

This power series represents a function, which assigns to each 𝑟 (|𝑟| < 1) a number: 1

1−𝑟. Thus, we say, the power series: 𝑥𝑛∞

𝑛=0 converges to a function: 𝑓 𝑥 =1

1−𝑥

on the interval: 𝑥 ∈ (−1, 1).

Each power series converges on a symmetric interval with the center in zero:

(−𝑟, 𝑟) a the parameter 𝑟 is called: radius of convergence.

Radius of convergence of a power series: 𝑎𝑛∞𝑛=0 𝑥𝑛 can be determined as:

𝑟 = lim𝑛→∞ 𝑎𝑛

𝑎𝑛+1

24

Example. Find a radius of convergence of power series:

𝑥

2

𝑛∞𝑛=0 = 1 +

1

2𝑥 +

1

4𝑥2 +

1

8𝑥3 + … +

1

2𝑛𝑥𝑛 + ….

Radius: 𝑟 = lim𝑛→∞ 𝑎𝑛

𝑎𝑛+1 = lim𝑛→∞

1

2𝑛

1

2𝑛+1

= lim𝑛→∞ 2 = 2

The power series 𝑥

2

𝑛∞𝑛=0 converges on interval: (−2, 2).

Since it is a geometric series with quotient 𝑞 =𝑥

2, we can determine its sum and find

a function to which it converges on (−2, 2):

𝑆 = 𝑥

2

𝑛∞𝑛=0 =

1

1−𝑞=

1

1−𝑥

2

=2

2−𝑥= 𝑓(𝑥)

Power series exists, which have the interval of convergence of (−∞,∞), e.g.:

2𝑛

𝑛 !𝑥𝑛∞

𝑛=0 .

If we know the sum of a power series we can replace it by a function 𝑆(𝑥).

Conversely, in some situations it may be convenient to expand a function 𝑆 𝑥 into

the form of a power series:

𝑆 𝑥 = 𝑎𝑛𝑥𝑛∞𝑛=0 .