lecture1 slides 2014

8/19/2019 Lecture1 Slides 2014

1/50

INFINITE SERIES

Luiza Bădin

FABIZ I, Fall 2014

Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 1 / 18

http://find/


2/50

Outline

1 Infinite series of numbers. Zeno’s paradox.


http://find/http://goback/


3/50

Outline


2 Sequences of partial sums.




4/50

Outline



3 The Geometric series.


http://find/


5/50


6/50

Outline




4 The Harmonic series.

5 Properties of infinite series. Series of positive terms.


http://find/


7/50

Outline




4 The Harmonic series.

5 Properties of infinite series. Series of positive terms.

6 Examples.


Infinite series of numbers Zeno’s paradox

http://find/


8/50

Infinite series of numbers. Zeno s paradox.

Infinite series of numbers

DefinitionLet (u n )n ≥1 be an infinite sequence. The formal expression

∞n =1

u n (1.1)

is called an infinite series.


Infinite series of numbers Zeno’s paradox

http://find/


9/50


Infinite series of numbers

DefinitionLet (u n )n ≥1 be an infinite sequence. The formal expression

∞n =1

u n (1.1)

is called an infinite series.

An infinite series will be equivalently denoted by:

n ≥1

u n , n

u n , u n . (1.2)

What is the meaning of adding infinitely many things together?

How can one divide the whole into infinitely many parts?


Infinite series of numbers. Zeno’s paradox.

http://find/


10/50


Zeno’s Paradox:

The Old Tale of Achilles and the Tortoise.

Example

Achilles, a fast runner was asked to race against a tortoise on a track 100 mlong. Being a fair sportsman, Achilles gives the tortoise 10 m advance.Knowing that Achilles can run 10 m/sec while the tortoise only 5 m/sec, whowill win the race?





11/50

p

Zeno’s Paradox:


Solution.

Both start running, the tortoise being 10 m ahead.



http://find/


12/50

p

Zeno’s Paradox:


Solution.


After 1 sec, Achilles has run 10 m and has reached the spot where thetortoise started. The tortoise, in turn, has run 5 m.



http://find/


13/50

Zeno’s Paradox:


Solution.


After 1 sec, Achilles has run 10 m and has reached the spot where thetortoise started. The tortoise, in turn, has run 5 m.

Achilles runs again and reaches the spot the tortoise has just been. Thetortoise, in turn, has run 2.5 m.


http://find/


14/50



15/50

Zeno’s Paradox:


Whenever Achilles manages to reach the spot where the tortoise has justbeen a split-second ago, the tortoise has again covered a little bit of distance,being still ahead of Achilles.





16/50

Zeno’s Paradox:



Hence, as hard as he tries, Achilles only manages to cut the remainingdistance in half each time, implying, of course, that Achilles can actually neverreach the tortoise. So, the tortoise is the one who wins the race.



http://find/


17/50

Zeno’s Paradox:




Obviously, this is not true, but where is the mistake?



http://find/


18/50

Zeno’s Paradox:




Obviously, this is not true, but where is the mistake?

The answer to Zeno’s paradox will be given by means of what is called thegeometric series .



http://find/


19/50


Sequences of partial sums.

http://find/


20/50

Sequences of partial sums

Definition

(S n )n ≥1, S n = u 1 + u 2 + ... + u n =

n k =1 u k , is called the sequence of partial

sums or the n th partial sum of the series∞

n =1 u n .



http://find/


21/50


Definition

(S n )n ≥1, S n = u 1 + u 2 + ... + u n =



n =1 u n .

We have: S n +1 = S n + u n +1, for n ≥ 1. Conversely, given (S n )n ≥1, one candefine the series whose partial sums are the terms of (S n )n ≥1:

u 1 = S 1, u 2 = S 2 − S 1, ..., u n = S n − S n −1, ...



http://find/


22/50


Definition

(S n )n ≥1, S n = u 1 + u 2 + ... + u n =



n =1 u n .


u 1 = S 1, u 2 = S 2 − S 1, ..., u n = S n − S n −1, ...

Therefore, an infinite series can be characterized by its sequence of partialsums (S n )n ≥1 and consequently, the study of the series can be reduced to thestudy of (S n )n ≥1.



http://find/


23/50


Definition

(S n )n ≥1, S n = u 1 + u 2 + ... + u n =



n =1 u n .


u 1 = S 1, u 2 = S 2 − S 1, ..., u n = S n − S n −1, ...

Therefore, an infinite series can be characterized by its sequence of partialsums (S n )n ≥1 and consequently, the study of the series can be reduced to thestudy of (S n )n ≥1.

Definition

We say that the infinite series∞

n =1 u n converges if lim n →∞S n exists and isfinite. We write: S =

∞n =1 u n .

If the limit lim n →∞S n does not exist or is infinite, the series is said to diverge.



http://find/


24/50

Remark

It is not necessary for the indexing of the terms of a series to start with 1. It is almost as common to start with 0, i.e. ∞n =0 u n , whose sequence of partial sums is: u 0, u 0 + u 1, u 0 + u 1 + u 2 and so on.



http://find/


25/50

Remark

It is not necessary for the indexing of the terms of a series to start with 1. It is almost as common to start with 0, i.e. ∞n =0 u n , whose sequence of partial sums is: u 0, u 0 + u 1, u 0 + u 1 + u 2 and so on.

Example

1

∞

n =1

1

2

∞n =1

(−1)n

3

∞

n =0(

1

2 )

n

4

∞n =1

1

n (n + 1)


The Geometric series.

∞



26/50

The Geometric series of ratio q :n =0

q n , q ∈ R.

1 Identify the geometric series when q = 1. Is it convergent?



∞



27/50


q n , q ∈ R.

1 Identify the geometric series when q = 1. Is it convergent?2 For q = −1 find an explicit expression for the partial sums S n . Is the

geometric series with q = −1 convergent?



∞

http://find/


28/50


q n , q ∈ R.


geometric series with q = −1 convergent?3 For any q = 1 show that:

S n = 1 − q n

1 − q . (3.3)

Check if this formula agrees with the case q = −1 considered previously.



∞

http://find/


29/50


q n , q ∈ R.



S n = 1 − q n

1 − q . (3.3)

Check if this formula agrees with the case q = −1 considered previously.4 Conclude that the geometric series is convergent if and only if |q |


30/50


q n , q ∈ R.



S n = 1 − q n

1 − q . (3.3)



31/50


q n , q ∈ R.



S n = 1 − q n

1 − q . (3.3)



32/50

The present value of a sequence of perpetuities

ExampleWhat is the amount S 0 (present value) which has to be deposited at a certainmoment in order to receive in perpetuity (ad infinitum), at the end of eachyear, a constant rate T , given that the annual interest rate is i ?

An annuity is a series of payments required to be made or received overtime at regular intervals.

If the payments are made for an indefinite number of years (as in thePension case), then the annuities are also called perpetuities.

The most common payment intervals are yearly, semi-annually, quarterly,

and monthly.

Examples of annuities include Mortgages, Car payments, Rent, Pensionfund payments, Insurance premiums.




http://goforward/http://find/http://goback/


33/50


The present value corresponding to an amount T payed after k (k ∈ N∗) yearsis:

PV k = T

(1 + i )k (3.5)

Therefore, we have:

S 0 = T

1 + i +

T

(1 + i )2 +

T

(1 + i )3 + ... (3.6)

= T

1 + i

1 +

1

1 + i +

1

(1 + i )2 + ...

(3.7)

= T 1 + i

· 11 − 1

1+i

= T i . (3.8)


The Harmonic series.

∞ 1

http://find/


34/50

The Harmonic series:n =1

1

n

Consider the sequence of partial sums:

S n =n

k =1

1

k = 1 +

1

2 +

1

3 + ... +

1

n . (4.9)

We shall prove that S n admits a divergent subsequence.



∞ 1

http://find/


35/50


1

n


S n =n

k =1

1

k = 1 +

1

2 +

1

3 + ... +

1

n . (4.9)


S 2n = 1 + 1

2 +

1

3 + ... +

1

2n = (1 +

1

2) + (

1

3 +

1

4) + (

1

5 +

1

6 +

1

7 +

1

8) + ...

+ ( 12n −1 + 1

+ ... + 12n − 1

+ 12n

) > 12

+ 24

+ 48

+ ... + 2n −1

2n = n

2 → ∞.



∞ 1

http://find/


36/50


1

n


S n =n

k =1

1

k = 1 +

1

2 +

1

3 + ... +

1

n . (4.9)


S 2n = 1 + 1

2 +

1

3 + ... +

1

2n = (1 +

1

2) + (

1

3 +

1

4) + (

1

5 +

1

6 +

1

7 +

1

8) + ...

+ ( 12n −1 + 1

+ ... + 12n − 1

+ 12n

) > 12

+ 24

+ 48

+ ... + 2n −1

2n = n

2 → ∞.

Therefore, lim n →∞S 2n = ∞, so (S n ) is unbounded. We have lim n →∞S n = ∞and thus the harmonic series

∞n =1

1n

is divergent.


Properties of infinite series. Series of positive terms.

Properties of infinite series and

http://find/


37/50


sequences of partial sums

If∞

n =1 u n converges, then (S n )n ≥1 is a bounded sequence. Thereciprocal is not true.




http://find/


38/50



If∞


(The divergence test) If∞

n =1 u n is convergent, then lim n →∞u n = 0.Equivalently:If lim n →∞u n = 0, then

∞

n =1 u n is divergent.




http://find/


39/50



If∞




∞


Proof.∞n =1 u n is convergent, therefore (S n )n ≥1 is convergent.

Let S = lim n →∞S n . We have u n = S n − S n −1 and taking the limit we obtainlim n →∞u n = lim n →∞(S n − S n −1) = lim n →∞S n − lim n →∞S n −1 = S − S = 0.




http://find/


40/50



If∞




∞


Proof.∞n =1 u n is convergent, therefore (S n )n ≥1 is convergent.

Let S = lim n →∞S n . We have u n = S n − S n −1 and taking the limit we obtainlim n →∞u n = lim n →∞(S n − S n −1) = lim n →∞S n − lim n →∞S n −1 = S − S = 0.

Remark

This test can never be used to show that a series converges. It can only

be used to show that a series diverges. The second version is the most important, applicable statement.



http://find/


41/50

Let∞

n =1 u n and∞

n =1 v n be two infinite series and α ∈ R an arbitraryreal number.



http://find/


42/50

Let∞

n =1 u n and∞


1 If∞n =1 u n and

∞n =1 v n are both convergent with

∞n =1 u n = S and∞

n =1 v n = T , then the series with general term (u n + v n )n ≥1 is alsoconvergent and

∞n =1(u n + v n ) = S + T .



http://find/


43/50

Let∞

n =1 u n and∞


1 If∞n =1 u n and


∞n =1 u n = S and∞


∞n =1(u n + v n ) = S + T .

2 If∞

n =1 u n is convergent and∞

n =1 u n = S then so is the series with generalterm (αu n )n ≥1 and

∞n =1 αu n = α

∞n =1 u n = αS .



http://find/


44/50

Let∞

n =1 u n and∞


1 If∞n =1 u n and


∞n =1 u n = S and∞


∞n =1(u n + v n ) = S + T .

2 If∞



∞n =1 αu n = α

∞n =1 u n = αS .

3 If∞


n =1 v n is divergent, then∞

n =1(u n + v n ) isdivergent.



http://find/


45/50

Let∞

n =1 u n and∞


1 If∞n =1 u n and


∞n =1 u n = S and∞


∞n =1(u n + v n ) = S + T .

2 If∞



∞n =1 αu n = α

∞n =1 u n = αS .

3 If∞


n =1 v n is divergent, then∞

n =1(u n + v n ) is

divergent.

Example

Find the nature of the series

∞

n =1

3n − 5n

15n

and in case of convergence,

calculate the sum.



Series of positive terms

http://find/


46/50

p

Definition

The series∞

n =1 u n is called series of positive terms if u n > 0 ∀n ∈ N.

Any series of positive terms has a sum because its sequence of partial sumsS n is increasing.The sum is finite if S n is bounded above, or equals ∞ otherwise.


Examples.

Examples of fundamental series

http://find/


47/50

1 The Geometric series

∞n =1

q n , q ∈ R is convergent iff |q | < 1.


Examples.


http://find/


48/50


∞n =1


2

The Harmonic series

∞

n =1

1

n is divergent.


Examples.


http://find/


49/50


∞n =1


2

The Harmonic series

∞

n =1

1

n is divergent.

3 The p -series (Riemann )

∞n =1

1

n p is convergent for p > 1 and

divergent otherwise.


Examples.

Exercises

http://find/


50/50

Study the nature of the following infinite series and in case ofconvergence, calculate the sum:

1

∞n =1

1

n (n + 1)(n + 2)

2

∞

n =1

ln n 2 + 3n + 2

n 2

+ 3n

3

∞n =1

n

(n + 1)!

4

∞n =0

(−2)n +3 + 32n +1

10n +2

5

∞n =1

(−1)n + n

2n

http://find/

lecture1 slides 2014

Documents