lecture1 slides 2014
TRANSCRIPT
-
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/
-
8/19/2019 Lecture1 Slides 2014
2/50
Outline
1 Infinite series of numbers. Zeno’s paradox.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 2 / 18
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
3/50
Outline
1 Infinite series of numbers. Zeno’s paradox.
2 Sequences of partial sums.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 2 / 18
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
4/50
Outline
1 Infinite series of numbers. Zeno’s paradox.
2 Sequences of partial sums.
3 The Geometric series.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 2 / 18
http://find/
-
8/19/2019 Lecture1 Slides 2014
5/50
-
8/19/2019 Lecture1 Slides 2014
6/50
Outline
1 Infinite series of numbers. Zeno’s paradox.
2 Sequences of partial sums.
3 The Geometric series.
4 The Harmonic series.
5 Properties of infinite series. Series of positive terms.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 2 / 18
http://find/
-
8/19/2019 Lecture1 Slides 2014
7/50
Outline
1 Infinite series of numbers. Zeno’s paradox.
2 Sequences of partial sums.
3 The Geometric series.
4 The Harmonic series.
5 Properties of infinite series. Series of positive terms.
6 Examples.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 2 / 18
Infinite series of numbers Zeno’s paradox
http://find/
-
8/19/2019 Lecture1 Slides 2014
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 3 / 18
Infinite series of numbers Zeno’s paradox
http://find/
-
8/19/2019 Lecture1 Slides 2014
9/50
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.
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?
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 3 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
10/50
Infinite series of numbers. Zeno s paradox.
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?
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 4 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
11/50
p
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
Solution.
Both start running, the tortoise being 10 m ahead.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 5 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
12/50
p
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
Solution.
Both start running, the tortoise being 10 m ahead.
After 1 sec, Achilles has run 10 m and has reached the spot where thetortoise started. The tortoise, in turn, has run 5 m.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 5 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
13/50
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
Solution.
Both start running, the tortoise being 10 m ahead.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 5 / 18
http://find/
-
8/19/2019 Lecture1 Slides 2014
14/50
Infinite series of numbers. Zeno’s paradox.
-
8/19/2019 Lecture1 Slides 2014
15/50
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 6 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
16/50
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
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.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 6 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
17/50
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
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.
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.
Obviously, this is not true, but where is the mistake?
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 6 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
18/50
Zeno’s Paradox:
The Old Tale of Achilles and the Tortoise.
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.
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.
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 .
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 6 / 18
Infinite series of numbers. Zeno’s paradox.
http://find/
-
8/19/2019 Lecture1 Slides 2014
19/50
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 7 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
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 .
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 8 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
21/50
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 .
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, ...
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 8 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
22/50
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 .
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, ...
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 8 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
23/50
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 .
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, ...
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 8 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 9 / 18
Sequences of partial sums.
http://find/
-
8/19/2019 Lecture1 Slides 2014
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)
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 9 / 18
The Geometric series.
∞
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
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?
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 10 / 18
The Geometric series.
∞
http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
27/50
The Geometric series of ratio q :n =0
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?
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 10 / 18
The Geometric series.
∞
http://find/
-
8/19/2019 Lecture1 Slides 2014
28/50
The Geometric series of ratio q :n =0
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?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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 10 / 18
The Geometric series.
∞
http://find/
-
8/19/2019 Lecture1 Slides 2014
29/50
The Geometric series of ratio q :n =0
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?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.4 Conclude that the geometric series is convergent if and only if |q |
-
8/19/2019 Lecture1 Slides 2014
30/50
The Geometric series of ratio q :n =0
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?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.4 Conclude that the geometric series is convergent if and only if |q |
-
8/19/2019 Lecture1 Slides 2014
31/50
The Geometric series of ratio q :n =0
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?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.4 Conclude that the geometric series is convergent if and only if |q |
-
8/19/2019 Lecture1 Slides 2014
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 11 / 18
The Geometric series.
The present value of a sequence of perpetuities
http://goforward/http://find/http://goback/
-
8/19/2019 Lecture1 Slides 2014
33/50
The present value of a sequence of perpetuities
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)
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 12 / 18
The Harmonic series.
∞ 1
http://find/
-
8/19/2019 Lecture1 Slides 2014
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 13 / 18
The Harmonic series.
∞ 1
http://find/
-
8/19/2019 Lecture1 Slides 2014
35/50
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.
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 → ∞.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 13 / 18
The Harmonic series.
∞ 1
http://find/
-
8/19/2019 Lecture1 Slides 2014
36/50
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.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 13 / 18
Properties of infinite series. Series of positive terms.
Properties of infinite series and
http://find/
-
8/19/2019 Lecture1 Slides 2014
37/50
Properties of infinite series and
sequences of partial sums
If∞
n =1 u n converges, then (S n )n ≥1 is a bounded sequence. Thereciprocal is not true.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 14 / 18
Properties of infinite series. Series of positive terms.
Properties of infinite series and
http://find/
-
8/19/2019 Lecture1 Slides 2014
38/50
Properties of infinite series and
sequences of partial sums
If∞
n =1 u n converges, then (S n )n ≥1 is a bounded sequence. Thereciprocal is not true.
(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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 14 / 18
Properties of infinite series. Series of positive terms.
Properties of infinite series and
http://find/
-
8/19/2019 Lecture1 Slides 2014
39/50
Properties of infinite series and
sequences of partial sums
If∞
n =1 u n converges, then (S n )n ≥1 is a bounded sequence. Thereciprocal is not true.
(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.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 14 / 18
Properties of infinite series. Series of positive terms.
Properties of infinite series and
http://find/
-
8/19/2019 Lecture1 Slides 2014
40/50
Properties of infinite series and
sequences of partial sums
If∞
n =1 u n converges, then (S n )n ≥1 is a bounded sequence. Thereciprocal is not true.
(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.
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 14 / 18
Properties of infinite series. Series of positive terms.
http://find/
-
8/19/2019 Lecture1 Slides 2014
41/50
Let∞
n =1 u n and∞
n =1 v n be two infinite series and α ∈ R an arbitraryreal number.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 15 / 18
Properties of infinite series. Series of positive terms.
http://find/
-
8/19/2019 Lecture1 Slides 2014
42/50
Let∞
n =1 u n and∞
n =1 v n be two infinite series and α ∈ R an arbitraryreal number.
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 .
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 15 / 18
Properties of infinite series. Series of positive terms.
http://find/
-
8/19/2019 Lecture1 Slides 2014
43/50
Let∞
n =1 u n and∞
n =1 v n be two infinite series and α ∈ R an arbitraryreal number.
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 .
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 .
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 15 / 18
Properties of infinite series. Series of positive terms.
http://find/
-
8/19/2019 Lecture1 Slides 2014
44/50
Let∞
n =1 u n and∞
n =1 v n be two infinite series and α ∈ R an arbitraryreal number.
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 .
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 .
3 If∞
n =1 u n is convergent and∞
n =1 v n is divergent, then∞
n =1(u n + v n ) isdivergent.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 15 / 18
Properties of infinite series. Series of positive terms.
http://find/
-
8/19/2019 Lecture1 Slides 2014
45/50
Let∞
n =1 u n and∞
n =1 v n be two infinite series and α ∈ R an arbitraryreal number.
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 .
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 .
3 If∞
n =1 u n is convergent and∞
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 15 / 18
Properties of infinite series. Series of positive terms.
Series of positive terms
http://find/
-
8/19/2019 Lecture1 Slides 2014
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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 16 / 18
Examples.
Examples of fundamental series
http://find/
-
8/19/2019 Lecture1 Slides 2014
47/50
1 The Geometric series
∞n =1
q n , q ∈ R is convergent iff |q | < 1.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 17 / 18
Examples.
Examples of fundamental series
http://find/
-
8/19/2019 Lecture1 Slides 2014
48/50
1 The Geometric series
∞n =1
q n , q ∈ R is convergent iff |q | < 1.
2
The Harmonic series
∞
n =1
1
n is divergent.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 17 / 18
Examples.
Examples of fundamental series
http://find/
-
8/19/2019 Lecture1 Slides 2014
49/50
1 The Geometric series
∞n =1
q n , q ∈ R is convergent iff |q | < 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.
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 17 / 18
Examples.
Exercises
http://find/
-
8/19/2019 Lecture1 Slides 2014
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
Luiza Bădin INFINITE SERIES FABIZ I, Fall 2014 18 / 18
http://find/