[Math 22] Lec 08 Integral Test and Comparison Test (Part 1 of 2)

Summary

This lecture video focuses on series convergence tests, particularly for series of non-negative terms. It begins by introducing the concept of convergence and divergence in series, then discusses various tests, including the integral test. Examples are provided, such as the series of 2n/(1+n^2) and others involving sine and cosine functions. The video also explores the p-series test, detailing conditions for series convergence or divergence based on the value of p. The instructor emphasizes applying integral tests and concludes with solving practical problems to determine series behavior.

Takeaways

• 📚 In the previous lecture, we learned about a series test which determines whether a series is divergent, but it cannot show convergence.
• 🔍 Today's focus is on convergence tests, specifically for series of non-negative terms.
• ➕ A series of non-negative terms is in the form of the summation of a_n where a_n ≥ 0 for all n.
• 🧮 Example 1: The series from n=1 to infinity of 2n/(1+n²) is non-negative for all n.
• 🧩 Example 2: The series from n=1 to infinity of (2^n * sin²(n))/(3^n + 1) is also non-negative for all n.
• ❌ Example 3: The series from n=2 to infinity of (-1)^(n-1)/n! includes negative terms and is therefore not non-negative.
• 🌐 The integral test for series convergence involves checking if the integral of a continuous, positive, and decreasing function over an interval converges.
• 📈 Applying the integral test: If the integral from k to infinity of f(x)dx converges, the series converges; if it diverges, the series diverges.
• 📝 Example application of the integral test: The series from n=1 to infinity of 2n/(1+n²) diverges.
• 🔍 Another example using the integral test: The series from n=2 to infinity of 1/(n * ln(n)²) converges.
• 🔍 Definition and analysis of p-series: A p-series of the form summation of 1/n^p converges if p > 1 and diverges if 0 < p ≤ 1.
• ✅ Conclusion: The sum of two convergent series is also convergent.

Q & A

What is the main focus of the lecture video?

The main focus of the lecture video is convergence tests for series of non-negative terms and the integral test used to determine whether a series converges or diverges.

What is a series of non-negative terms?

A series of non-negative terms is a summation of terms where each term (a sub n) is greater than or equal to zero for all values of n.

What is the integral test for convergence?

The integral test states that if a function f is continuous, positive, and decreasing on an interval [k, ∞) and f(n) equals a sub n for all n ≥ k, then the behavior of the integral from k to infinity of f(x) dx will determine if the series converges or diverges.

What are the conditions for applying the integral test?

The function must be continuous, positive-valued, and decreasing on the interval [k, ∞) for some natural number k.

Can the divergence test prove if a series is convergent?

No, the divergence test can only determine if a series is divergent but cannot prove convergence.

How is the integral test applied in the first example?

In the first example, the series is summation of 2n / (1 + n²). The function f(x) = 2x / (1 + x²) is continuous, positive, and decreasing on [1, ∞), allowing the application of the integral test. After calculating the integral, it is found that the series diverges.

What does it mean if the integral from k to infinity of f(x) dx is infinite?

If the integral from k to infinity of f(x) dx is infinite, the series diverges.

What is a p-series, and when does it converge?

A p-series is a series of the form summation of 1 / n^p, where p is a positive real number. It converges if p > 1 and diverges if 0 < p ≤ 1.

What is the behavior of the harmonic series?

The harmonic series, which is the p-series with p = 1 (summation of 1 / n), is divergent.

How does the integral test help determine the behavior of a p-series?

The integral test shows that a p-series converges when p > 1 because the corresponding integral results in a finite real number. For p ≤ 1, the integral diverges, indicating that the series also diverges.

Outlines

00:00

📚 Introduction to Convergence Tests

The first paragraph introduces the concept of convergence tests for series, focusing specifically on series with non-negative terms. It reiterates that the divergence test can determine if a series is divergent but not if it is convergent. The paragraph explains that a series with non-negative terms takes the form of a summation where all terms are greater than or equal to zero. Several examples of such series are provided, including an analysis of their non-negativity. It concludes by introducing the first test called the 'Integral Test,' which is applicable to continuous, positive, and decreasing functions over a specified interval.

05:01

🔍 Applying the Integral Test

The second paragraph delves into the application of the Integral Test for series. It emphasizes the need to verify that a function satisfies three conditions: positivity, continuity, and decreasing behavior on an interval [k, ∞). The paragraph provides a detailed example using the series 2n / (1 + n²) to determine whether it converges or diverges. It demonstrates the calculation of the integral and checks all necessary conditions, showcasing the procedure to apply the Integral Test accurately. The example concludes by determining whether the given series converges based on the evaluation of the integral.

10:02

🧮 Evaluating P-Series Using the Integral Test

The third paragraph introduces the concept of a p-series, which is a series in the form of 1 / n^p for a positive real number p. The goal is to identify values of p for which the series converges or diverges. Two cases are considered: when p equals 1 (harmonic series) and when p is not equal to 1. Detailed calculations demonstrate how the Integral Test is used to evaluate the integral for these cases. It concludes that a p-series is convergent if p > 1 and divergent if 0 < p ≤ 1.

15:11

🔗 Convergence of Various Series Types

The final paragraph provides additional examples of applying convergence tests to different series. It examines multiple series with different terms to determine whether each is convergent or divergent. The paragraph highlights that certain series types, such as geometric series and p-series with p > 1, are convergent. Furthermore, it explains that the sum of two convergent series remains convergent, reinforcing the concept through practical examples.

Keywords

💡Series

A series refers to the sum of the terms of a sequence. In this context, the video is focused on mathematical series, specifically examining whether they converge (approach a limit) or diverge. The study of convergence and divergence of series is fundamental to understanding the behavior of infinite sums.

💡Convergence

Convergence occurs when the sum of an infinite series approaches a specific, finite number as more terms are added. The video discusses various tests to determine whether a series of non-negative terms converges. For example, the series involving the function 2n/(1 + n^2) is analyzed to check for convergence.

💡Divergence

Divergence means that the sum of the terms of a series increases without bound, or does not approach any particular limit. The video emphasizes that while the divergence test can confirm if a series diverges, it cannot be used to show convergence. An example of a divergent series in the video is the harmonic series when p = 1.

💡Integral Test

The Integral Test is a method for determining whether a series converges or diverges by comparing it to an improper integral. If the integral of a related function converges, the series converges, and if the integral diverges, so does the series. This test is applied to series of non-negative terms in the video, such as 2n/(1 + n^2).

💡Non-negative Terms

Series of non-negative terms are those where all terms are greater than or equal to zero. These series are specifically examined for convergence in the video, as they form the basis for using tests like the Integral Test. An example provided is the series involving 2n/(1 + n^2), which contains only non-negative terms.

💡Harmonic Series

The harmonic series is a series where the terms are of the form 1/n. The video mentions that the harmonic series is divergent when p = 1. This is a classic example in the study of series, demonstrating a series that grows without bound as more terms are added.

💡Improper Integral

An improper integral is an integral that has one or both limits of integration as infinity or where the integrand becomes infinite within the interval. In the video, improper integrals are used within the Integral Test to determine if a series converges or diverges. For instance, the video computes the integral of 2x/(1 + x^2) from 1 to infinity.

💡P-Series

A P-series is a series of the form 1/n^p, where p is a positive real number. The video explores the conditions under which a P-series converges or diverges. Specifically, a P-series converges if p > 1 and diverges if 0 < p ≤ 1. This is illustrated in the video through examples involving different values of p.

💡Continuity

Continuity refers to a function being smooth, without breaks or jumps, over a given interval. The video explains that for the Integral Test to apply, the function f(x) representing the terms of the series must be continuous, positive, and decreasing over the interval [k, ∞). Rational functions like 2x/(1 + x^2) are given as examples of continuous functions.

💡Decreasing Function

A decreasing function is one that becomes smaller as its input increases. In the context of the Integral Test, the video states that the function f(x) must be decreasing on the interval [k, ∞) for the test to apply. For example, the function 2x/(1 + x^2) is shown to be decreasing, which allows the application of the Integral Test.

Highlights

The divergence test cannot be used to show that a series is convergent.

A series of non-negative terms is of the form summation of a sub n where a sub n is greater than or equal to zero for all n.

First example: the series summation from n equals 1 to infinity of 2n over 1 + n squared.

Second example: the series summation from n equals 1 to infinity of 2^n sin^2(n) over 3^n + 1.

The integral test can determine whether a series converges or diverges if the function is continuous, positive, and decreasing.

If the integral from k to infinity of f(x) dx converges, the series summation from n equals k to infinity of a sub n converges.

If the integral from k to infinity of f(x) dx diverges, the series summation from n equals k to infinity of a sub n diverges.

In the first example, the function f(x) = 2x / (1 + x^2) is used to apply the integral test.

The first example series summation of 2n over 1 + n^2 diverges according to the integral test.

In the second problem, the series summation from n equals 2 to infinity of 1 / (n ln^2(n)) converges by the integral test.

A p-series is defined as summation from n equals 1 to infinity of 1 / n^p, where p is a positive real number.

The p-series converges if p > 1 and diverges if 0 < p <= 1, according to the integral test.

In the third problem, the p-series with p = 3/2 converges since p > 1.

The geometric series summation from n equals 1 to infinity of 3^n converges when the common ratio r is less than 1.

Summing two convergent series results in a convergent series.

Transcripts

welcome to another lecture video on

series

in the previous lecture we learned about

a series test

which can determine whether a particular

series is divergent

or not as a reminder we cannot use the

divergence test

to show that a series is convergent

so our study for today is on convergence

tests

specifically for series of non-negative

terms

a series of non-negative terms is of the

form

summation of a sub n where a sub n

is greater than or equal to zero for all

n

let's take a look at some examples of

series of non-negative terms

our first example is the series

given by summation from n equals one to

infinity

of two n over one plus n squared

in the nut in combat non-negative on

terms niton series

dial n greater than zero

it follows that two n is also greater

than zero

moreover you know one plus n squared

greater than zero

because the n squared type a logging

non-negative

two n over one plus n squared is greater

than zero for all

m the pixel terms i hindi

negative

another example is given by the

summation from n equals one to infinity

of two to the n sine squared and over

three to the n plus one

example kappa given an exponent

greater than zero at three to the n

greater than zero

for all m the positive long base

a big sub in 2 to the n sine squared n

over 3 to the n plus 1 is greater than

or equal to 0 for

all my monthly

with negative terms

for the first example we have the

summation

of negative 1 to the n minus 1

over n factorial malinoditos

example naha pug n equals two

and second term i negative one over two

factorial

which is equal to negative one half

another example we have the summation

from n equals two

to infinity of cosine n over ln

so nothing in non-negative terms

our first term when n equals 2

dial 2 is greater than 1 at l n

i increasing it follows that ln2

is greater than ln1 which is equal to 0.

pi over

therefore cosine two i negative

cosine two over ln two is also negative

we now proceed with our first test for

series of non-negative terms

we call this test the integral test

and the integral test is given by the

following theorem

let f be a function that is continuous

positive valued and decreasing on the

interval

closed bracket k comma infinity

for sum k in the set of natural numbers

suppose that f of n equals a sub n

for all n in the set of natural numbers

such that n is greater than equal to k

given these conditions the following two

possibilities may happen

first if the integral from k to infinity

of f of x dx converges

or in other words has a real value

then the series summation from n equals

k

to infinity of a sub n converges

on the other hand if the integral k to

infinity

f of x dx is equal to infinity

then the series summation n equals k to

infinity

of a sub m diverges

as a reminder before using the integral

test

we have to check if our function f

satisfies

the three conditions namely f must be

positive valued

f must be continuous and f must be

decreasing on some interval

k to infinity usually

we choose k to be the smallest value of

n

in the given series let's take a look at

some examples that use the integral test

the problem is to determine whether the

series

given by summation of two n

all over one plus n squared converges

or diverges as stated in our theorem

such that f of n equals a sub n

in this case f of x equals 2x over

one plus x squared

the f is positive on the interval

bracket one to infinity

note that f is a rational function and

rational functions are continuous

everywhere except at the points where

the denominator

is zero there is a example nato

and one plus x squared hyperlogging

greater than zero

for any value of x

k f is continuous in particular

in the interval 1 up to infinity

the f is decreasing on the interval 1 to

infinity

as an exercise you can show that f prime

of

x is equal to 2 times 1 minus x squared

all over 1 plus x squared quantity

squared

note that the denominator of f prime

is greater than zero for any real value

of

x

use a numerator naman for all x greater

than equal to one

one minus x squared is less than or

equal to zero

here f prime is less than or equal to

zero

for all x greater than or equal to one

in other words

f is decreasing on one up to infinity

since all of the conditions for f are

satisfied

we can now apply the integral test

we now compute the integral from one to

infinity

of two x over one plus x squared

first we rewrite this improper integral

as the limit

where t goes to infinity

now u equals one plus x squared

d u equals two x d x

integral d u over u i

l n u damn evaluating

ln of 1 plus x squared gives us

the limit as shown in our

slide

limit oh i positive infinity

the hill constantly ln2

by integral test a conclusion nothing

the series diverges

for our second problem we want to

determine whether the series

summation from n equals 2 to infinity

of 1 over n times quantity ln

n squared converges or diverges

again the first step not

f of x equal to one of over

x times ln x squared

f i positive continuous

at decreasing on the interval 2

up to infinity as an

exercise

we now compute the integral from 2 to

infinity

of 1 all over x times quantity ln x

squared

again it's not an improper thing

integral as the following limit

user u equals lnx

negative one over u

evaluating from x

equals 2 to x equals t gives us the

following

limit

this limit goes to the real number

1 over ln 2 thus by integral

test the series converges

before we solve our third problem we

define

munanaten ang p series

let p be a positive real number

a p series is a series of the form

summation from n equals 1 to infinity

of 1 over n to the p

our goal in this problem is to find all

values of p

for which a p series converges

into two cases for case one

suppose p equals one

then the p the p series is just the

harmonic series

summation of one over m

that from previous lectures we know that

the harmonic series

is divergent

equals 1 problem

for case 2 we suppose that p is not

equal to one

again is not in a function f of x equal

to one over

x to the p or x equal

x raised to negative p

four x in the interval 1 to infinity

but f is positive

at continuous c interval 1 to infinity

moreover f is decreasing on

the interval one to infinity since f

prime of x

equal to negative p times x raised to

negative p minus one which is always

less than zero

for any x on the interval one to

infinity

diagonal satisfy our conditions now

i think theorem but nothing committing

we now compute the integral one to

infinity

of one over x to the p dx where p is not

equal to one

as t approaches infinity

of the following integral at the heel p

not equal to 1

an integral now x to the minus p

i x to the minus p plus one over minus p

plus one evaluating from x equals one to

x equals t

gives us the following limit

take note that for zero less than p less

than one

term now minus p plus one i positive

nothing limits

positive infinity

kappa p greater than one greater than

one

young negative p plus one i less than

zero

okay unlimited

one over p minus one

which is a real number thus by integral

test

a p series is convergent if p is greater

than one

and divergent if zero less than p

less than or equal to one

from now on point nothing in the top

concept

we want to determine whether the given

series is convergent

or divergent

islam where p is equal to one half

which is less than one

of n point nothing is to let us n raised

to

three over two here p equals three over

two which is greater than one

and so the series is convergent

to the n plus 1 over n cubed

[Music]

series

3 to the n convergence

is some geometric series with r equal to

one third the only sampan series

is an amount p series where

p is equal to 3 which is greater than

1. i

convergent since the sum

of two convergent series is convergent

then item number three is also

convergent