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00:00:01

study alternating series and study their

00:00:03

absolute and conditional convergence let's

00:00:06

look at some what u n series is

00:00:13

called variable sign if it

00:00:16

contains an infinite number of

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positive and infinite number of

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negative terms such a series is called

00:00:21

alternating sign for example the following series

00:00:26

minus 1 plus 1 2 minus 1 3 plus 1 4 and

00:00:31

so on will be alternating in sign, it

00:00:35

contains an infinite number of

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positive numbers, an infinite number of

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negative numbers of sign, a variable series, the

00:00:41

concept of absolute convergence is introduced for it,

00:00:44

if the series is made up of modules of

00:00:49

modules rn, that is, a series of this type converges,

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then they say that the series

00:00:57

Absolutely converges, this is absolute

00:01:02

convergence, if a series made up of

00:01:05

modules does not converge, the series converges,

00:01:09

that is, this one diverges, then

00:01:15

they talk about conditional convergence, that is, the

00:01:17

series converges conditionally,

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by the way, if the series converges absolutely, then it

00:01:22

will converge

00:01:23

conditionally, but the opposite is incorrect, let’s

00:01:25

consider the example let us have a series of

00:01:28

this type from 1 to infinity,

00:01:32

say sine n divided by the square due to the

00:01:37

fact that we have the expression sine now we have

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1 to infinity, sine can

00:01:42

take on both positive and

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negative values, that is, in this

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series we will have an infinite number of

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positive numbers an infinite number of

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negative numbers

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means this series will be alternating

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let's examine it for convergence

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first of all let's look at the

00:01:59

expression modulus y what it equals

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is the modulus of this we

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can make the expression as modulus sine n n

00:02:07

square no longer needs a modulus

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either it is easy to conclude that this expression

00:02:13

is less than equal to 1 divided by n square,

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indeed sine is limited to

00:02:18

one, so this expression is less than

00:02:20

equal to 1, we have a square and let's

00:02:23

now look at the series already compiled from

00:02:25

such expressions, this is our Dirichlet series, the

00:02:31

degree of alpha is greater than one and we know

00:02:34

such a series converges, if such a series

00:02:38

converges, it converges and the series is composed

00:02:41

of such expressions, it turns out that

00:02:43

our series converges absolutely because the

00:02:51

series is composed of modules,

00:02:53

the expression n is convergent, if it is

00:02:56

absolutely convergent, then naturally it

00:02:57

converges and conditionally, thus ours

00:03:00

is an absolutely convergent

00:03:02

one one of the special cases of the sign of an

00:03:03

alternating series is the sign of an

00:03:05

alternating series, let us have

00:03:07

some numbers n greater than equal to 0,

00:03:13

then the series of the following form y 1 -2 plus y

00:03:17

three and so on is called a sign

00:03:20

alternating series and we can

00:03:22

denote it as the sum minus 1 to the power of m

00:03:25

minus 1st and now to infinity in

00:03:31

order to investigate the convergence of the

00:03:32

sign of an alternating series, the

00:03:34

Leibniz criterion is used,

00:03:42

it consists of the following: let the elements of the

00:03:44

sign of an alternating

00:03:45

series be monotone, that is,

00:03:48

monotonically decreasing u1 is greater than u2

00:03:51

is greater than y three and so on, it is u1 u2

00:03:54

to pay attention to we take exactly

00:03:55

positive part if it is

00:03:59

monotone and the necessary

00:04:02

condition is met, that is, the limit tends to

00:04:04

infinity while n is equal to zero,

00:04:08

then if these conditions are met,

00:04:11

then our series

00:04:12

converges, this is Leibniz’s test, let’s look at an

00:04:17

example, let’s say we have a series of the

00:04:19

form

00:04:20

one minus 1 2 + 1 3 minus 1 4 and so

00:04:27

on pay attention this is almost a

00:04:29

harmonic series

00:04:30

but only the sign is alternating we can

00:04:32

write it as the sum minus 1 to the power of

00:04:35

m minus 11 by n and let's follow it for

00:04:41

convergence

00:04:42

first check the monotonicity

00:04:44

indeed one is greater than 1 2 is greater

00:04:48

than 1 3 is greater than 1 4 and so on, that

00:04:51

is, the terms of this series are monotonic, that is, the

00:04:53

first condition is satisfied, but the second

00:04:56

condition is satisfied, especially since the

00:04:58

limit from 1 to n a queen is equal to zero,

00:05:05

so according to Leibniz’s criterion,

00:05:08

it turns out that this series is

00:05:10

convergent,

00:05:12

however it is not absolutely

00:05:15

convergent because if we consider the

00:05:17

moduli of these expressions we will get a

00:05:18

harmonic series that diverges,

00:05:20

so this series conditionally

00:05:22

converges and finally we will consider one

00:05:24

interesting thing,

00:05:25

let us have some absolutely

00:05:29

convergent series

00:05:36

absolutely convergent from so if our

00:05:38

series is

00:05:39

absolutely convergent then, considering this sum,

00:05:41

we can rearrange the terms

00:05:43

and then the series the sum will not change,

00:05:46

that is, an absolutely convergent series

00:05:49

behaves like an ordinary finite sum, but if

00:05:52

our series converges, let’s say there is a

00:05:55

series v.n. it only conditionally converges, but

00:05:58

does not converge absolutely, then rearranging

00:06:03

the terms can give amazing,

00:06:05

amazing values, let's look at

00:06:07

an example that we have already looked at, consider

00:06:11

this series 1 minus 1 2 + 1 3 minus 1

00:06:15

4 and so on, this is our

00:06:20

alternating sign series -1 degree m minus 11

00:06:24

by n from 1 to infinity, we have proven

00:06:28

that such a series is conditionally convergent, that is,

00:06:30

this sum is equal to a certain

00:06:31

number c and let’s denote it now let’s do the

00:06:35

following, let’s try to rearrange the terms in this

00:06:37

sum as follows,

00:06:39

make one, take a positive number and

00:06:43

put two in its post negative

00:06:45

1 2 1 4 now take the next

00:06:50

positive number and post it again

00:06:53

put two negative ones and this is exactly

00:06:56

minus 1 6 minus 1 8

00:07:02

then comes positive 1 5 minus and

00:07:04

so on according to this principle

00:07:06

positive two negative

00:07:08

positive two negative and now let’s

00:07:12

group it as follows So let's

00:07:15

take these two terms, then

00:07:19

these two terms, and so on, that is, we

00:07:23

take two terms, skip them, take two terms,

00:07:25

skip them, now let's calculate 1

00:07:27

2 minus 1 2 1 2 minus 1 4, so that leaves

00:07:32

1 3 minus 1 6,

00:07:35

this is 1 6 minus 1 8 remains so, and so

00:07:40

on, to

00:07:41

draw our attention, it will turn out to be of this

00:07:43

type, taken out of the bracket 1 2,

00:07:48

we get 1 2, here there will be one minus 1 2 +

00:07:58

1 3 and so on ad infinitum, but

00:08:03

pay attention to this bracket, this is

00:08:06

how much is nothing more than again this is the

00:08:08

same series and we denoted its sum with c, we

00:08:11

get the result 1 2 divided

00:08:14

by c, so pay attention that

00:08:19

by rearranging the terms in our

00:08:21

seemingly convergent series, we were able to

00:08:23

reduce the sum by half, although the

00:08:26

terms remained the same, generally

00:08:28

speaking we can rearrange and terms

00:08:31

in a conditional convergent series, get

00:08:33

any number that we specify in advance,

00:08:35

which means that by rearranging the places of the

00:08:37

terms, the sum can change if

00:08:39

we are talking about a conditionally convergent series, and

00:08:41

this ends this video lesson

00:08:43

[music]

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