All language subtitles for 303 Limit comparison test-subtitle-en

af Afrikaans
ak Akan
sq Albanian
am Amharic
ar Arabic
hy Armenian
az Azerbaijani
eu Basque
be Belarusian
bem Bemba
bn Bengali
bh Bihari
bs Bosnian
br Breton
bg Bulgarian
km Cambodian
ca Catalan
ceb Cebuano
chr Cherokee
ny Chichewa
zh-CN Chinese (Simplified)
zh-TW Chinese (Traditional)
co Corsican
hr Croatian
cs Czech
da Danish
nl Dutch
eo Esperanto
et Estonian
ee Ewe
fo Faroese
tl Filipino
fi Finnish
fr French
fy Frisian
gaa Ga
gl Galician
ka Georgian
de German
el Greek
gn Guarani
gu Gujarati
ht Haitian Creole
ha Hausa
haw Hawaiian
iw Hebrew
hi Hindi
hmn Hmong
hu Hungarian
is Icelandic
ig Igbo
id Indonesian
ia Interlingua
ga Irish
it Italian
ja Japanese
jw Javanese
kn Kannada
kk Kazakh
rw Kinyarwanda
rn Kirundi
kg Kongo
ko Korean
kri Krio (Sierra Leone)
ku Kurdish
ckb Kurdish (Soranรฎ)
ky Kyrgyz
lo Laothian
la Latin
lv Latvian
ln Lingala
lt Lithuanian
loz Lozi
lg Luganda
ach Luo
lb Luxembourgish
mk Macedonian
mg Malagasy
ms Malay
ml Malayalam
mt Maltese
mi Maori
mr Marathi
mfe Mauritian Creole
mo Moldavian
mn Mongolian
my Myanmar (Burmese)
sr-ME Montenegrin
ne Nepali
pcm Nigerian Pidgin
nso Northern Sotho
no Norwegian
nn Norwegian (Nynorsk)
oc Occitan
or Oriya
om Oromo
ps Pashto
fa Persian
pl Polish
pt-BR Portuguese (Brazil)
pt Portuguese (Portugal)
pa Punjabi
qu Quechua
ro Romanian
rm Romansh
nyn Runyakitara
sm Samoan
gd Scots Gaelic
sr Serbian
sh Serbo-Croatian
st Sesotho
tn Setswana
crs Seychellois Creole
sn Shona
sd Sindhi
si Sinhalese
sk Slovak
sl Slovenian
so Somali
es Spanish
es-419 Spanish (Latin American)
su Sundanese
sw Swahili
sv Swedish
tg Tajik
ta Tamil
tt Tatar
te Telugu
th Thai
ti Tigrinya
to Tonga
lua Tshiluba
tum Tumbuka
tr Turkish
tk Turkmen
tw Twi
ug Uighur
uk Ukrainian
ur Urdu
uz Uzbek
vi Vietnamese
cy Welsh
wo Wolof
xh Xhosa
yi Yiddish
yo Yoruba
zu Zulu

Original subtitles

1

In this video we're talking about how to use the limit comparison test to say whether a series converges

2

or diverges.

3

And in this particular problem we've been given the series from equals 1 to infinity of end squared

4

divided by quantity and cute plus two.

5

Now as a reminder the limit comparison test relies on a couple of conditions.

6

First of all both the original series A7 and the comparison series be seben have to be greater than

7

zero.

8

So all the terms of the series have to be positive and in addition to that the limit as and goes to

9

infinity of a sideband divided by B-7.

10

So we're literally going to take the original series and divide it by the comparison series.

11

If that limit the value of that limit is greater than zero then we can draw conclusions.

12

We can say if the comparison series B-7 converges then the original series a seben also converges.

13

Or if the comparison series B-7 diverges than the original series A7 also diverges.

14

So we're going to first do is make sure that our original series and our comparison series meet these

15

conditions before we go on to drawing any conclusion.

16

So what will be our comparison series.

17

Well what we usually want to do if we have a rational function or a fraction is we want to take the

18

term from the numerator and denominator that's going to have the greatest effect on the numerator and

19

denominator or the term with the greatest magnitude.

20

So in this case there's only one term in the numerator it's end squared.

21

So we're going to go ahead and take and squared for the new numerator and then for the denominator and

22

cubed has a greater magnitude or greater degree than just the constant 2.

23

So we're going to take and cubed only for the denominator.

24

This is going to simplify we're going to be able to cancel and squared from both the numerator and denominator

25

and this will simplify to 1 over n that means we're going to call our comparison series 1 over and so

26

let's go ahead and say that this series from N equals 1 to infinity of 1 over n and we'll go ahead and

27

call the original series a 7 and we'll call the comparison Series B 7.

28

So now we want to do is show that both of these series are always going to be positive.

29

So if we look at the original series we know we're starting at and equals 1 and we're counting up and

30

equals 2 equals 3 and equals four.

31

All the way up to infinity.

32

Well if we plug in N equals 1 in the numerator here we can get 1 squared which is 1.

33

Here we're going to get one plus two is three so we're going to end up with one third.

34

If we plug in and equals 2 we're going to get 4 over 8 plus 2 is 10 for over 10.

35

Or two over five.

36

If we plug in N equals three we're going to get eight over twenty seven plus two is 29.

37

So we're going to get 8 over 29 and we could keep going.

38

Let's go ahead and look at though the comparison Series B 7.

39

We're also starting at any one equals one we get 1 over 1 or 1 at N equals 2 we get 1 1/2 and equals

40

3 we get one third and we could see that we could keep going.

41

But in either case there's no value of.

42

And then we could plug in if we're counting up from N equals 1 2 3 4 et cetera that we could plug into

43

either series that would make any of these terms negative.

44

And so we can say then that a 7 is always going to be greater than zero and we can say that B-7 is always

45

going to be greater than zero.

46

So those two conditions have been met and the next thing we want to show is that the limit is and goes

47

to infinity of the ratio of these series is going to be greater than zero.

48

So if we take the limit as and approaches infinity of a sob in our original series and squared over

49

and cubed plus two and we divide it by our comparison series we're going to divide this whole thing

50

by one over and.

51

But keep in mind that that's exactly the same thing as multiplying by an over 1.

52

The result then here is going to be the limit as and goes to infinity will multiply and squared by and

53

we'll get and cubed over and cubed plus two.

54

Now in order to take the limit what we want to do is we want to multiply both the numerator and denominator

55

by the highest degree variables so we have one over and cubed and one over and cubed like this.

56

What that's going to do when we say one over and cubed times and cubed we're going to get one in the

57

numerator so we're going to have limit as and approaches infinity we're going to have one in the numerator

58

and denominator one over and cubed times and cubed is going to give us 1.

59

Then we're going to have here plus two over and cubed that way when we take the limit as and goes to

60

infinity the denominator right here is going to get extremely large to divided by an infinitely large

61

number is going to be zero.

62

So this term is going to go to zero which means it's going to become zero there.

63

And we're just going to end up with one over 1 or that this limit here if we call this limit L that

64

L is going to be equal to 1 and one is greater than zero.

65

So we've shown that this condition here has also been met because Al is greater than zero.

66

So now that those conditions have been met we can go on to our conclusions.

67

What we want to do is evaluate be stubborn and say whether B-7 converges or diverges.

68

If we look here at B-7 we notice that it's the harmonic series one divided by and is a harmonic series

69

and we know that the harmonic series diverges.

70

So since B Subban diverges we're looking at this conclusion right here.

71

If B-7 diverges then A7 also diverges.

72

So since we've shown that B Subban diverges we can conclude that a Sabun also diverges and that's how

73

you use the limit comparison test to say whether a series converges or diverges.

Can't find what you're looking for?
Get subtitles in any language from opensubtitles.com, and translate them here.