Question

Jim: ALTERNATING SERIES QUESTION

Answer 1

ill take a picture of the limit comparison also since they do it there also

Answer 2

alright it's up

Answer 3

problem #9 a?

Answer 4

9a and 8b are much of the same.. for 8b you'd have to use bn= 1/n

Answer 5

which one are you looking at?

Answer 6

both 9a and 8b

Answer 7

ok

Answer 8

\[\frac{n}{\sqrt{n^2-4}}\]

Answer 9

they say this equals one

Answer 10

I see a 3 in the numerator for 8b not a an 'n'

Answer 11

that might be a typo.. i've never seen him do hta before

Answer 12

as lal the other ones have 1 and that would just mean you'd have to m it back 2

Answer 13

well anyway, you're asking why \[\large \lim_{x\rightarrow \infty}\left(\frac{n}{\sqrt{n^2-4}}\right)=1\]
correct?

Answer 14

yes

Answer 15

i mean i know the sqroot(n^2) will bring a n out which will be n/n which will get you 1

Answer 16

but i don't believe thats exactly how they did it

Answer 17

if you simply "plug in" infinity into 'n', you'll see that an indeterminate form results in the form of \[\large \frac{\infty}{\infty}\]
so L'Hospital's rule will be used here. Are you familiar with it?

Answer 18

yeah

Answer 19

so derive the numerator to get 1 and derive the denominator to get \[ \frac{n}{\sqrt{n^2-4}} \], keep in mind that this fraction is in the denominator now

Answer 20

how is this fraction in the denominator?

Answer 21

because the derivative of the denominator \[\sqrt{n^2-4}\] is \[\frac{n}{\sqrt{n^2-4}}\]

Answer 22

yes

Answer 23

i skipped a rule in the chain rule gotcha now

Answer 24

hmm L'Hospital's rule isn't going to work here, I ended up where I started, so that's no help.

Answer 25

yeah it says use limit comparison

Answer 26

ok

Answer 27

could have to do with the n=3 and limit comparison being for n=1

Answer 28

so you have to move the series back 2 so n^2 =n-2)^2

Answer 29

and that cancels out the 4

Answer 30

I mean i've never seen my teacher throw a curve ball like this ... maybe when we got to th power series but not like during htis part

Answer 31

one sec while I look something up real quick

Answer 32

seen as though they really ahdn't taught moving back anr forth a series until power series

Answer 33

i'll write my answer and tell me i this is logically right haha

\[\sum_{n=3}^{\infty}\frac{3}{\sqrt{n^2-4}}=\sum_{n=1}^{\infty}\frac{3}{\sqrt{(n-2)^2-4}} =\sum_{n=1}^{\infty} \frac{3}{\sqrt{n^2-4n+4-4}}\]

Answer 34

that doesn't work either....

Answer 35

in my solution manual it just seems like they assume an "n" will come out due to the n^2

Answer 36

oh right, why didn't I think of that, guess I'm tired lol

Answer 37

anyway the limit comparison test works like this

Answer 38

if the limit of (a sub n over b sub n) = some positive number, then either a and b converge or they both diverge

Answer 39

that sound familiar?

Answer 40

true

Answer 41

so you pick another sequence b sub n to compare the original sequence to. In this case, you chose n/sqrt(n^2-4) because this will cancel out with the first denominator

Answer 42

So using the limit comparison test, we get

\[\lim_{x\rightarrow \infty}\left(\frac{a_{n} }{b_{n} }\right)=\lim_{x\rightarrow \infty}\left(\frac{ \frac{3}{\sqrt{n^2-4}} }{ \frac{n}{\sqrt{n^2-4}} }\right)=\lim_{x\rightarrow \infty}\left(\frac{ 3}{ n}\right) = 0\]

Answer 43

oops those xs are ns

Answer 44

so because this limit is zero, this means that if b sub n converges, then a sub n converges

Answer 45

0 is not a finite number

Answer 46

it's not positive

Answer 47

book says where L is finite and positive

Answer 48

I should have mentioned that if the limit is zero, then if b sub n converges then a sub n converges

Answer 49

so you cannot deduct a reasoning that it converges

Answer 50

snce if it is not a finite number and positive the test fails

Answer 51

i'll show you a problem like this in my book

Answer 52

that's one part of the theorem, if L is finite and positive, then they both converge/diverge together, but I'm talking about another branch of the theorem

Answer 53

see what i mean this is really simple to our problem here... and they say it = 1

Answer 54

as for the other part of the theorem ... it's not even in my book >_<

Answer 55

but that's 23 not 8b or 9a?

Answer 56

http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

Answer 57

example four he pulls it out also... like but how if it doesn't obey the sqroot rules

Answer 58

he factors out the n^7 and to seperate it and then takes it out

Answer 59

is this what is actually going on?

Answer 60

that's so it cancels with the other n^(7/3)

Answer 61

yes so could youpull out n^2 and make the -4/n^2

Answer 62

well what he's doing is he's saying that \[n^7+n^4=n^7(1+\frac{1}{n^3})\]

Answer 63

yeah so couldn't you pull out the n^2 and and have -4/n^2 which would also equal 4.. since cancelation

Answer 64

which is perfectly fine, he's doing this because he wants to be left with \[1+\frac{1}{n^3}\], so when he evaluates the limit of this when n approaches infinity, the last term drops off

Answer 65

so you mean \[n^2-4=n^2(1-\frac{4}{n^2})\]

Answer 66

you'd get 3sqroot/1

Answer 67

if you did it his way

Answer 68

which would simply be 3

Answer 69

yeah so \[\sqrt{n^2-4}=\sqrt{n^2(1-\frac{4}{n^2})}=n\sqrt{1-\frac{4}{n^2}}\]

Answer 70

alright so that is what actually is going on.. my teacher always just took the n out and said alright those cancel.. .he never really explained in depth what was going on ... that the constant would go to 0

Answer 71

or would be 0

Answer 72

as for 9a

Answer 73

best way to say ] a sub n+1 <= a sub is saying that the denominator is growing exponentially fast

Answer 74

best way to say ] a sub n+1 <= a sub is saying that the denominator is growing exponentially fast

Answer 75

I think i killed jim >_<

Answer 76

lol sry just reading something

Answer 77

can you see that the sequence (minus the (-1)^n term) is going to 0?

Answer 78

well that would be another story you'd have to figure that out using another way like l'hopitals or something

Answer 79

if not, evaluate the limit and use L'Hospital's rule (which works this time) to get the limit to be zero

Answer 80

but how would i explain that an+1 < an+1 is getting exponentially bigger /faster than the numerator

Answer 81

after deriving and simplifying you should get \[\frac{3n^2}{5n^4\sqrt{n^3+1}}=\frac{3}{5n^2\sqrt{n^3+1}}\]

Answer 82

you just need to show that the sequence b sub n is a decreasing sequence

Answer 83

so all you need to show is that \[b_n < b_{n+1}\]

Answer 84

isn't it the other way aroun?

Answer 85

oh should mention that \[a_n=(-1)^n b_n\]

Answer 86

oh right, you got it, should be \[b_n > b_{n+1}\]

Answer 87

so just say that n^5 is going to grow bigger when added 1 or would i just test with a value.... to show like my teacher

Answer 88

well you could use explicit values, but that doesn't always work

Answer 89

I would compare \[\frac{\sqrt{n^3+1}}{n^5}\] with \[\frac{\sqrt{(n+1)^3+1}}{(n+1)^5}\]

Answer 90

alright

Answer 91

this exam is going to be a doozy lol

Answer 92

thanfully we don't have to memorize integraton tables... i would die

Answer 93

lol

Answer 94

I gotta figure out a wya to memorize the centroid fomulaes

Answer 95

ok it's best to remember back to calc 1 about finding the intervals where a function increases or decreases