- #1

- 9

- 0

PROOF:

#1 Limit Law: lim (anbn) = (lim an ) (lim bn)

n->infinity n->infinity n->infinity

#2 an = (-1)^n (1,-1,1,-1,1,-1....) when n=0,1,2,3.... Proof that it diverges

THANKS A LOT IF YOU COULD HELP ME!@@"

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- Thread starter fangtu
- Start date

- #1

- 9

- 0

PROOF:

#1 Limit Law: lim (anbn) = (lim an ) (lim bn)

n->infinity n->infinity n->infinity

#2 an = (-1)^n (1,-1,1,-1,1,-1....) when n=0,1,2,3.... Proof that it diverges

THANKS A LOT IF YOU COULD HELP ME!@@"

- #2

- 508

- 0

Meaning, you use the standard definition of a limes:

a = lim(n->[oo])a

if and oly if

for each epsilon > 0

there exists n

|a

for each n > n

- #3

- 9

- 0

- #4

- 9

- 0

I've really never learned that...please help me

- #5

- 508

- 0

OK, as for #1:

You want to show that |a

goes to zero, OK?

Here's a trick: Let's insert - ab

|a

= |a

= |(a

since (a

- #6

- 9

- 0

oh...ok, thanxz a lot arcnets...actually I did take notes...but the problem is that the professor just went through the whole thing whithout saying much about it...anyways...thank you!

uhmmm...do you have any clue on how to proof number two by chance? Anybody!?@@"

uhmmm...do you have any clue on how to proof number two by chance? Anybody!?@@"

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- #7

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does anyone know how to proof it@@? please?! I desperately wanna know it...thanxz

- #8

- 9

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PLEASEEEEEEEEEEEEE!!!!!

- #9

- 70

- 2

2. A sequence converges to a number L if for every ε > 0 there exists an N such that n > N implies |x

- #10

- 9

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thank you grady!

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