Prove some laws of sequencing limits

  • #1
Hi, I got some problems with the homework that I was assigned last week. I really got some problems to even start doing's too difficult that I don't even know how to start it...the professor asked us to proof some laws...of sequencing limits when I don't even understand anything...I've discussed the problem with lots of classmates but non of them know how to do them either...please help me PROOF these problems><" thanxz in advance!
#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

Answers and Replies

  • #2
I think this is an exercise in 'epsilontics'.
Meaning, you use the standard definition of a limes:

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

if and oly if

for each epsilon > 0
there exists n0, so that
|an - a| < epsilon
for each n > n0.
  • #3
yeah, I saw my professor writing something like thaht on the board, but I have no idea how and what that is@@"...could you help me doing the problem further?? please?
  • #4
I've really never learned that...please help me
  • #5
You just saw it? And didn't take notes?

OK, as for #1:
You want to show that |anbn - ab|
goes to zero, OK?
Here's a trick: Let's insert - abn + abn (which is zero):
|anbn - ab|
= |anbn - abn + abn- ab|
= |(an - a)bn + (bn - b)a|

since (an - a) and (bn - b) go to zero, the whole expression goes to zero. It's a matter of writing it down properly.
  • #6
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! you have any clue on how to proof number two by chance? Anybody!?@@"
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  • #7
the first problem looks something like this:

does anyone know how to proof it@@? please?! I desperately want to know it...thanxz
  • #8
  • #9
1. arcnets already showed you how to do this, why do you keep asking about it?

2. A sequence converges to a number L if for every &epsilon; > 0 there exists an N such that n > N implies |xn-L|<&epsilon;. Given xk=(-1)k, for any L there exists infinitely many m such that |xm-L| >= 1, so the definition of convergence can't be satisfied satisfied. This prove is a little more slick if you can use the concept of a cauchy sequence which is not a difficult concept.
  • #10
thank you grady!

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