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Prove the limit theorem: a_n < b_n > AB 
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#1
Sep3009, 12:29 PM

P: 15

1. The problem statement, all variables and given/known data
Given the sequence {a_n} converges to A and {b_n} converges to B, and a_n <_ b_n for all n>_ n*, prove A <_ B 2. Relevant equations x+ epsilon < y for every positive real epsilon, implies x <_ y A  B = (A  a_n) + (b_n  B) + (a_n  b_n) 3. The attempt at a solution I want to show A  B is not positive, I know a_n  b_n is not positive by hypothesis, and A  a_n and b_n  B are bounded by (e/2, e/2) where e is any postive number, hence their sum is bounded by (e,e). But since their sum is not bounded by zero, I can't figure out how to apply (x + e)  y negative implies (x  y) non positive. 


#2
Sep3009, 01:00 PM

HW Helper
Thanks
PF Gold
P: 7,581

If you suppose A > B and take epsilon (1/3)(AB) can you find a contradiction?



#3
Sep3009, 01:03 PM

P: 15




#4
Sep3009, 01:21 PM

HW Helper
Thanks
PF Gold
P: 7,581

Prove the limit theorem: a_n < b_n > AB
If you see how to do the indirect argument I am hinting at, you can easily modify it to be a direct argument. (Shhh... I won't tell anybody if you look at the indirect argument first).



#5
Sep3009, 01:33 PM

P: 15

EDIT: hah i think i got it! EDIT: just kidding... 


#6
Sep3009, 02:16 PM

HW Helper
Thanks
PF Gold
P: 7,581

Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?



#7
Sep3009, 02:22 PM

P: 15

so if I use
a_n  A < f(A,B) b_n  A < g(A,B) I can split up each inequality and ultimately end up with something like a_n > f(A,B) + A b_n < G(A,B) + B so a_n  b_n > f(A,B)  g(A,B) + A  B so if i want to show a_n  b_n > A  B, the function f + g must be positive. But this is impossible since both of these functions must be strictly positive, hence negative f plus negative g is also negative. So i'm still stuck 


#8
Oct109, 03:15 AM

P: 84

If you know liminf, limsup, then this is easy.
Since [tex]a_n \leq b_n[/tex], or [tex]0 \leq b_n  a_n[/tex]. Then it follows that [tex]0 \leq \liminf (b_n  a_n) \leq \limsup (b_n  a_n)[/tex]. But since [tex]\lim a_n = A[/tex] and [tex]\lim b_n = B[/tex], it follows that [tex]\lim (b_n  a_n) = B  A[/tex], which implies the limit inferior and limit superior must equal. That is, [tex]\limsup (b_n  a_n) = \liminf (b_n  a_n) = B  A[/tex]. Thus, we have that [tex]0 \leq B  A[/tex], rearrange then you're done. 


#9
Oct109, 06:51 PM

Sci Advisor
P: 905

For every e>0 there is N such that n>N implies a_nA<e and b_nB<e.
In other words: Ae<a_n<A+e and Be<b_n<B+e. Try to put these two together. \\edit: as LCKurtz remarked, I gave away too much, so I removed the last part. 


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