Proof: Limit Comparison for a < b in Convergent Sequences

[khdani]

Hello,
Please help me prove the following:
given lim(an)=a and lim(bn)=b if a<b prove that an < bn.

can i say that if a/b < 1 than an<bn ?

 

[e(ho0n3]

What if b = 0?

 

[khdani]

if b=0 than a is negative and probably an < bn
so it still holds, but how do i prove that?

 

[e(ho0n3]

Consider the sequence an - bn. What can you say about this sequence?

 

[khdani]

can I say that
lim(an-bn) = (a-b) < 0
hence an < bn ?

 

[e(ho0n3]

You can, but it doesn't convince me that an < bn. Also, I think that what you're trying to prove is false.

 

[HallsofIvy]

Hello,
Please help me prove the following:
given lim(an)=a and lim(bn)=b if a<b prove that an < bn.

You can't prove it- it isn't true. What you can prove is that for n large enough, a_n< b_n. Use the definition of limit with \epsilon less that half the difference between a and b.

But you cannot say anything about a_n and b_n for smaller values of n.

can i say that if a/b < 1 than an<bn ?

No, that's not true either. Again, it is only true for "sufficiently large" n.

Got example, a_n= 1/n converges to 1 while b_n= 1/2n for n= 1 to 1000000, b_n= 2- 1/n for n> 1000000 converges to 2 (so a= 0< 2= b and a/b= 1/2< 1) but a_n< b_n only for n> 1000000. And you should be able to see how to make examples where that is true only for n> whatever number you want.