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mmmboh
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[PLAIN]http://img839.imageshack.us/img839/2670/hardlimitproblem.jpg
I am having trouble showing this. I figured I would divide both sides by a, and show that
lim n-> infinity [(a1/a)*b1+...+(an-1/a)*bn-1+(an/a)*bn]/(b1+...+bn) = 1.
I can explain in words why I believe this to be true; as n goes to infinity, there are infinitely many terms of ai/a = 1, because as i gets really big it approaches a. Since there are infinitely many of these, and the denominator is infinite, the ai/a terms that don't approach one are negligible. The problem is I can't figure out how to translate this into math. I was thinking of somehow doing an epsilon argument, but I'm not sure how I would do that.
Help please :)
Edit: I think I have managed to bound it below...I'm not sure its legitimate though..
Let an-a = e such that |an-a| is the smallest of the |ai-a| values. Then with some algebra, we find an/a = 1+e/a...replacing all of the ai/a's in my equation with 1+e/a, we get
(1+e/a)*(b1+..+bn)/(b1+..+bn)) = 1+e/a, taking the limit as n goes to infinity, this equals 1, because an converges to a. Since 1+e/a is the smallest of the ai/a values, 1 < lim n-> infinity [(a1/a)*b1+...+(an-1/a)*bn-1+(an/a)*bn]/(b1+...+bn)...
If this is right, then I still need to bound this above by 1, which I can't figure out how to do, since taking the largest of ai/a doesn't work.
I don't think what I did works though, because I think my argument gets messed up if some of the ai/a are negative.
I am having trouble showing this. I figured I would divide both sides by a, and show that
lim n-> infinity [(a1/a)*b1+...+(an-1/a)*bn-1+(an/a)*bn]/(b1+...+bn) = 1.
I can explain in words why I believe this to be true; as n goes to infinity, there are infinitely many terms of ai/a = 1, because as i gets really big it approaches a. Since there are infinitely many of these, and the denominator is infinite, the ai/a terms that don't approach one are negligible. The problem is I can't figure out how to translate this into math. I was thinking of somehow doing an epsilon argument, but I'm not sure how I would do that.
Help please :)
Edit: I think I have managed to bound it below...I'm not sure its legitimate though..
Let an-a = e such that |an-a| is the smallest of the |ai-a| values. Then with some algebra, we find an/a = 1+e/a...replacing all of the ai/a's in my equation with 1+e/a, we get
(1+e/a)*(b1+..+bn)/(b1+..+bn)) = 1+e/a, taking the limit as n goes to infinity, this equals 1, because an converges to a. Since 1+e/a is the smallest of the ai/a values, 1 < lim n-> infinity [(a1/a)*b1+...+(an-1/a)*bn-1+(an/a)*bn]/(b1+...+bn)...
If this is right, then I still need to bound this above by 1, which I can't figure out how to do, since taking the largest of ai/a doesn't work.
I don't think what I did works though, because I think my argument gets messed up if some of the ai/a are negative.
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