# Hard Limit Problem (1 Viewer)

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#### mmmboh

[PLAIN]http://img839.imageshack.us/img839/2670/hardlimitproblem.jpg [Broken]

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.

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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#### ystael

You're right that the argument you give doesn't work, because there's no reason why there should be a smallest value of $$|a_n - a|$$ -- in fact, there won't be one, unless it's zero, because $$a_n \to a$$ means $$|a_n - a| \to 0$$.

The pattern of argument you need to use here is a common one in analysis: divide your sequence into a part which is bad one way and a part which is bad a different way, then allocate your available error between the two parts.

Imagine a term $$c_N = \frac{a_1 b_1 + \cdots + a_N b_N}{b_1 + \cdots + b_N}$$ and some cutoff point $$n < N$$. In the beginning of the sequence, that is $$j < n$$, $$a_j$$ is not very close to $$a$$, but if you pile enough terms onto the end, you can balance that by the fact that $$a_1 b_1 + \cdots + a_n b_n$$ is small compared to $$a_1 b_1 + \cdots + a_N b_N$$. In the part where $$j > n$$, you have a lot of absolute magnitude, but on the other hand you can make it so $$|a_j - a|$$ is small.

#### mmmboh

In the beginning of the sequence, that is $$j < n$$, $$a_j$$ is not very close to $$a$$, but if you pile enough terms onto the end, you can balance that by the fact that $$a_1 b_1 + \cdots + a_n b_n$$ is small compared to $$a_1 b_1 + \cdots + a_N b_N$$. In the part where $$j > n$$, you have a lot of absolute magnitude, but on the other hand you can make it so $$|a_j - a|$$ is small.
Sorry you lost me there. I get that if j<n, then aj isn't close to a compared to if j>>n, and that to make the side where j>n to equal to the other side we need a lot more terms...but I don't totally get what you're saying, why these sides even have to be equal, or how I should continue.

Anything?

#### ystael

Sorry, I think my phrasing wasn't clear. You're not trying to make the two parts of the term equal. You're trying to distribute some error evenly between them. Think of wanting to find $$\left| \frac{a_1 b_1 + \cdots + a_N b_N}{b_1 + \cdots + b_N} - a \right| < \epsilon$$, and then splitting this up into two differences, each of which is bounded by $$\frac\epsilon2$$.

#### mmmboh

I think I sort of get what you're saying. I thought of doing something like this, but I'm not SURE that it works.

Suppose $$\left| \frac{a_1 b_1 + \cdots + a_N b_N}{b_1 + \cdots + b_N} - a \right| < \epsilon$$ then this equals $$\left| \frac{(a_1-a) b_1 + \cdots + (a_j-a) b_j}{b_1 + \cdots + b_N} \right + \left \frac{(a_k-a) b_k + \cdots + (a_N-a) b_N}{b_1 + \cdots + b_N} \right| < \epsilon$$
Choose j such that when n>j, |an-a|<$$\delta$$ (infinitely many terms),
then when n<j |an-a|>$$\delta$$ (finite number of terms). Plugging in $$\delta$$ into our equation the right side of the absolute value, we get $$\left| \frac{(a_1-a) b_1 + \cdots + (a_j-a) b_j}{b_1 + \cdots + b_N} \right + \left \frac{(a_k-a) b_k + \cdots + (a_N-a) b_N}{b_1 + \cdots + b_N} \right|< \left|\frac{(a_1-a) b_1 + \cdots + (a_j-a) b_j}{b_1 + \cdots + b_N} \right + \left \frac{(\delta) b_k + \cdots + (\delta) b_N}{b_1 + \cdots + b_N} \right| < \epsilon$$. Since on the left side of the absolute value there are finitely many terms, and the denominator goes to infinity, that side is zero, while the limit of the right side is
$$\delta$$. So just choose $$\delta < \epsilon$$.
Is this kind of close?

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