Proving the existence of the arithmetic-geometric mean

[NascentComp]

Homework Statement

Given:
1. a_{1} < b_{1}
2. a_{n} = \sqrt{a_{n-1}b_{n-1}}
3. b_{n} = \frac{a_{n-1} + B_{n-1}}{2}
4. The sequences a_{n} and b_{n} are convergent.

Prove: The sequences a_{n} and b_{n} have the same limit.

The Attempt at a Solution

Assume by contradiction that \lim_{n\to\infty}a_{n} = l_{a} and \lim_{n\to\infty}b_{n} = l_{b} and l_{a} \neq l_{b}.

Because b_{n} = \frac{a_{n-1} + B_{n-1}}{2}, we can write \lim_{n\to\infty}b_{n} = \lim_{n\to\infty}\frac{a_{n-1} + B_{n-1}}{2}.

From arithmetic of limits, we have \lim_{n\to\infty}\frac{a_{n-1} + B_{n-1}}{2} = \frac{1}{2} (\lim_{n\to\infty}a_{n-1} + \lim_{n\to\infty}B_{n-1}).

For all n \ge 2, the sequences a_{n} and a_{n-1} are identical. Thus their limit is identical, l_{a}.

The same is true of the sequences b_{n} and b_{n-1}. Thus their limit is identical too, l_{b}.

Thus, \frac{1}{2} (\lim_{n\to\infty}a_{n-1} + \lim_{n\to\infty}B_{n-1}) = \frac{1}{2} (l_{a} + l_{b}).

We assumed, by way of contradiction, that l_{a} \neq l_{b}, that is either l_{a} < l_{b} or l_{a} > l_{b}.

Assuming l_{a} < l_{b}, we get \lim_{n\to\infty}b_{n} = \frac{1}{2} (l_{a} + l_{b}) < \frac{2l_{b}}{2} = l_{b}.

Assuming l_{a} > l_{b}, we get \lim_{n\to\infty}b_{n} = \frac{1}{2} (l_{a} + l_{b}) > \frac{2l_{b}}{2} = l_{b}.

Thus, if l_{a} \neq l_{b}, then \lim_{n\to\infty}b_{n} \neq l_{b}. Contradiction.

Therefore, l_{a} = l_{b}.

---

Here's my question. I wasn't sure that I didn't make some logical leap along the way. Is my attempt correct? Did I manage to prove what I set out to prove? Thanks!

 

[gopher_p]

It looks fine. I do have a couple of remarks on style though. First, call the limits $a$ and $b$. Second, try to see if you can get a direct proof. Proofs by contradiction are perfectly fine, but when there is a very easy way to turn a proof by contradiction into a direct proof (and there is here), I've always been told that the direct proof is preferable. In this case, when you use my tips with the key part of your argument, you get $$b=\lim b_n=\lim\frac{1}{2}(a_n+b_n)=\frac{1}{2}(a+b)$$ and the desired result is a couple steps of elementary algebra away.

Again, though, your version works. It's just a little messier.