- #1
NascentComp
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Homework Statement
[itex][/itex]
Given:
1. [itex]a_{1} < b_{1}[/itex]
2. [itex]a_{n} = \sqrt{a_{n-1}b_{n-1}}[/itex]
3. [itex]b_{n} = \frac{a_{n-1} + B_{n-1}}{2}[/itex]
4. The sequences [itex]a_{n}[/itex] and [itex]b_{n}[/itex] are convergent.
Prove: The sequences [itex]a_{n}[/itex] and [itex]b_{n}[/itex] have the same limit.
The Attempt at a Solution
Assume by contradiction that [itex]\lim_{n\to\infty}a_{n} = l_{a}[/itex] and [itex]\lim_{n\to\infty}b_{n} = l_{b}[/itex] and [itex]l_{a} \neq l_{b}[/itex].
Because [itex]b_{n} = \frac{a_{n-1} + B_{n-1}}{2}[/itex], we can write [itex]\lim_{n\to\infty}b_{n} = \lim_{n\to\infty}\frac{a_{n-1} + B_{n-1}}{2}[/itex].
From arithmetic of limits, we have [itex]\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})[/itex].
For all [itex]n \ge 2[/itex], the sequences [itex]a_{n}[/itex] and [itex]a_{n-1}[/itex] are identical. Thus their limit is identical, [itex]l_{a}[/itex].
The same is true of the sequences [itex]b_{n}[/itex] and [itex]b_{n-1}[/itex]. Thus their limit is identical too, [itex]l_{b}[/itex].
Thus, [itex]\frac{1}{2} (\lim_{n\to\infty}a_{n-1} + \lim_{n\to\infty}B_{n-1}) = \frac{1}{2} (l_{a} + l_{b})[/itex].
We assumed, by way of contradiction, that [itex]l_{a} \neq l_{b}[/itex], that is either [itex]l_{a} < l_{b}[/itex] or [itex]l_{a} > l_{b}[/itex].
Assuming [itex]l_{a} < l_{b}[/itex], we get [itex]\lim_{n\to\infty}b_{n} = \frac{1}{2} (l_{a} + l_{b}) < \frac{2l_{b}}{2} = l_{b}[/itex].
Assuming [itex]l_{a} > l_{b}[/itex], we get [itex]\lim_{n\to\infty}b_{n} = \frac{1}{2} (l_{a} + l_{b}) > \frac{2l_{b}}{2} = l_{b}[/itex].
Thus, if [itex]l_{a} \neq l_{b}[/itex], then [itex]\lim_{n\to\infty}b_{n} \neq l_{b}[/itex]. Contradiction.
Therefore, [itex]l_{a} = l_{b}[/itex].
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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!