Determine Convergence/Divergence Of A Sequence

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  • #1
Bashyboy
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Homework Statement


I attached the problem and solution as one file.


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The Attempt at a Solution


I just can't quite follow the solution. Could someone perhaps explain what the author is doing?
 

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  • #2
jbunniii
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[tex]a_n = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{(2n)^n}[/tex]
I assume you have no problem with the first step, which is simply pairing each factor in the numerator with a factor in the denominator:
[tex]a_n = \frac{1}{2n} \cdot \frac{3}{2n} \cdot \frac{5}{2n} \cdots \frac{2n-1}{2n}[/tex]
Now all you have to do is recognize that each of these fractions is less than 1. Therefore the product of the fractions is smaller than any of the individual fractions, and in particular the product of the fractions is smaller than 1/(2n). Does that clarify the next step?
 
  • #3
Bashyboy
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Oh, I think I see. So, essentially, we are breaking the fraction up into products, and then, in a way, evaluating the limit at that one term, because we know how the rest of the terms behave?
 
  • #4
jbunniii
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Oh, I think I see. So, essentially, we are breaking the fraction up into products, and then, in a way, evaluating the limit at that one term, because we know how the rest of the terms behave?

We are breaking the fraction into products so we can more easily obtain an upper bound for it:

Clearly [itex]\frac{3}{2n} < 1, \frac{5}{2n} < 1, ..., \frac{2n-1}{2n} < 1[/itex], so [itex]a_n < \frac{1}{2n}[/itex].
 
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