# Determine Convergence/Divergence Of A Sequence

1. Sep 28, 2012

### Bashyboy

1. The problem statement, all variables and given/known data
I attached the problem and solution as one file.

2. Relevant equations

3. 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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Last edited: Sep 28, 2012
2. Sep 28, 2012

### jbunniii

$$a_n = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{(2n)^n}$$
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:
$$a_n = \frac{1}{2n} \cdot \frac{3}{2n} \cdot \frac{5}{2n} \cdots \frac{2n-1}{2n}$$
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. Sep 30, 2012

### Bashyboy

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. Sep 30, 2012

### jbunniii

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

Clearly $\frac{3}{2n} < 1, \frac{5}{2n} < 1, ..., \frac{2n-1}{2n} < 1$, so $a_n < \frac{1}{2n}$.

Last edited: Sep 30, 2012