# Determine Convergence/Divergence Of A Sequence

Bashyboy

## Homework Statement

I attached the problem and solution as one file.

## 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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Homework Helper
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$$a_n = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{(2n)^n}$$
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?

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?

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