Proof that ratio of the products of odd and even numbers converges

In summary: Monotone sequences converge to a limit if and only if the sequence is bounded and the limit is also a monotone sequence.
  • #1
Rasmus
8
0

Homework Statement


Show that

[itex]a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}[/itex]

converges when n [itex]\rightarrow[/itex]∞

and n is a natural number

Homework Equations


None that I can think of.

The Attempt at a Solution



This was from an exam and I was pretty much stumped.
 
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  • #2
What can you say about [itex]a_{n+1}[/itex] vs. [itex]a_{n}[/itex]? Which one is larger?
 
  • #3
Thanks for the hint!

As far as I understand the problem we start by 1/2 and multiply another rational for each increment increase in n. Meaning that [itex]a_{n+1}[/itex] = [itex]a_{n}[/itex] * (2n)/(2n +1). Since n is a natural number that menas the latter factor is less than one, therefore [itex]a_{n}[/itex] must be greater than [itex]a_{n+1}[/itex]

I'm pretty sure I can work with this, but right now I need to be away from the computer for an hour or so.

So far I'm thinking maybe proof by induction to show that each n must be smaller than the next and to use the fact that they're natural numbers to put a lower limit of zero on the product. And then squeeze [itex]a_{n}[/itex] to zero.

Thanks a lot
 
  • #4
Rasmus said:

Homework Statement


Show that

[itex]a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n}[/itex]

converges when n [itex]\rightarrow[/itex]∞

and n is a natural number

Homework Equations


None that I can think of.


The Attempt at a Solution



This was from an exam and I was pretty much stumped.

The denominator is [itex] 2 \cdot 4 \cdot 6 \cdots 2n = 2^n n! [/itex] and the numerator is
[tex] 1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n-1)!}{2 \cdot 4 \cdots (2n-2)}
= \frac{(2n-1)!}{2^{n-1} (n-1)!}.[/tex] Thus,
[tex] a_n = \frac{(2n-1)!}{2^{n-1} (n-1)!} \cdot \frac{1}{2^n n!}.[/tex] Now use Stirling's formula.

RGV
 
  • #5
Rasmus said:
So far I'm thinking maybe proof by induction to show that each n must be smaller than the next and to use the fact that they're natural numbers to put a lower limit of zero on the product. And then squeeze [itex]a_{n}[/itex] to zero.
You don't need to "squeeze [itex]a_n[/itex] to zero." All the problem asks you to do is to show that the sequence converges. This is easier than finding what value it converges to.

You established that [itex]a_n[/itex] is a monotonically decreasing sequence. What do you know about convergence of monotone sequences?
 
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1. What is the proof that the ratio of the products of odd and even numbers converges?

The proof is based on the concept of infinite series and involves using the properties of limits and convergence. It states that as the number of terms in the series increases, the ratio of the products of odd and even numbers approaches a constant value.

2. How does the proof work?

The proof involves taking the infinite series of odd and even numbers and manipulating it using algebraic techniques and the properties of limits. By taking the limit of the series as the number of terms approaches infinity, we can determine the value of the constant to which the ratio converges.

3. What is the significance of this proof?

This proof has important implications in various fields of mathematics and science. It helps us understand the behavior of infinite series and provides a way to calculate the value of certain infinite series. It also has applications in areas such as probability, statistics, and physics.

4. Can you provide an example of the proof in action?

Sure, let's consider the series 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... In this series, the ratio of the products of odd and even numbers would be (1*3*5*7*9*...) / (2*4*6*8*10*...). As we take more and more terms, the ratio will approach a constant value, in this case, 3/2.

5. Are there any limitations to this proof?

Like any mathematical proof, there are certain assumptions and limitations to consider. This proof assumes that the series is infinite and that the terms follow a specific pattern. Additionally, it may not be applicable to all types of infinite series, so it is important to carefully consider the context and conditions of the problem at hand.

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