# Factorial Sequences

1. Aug 25, 2013

### Jimbo57

1. The problem statement, all variables and given/known data

These trickly little buggers always seem to confuse me. I need to find out whether or not the sequence is increasing, decreasing or neither.

An=(n!)2/(2n)!

2. Relevant equations

3. The attempt at a solution

I'm pretty sure that it's a decreasing sequences but when I expand and reduce, it's not so obvious:

(n!)2/(2n)! = (n!)2/2n*(n!)

Which leaves me with,

= (n!)/2n since one of the factorials in the numerator cancels out with the factorial in the denominator. Did I expand this wrong or something?

Jim

2. Aug 25, 2013

### MrAnchovy

Yes you expanded it wrong. (2n)! ≠ 2n*(n!).

Try looking at An+1.

3. Aug 25, 2013

### MrAnchovy

Oops, double posted.

4. Aug 25, 2013

### Jimbo57

Hey Mr. Anchovy,

I stumbled upon what it expands to,

(2n)! = 2ⁿ*(n!)*[ 1.3.5...(2n-1) ]

And see hows it's decreasing.

But how (2n)! expands out to that, seems unintuitive at this point.

Using An+1 = ((n+1)!)^2/(2n+2)! Which still leaves me lost. I feel like I just don't understand factorials that well...

5. Aug 25, 2013

### haruspex

2ⁿ*(n!) = 2.4.6....(2n-2).2n

6. Aug 26, 2013

### MrAnchovy

Write out the first few and the last few terms of each series and match them up. But that is not going to help answer the question.

Well (n+1)! = (n+1)n! and (2n+2)! = (2n+2)(2n+1)(2n)!, so write that expression with these substitutions and see if you can group terms together to get An+1 = (blob)An. If (blob) is less than 1 then the series is decreasing.

7. Aug 26, 2013

### CompuChip

I love that, I will start using (blob) instead of c or n or x whenever I need a mathematical symbol!
I'll probably even take out my old-fashioned ink pen just to be able to write (blob) by shaking it up and down.

8. Aug 26, 2013

### MrAnchovy

I wouldn't go that far, but in this case the more conventionally appropriate $f(n)$ can easily cause confusion IME.

9. Aug 26, 2013

### Jimbo57

Okay. I think I'm starting to see something, but am unsure if what I did was correct.

((n+1)n!)2/(2n+2)(2n+1)(2n) = (n!)2/(2n!)

((n+1)n!)2/(n!)2 = (2n+2)(2n+1)(2n!)/(2n!)

(n+1)2 = (2n+2)(2n+1)

n2+2n+1/4n2+5n+2 = 1

I see that the denominator is greater from here and this will be less than 1. Did I do this right? Sorry if it's hard to read.

10. Aug 26, 2013

### MrAnchovy

Not really I'm afraid. It should go like this...$$A_n = \frac{n!^2}{(2n)!} \\ A_{n+1} = \frac{(n+1)!^2}{(2(n+1))!} \\ = \frac{((n+1)n!)^2}{(2n+2)(2n+1)(2n)!} \\ = \frac{(n+1)^2 n!^2}{(2n+2)(2n+1)(2n)!} \\ = \frac{(n+1)^2}{(2n+2)(2n+1)} . \frac{n!^2}{(2n)!} \\ A_{n+1} = \frac{(n+1)^2}{(2n+2)(2n+1)}A_n$$

Now you can multiply out the brackets, and compare the numerator and denominator.

I have taken the time to set this out because this is in general how you need to approach this kind of problem - write down an expression for $A_{n+1}$, expand it a bit, regroup terms to isolate $A_n$ and see what you have got.

11. Aug 26, 2013

### Jimbo57

Okay, that's some great knowledge that I can take forward. I'll just keep practicing on different problems and see what I get using what you showed me. Thanks!

12. Aug 26, 2013

### MrAnchovy

No problem. I can't stress enough how important this technique is - not just in its own right, but it forms the basis both of proof by induction and of analysis (the study of limits) which together probably make up about a quarter of the maths you will ever do!

13. Aug 26, 2013

### Jimbo57

I'll be studying engineering and am thinking I won't do too much in terms of analysis, unless of course I want to.

Another method in my text, which I somehow missed, was to take the ratio of A_n/A_(n+1) then simplify and take the limit as n approaches infinite. If the limit is infinite then it's eventually decreasing. Which seems like it's along the lines of what you had me do. Both have worked great so far!