# Factorial Sequences

## Homework Statement

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)!

## 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

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

Try looking at An+1.

pbuk
Gold Member
Oops, double posted.

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

Try looking at An+1.
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...

haruspex
Homework Helper
Gold Member
(2n)! = 2ⁿ*(n!)*[ 1.3.5...(2n-1) ]
But how (2n)! expands out to that, seems unintuitive at this point.
2ⁿ*(n!) = 2.4.6....(2n-2).2n

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

Using An+1 = ((n+1)!)^2/(2n+2)! Which still leaves me lost..
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.

CompuChip
Homework Helper
An+1 = (blob)An. If (blob) is less than 1
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.

pbuk
Gold Member
I love that, I will start using (blob) instead of c or n or x whenever I need a mathematical symbol!
I wouldn't go that far, but in this case the more conventionally appropriate ## f(n) ## can easily cause confusion IME.

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.
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.

pbuk
Gold Member
Did I do this right?
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.

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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.
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!

pbuk