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Factorial Sequences

  • Thread starter Jimbo57
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  • #1
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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)!

Homework Equations





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
 

Answers and Replies

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

Try looking at An+1.
 
  • #3
pbuk
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Oops, double posted.
 
  • #4
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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...
 
  • #5
haruspex
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(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
 
  • #6
pbuk
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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.
 
  • #7
CompuChip
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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.
 
  • #8
pbuk
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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.
 
  • #9
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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.
 
  • #10
pbuk
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Did I do this right?
Not really I'm afraid. It should go like this...[tex]
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[/tex]

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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  • #11
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Not really I'm afraid. It should go like this...[tex]
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[/tex]

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!
 
  • #12
pbuk
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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!
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
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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!
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!
 

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