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Ratio Test for series convergence factoring problems

  • #1

Homework Statement


[tex]\Sigma[/tex]2nn!/(n+2)!


Homework Equations


I'm using the ratio test because there are factorials but I'm a little stuck on whether or not to factor out


The Attempt at a Solution


lim 2n+1(n+1)!/(n+3)!*(n+2)!/2n(n)! After I set it up here I'm not sure of how to factor out the factorials. would it end up looking like 2(n+1)(n+2)/(n+3) ?
 

Answers and Replies

  • #2
139
0
Close, it should simplify to [2(n+1)]/(n+3)
 
  • #3
Alright so the (n+2)! would factor out. I wasn't sure whether or not I would just factor out the factorial or since the (n+3)! is bigger it would factor out the (n+2) as well. Thanks for your help.
 
  • #4
139
0
(n+3)! factors out to (n+3)[(n+2)!]. You only have to factor it out that far since you want to cancel the (n+2)! in the numerator.

Generally, you can factor any factorial the way I'll show you, but we'll use (n+3)! as an example: factor out like this, until you can stop to cancel out something in either the numerator or denominator depending on where the larger factorial is:

ex.) (n+3)(n+2)(n+1)(n)(n-1)(n-2)...3(2)(1)
 
  • #5
33,632
5,287
You can make life easier on yourself by doing some simplification first, before applying any of the tests. n!/(n + 2)! = n!/[(n + 2)(n + 1)n!] = 1/[(n + 2)(n + 1)], so you series is identically equal to [itex]\sum 2^n/[(n + 2)(n + 1)][/itex],
 
  • #6
Aaaaahh Fantastic!! Thank you guys both so much you've been very helpful to a truly lost cause in Calculus.
 

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