anonymous
  • anonymous
i need to find if the series converges or diverges using ratio test: sum of all ((n^2)(n+2)!)/(n!*3^(2n)) when n=1 to info
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
by the ratio test, this series converges. Are you having trouble setting up the ratio test?
anonymous
  • anonymous
yes definitely having trouble seeing how stuff cancels once i set up An+1 over !n
anonymous
  • anonymous
i meant An instead of !n

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anonymous
  • anonymous
I keep trying to type it in the equation solver, but it doesnt take
anonymous
  • anonymous
I mean the equation formater
anonymous
  • anonymous
you can write things like (n+1)! as n! (n+1)
anonymous
  • anonymous
and (n+3)! as (n+2)*(n!)
anonymous
  • anonymous
i was doing that but i must be doing something wrong because I keep getting left with (n+1)(n+3)n! over 9n\[^{2}\]
anonymous
  • anonymous
9n^2. sorry im a noob
anonymous
  • anonymous
no problem
anonymous
  • anonymous
let me see what I am left with...
anonymous
  • anonymous
You will be left with a 9n^2 and (n+2) in the denominator
anonymous
  • anonymous
but remember, you are taking the limit as n goes to infinity
anonymous
  • anonymous
what i dont get is doesnt the (n+3)! over (n+2)! cancel out as just (n+3) in the numerator
anonymous
  • anonymous
never mind. after some factorial expansion, everything cancels out and you are left with (n+1)(n+3) over 9n^2 and do the nth term test, then take the limit and it eventually equals to (1/9) which is <1. BOOM roasted

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