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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    by the ratio test, this series converges. Are you having trouble setting up the ratio test?

  2. anonymous
    • 5 years ago
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    yes definitely having trouble seeing how stuff cancels once i set up An+1 over !n

  3. anonymous
    • 5 years ago
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    i meant An instead of !n

  4. anonymous
    • 5 years ago
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    I keep trying to type it in the equation solver, but it doesnt take

  5. anonymous
    • 5 years ago
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    I mean the equation formater

  6. anonymous
    • 5 years ago
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    you can write things like (n+1)! as n! (n+1)

  7. anonymous
    • 5 years ago
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    and (n+3)! as (n+2)*(n!)

  8. anonymous
    • 5 years ago
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    i was doing that but i must be doing something wrong because I keep getting left with (n+1)(n+3)n! over 9n\[^{2}\]

  9. anonymous
    • 5 years ago
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    9n^2. sorry im a noob

  10. anonymous
    • 5 years ago
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    no problem

  11. anonymous
    • 5 years ago
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    let me see what I am left with...

  12. anonymous
    • 5 years ago
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    You will be left with a 9n^2 and (n+2) in the denominator

  13. anonymous
    • 5 years ago
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    but remember, you are taking the limit as n goes to infinity

  14. anonymous
    • 5 years ago
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    what i dont get is doesnt the (n+3)! over (n+2)! cancel out as just (n+3) in the numerator

  15. anonymous
    • 5 years ago
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    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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