Question

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

Answer 1

by the ratio test, this series converges. Are you having trouble setting up the ratio test?

Answer 2

yes definitely having trouble seeing how stuff cancels once i set up An+1 over !n

Answer 3

i meant An instead of !n

Answer 4

I keep trying to type it in the equation solver, but it doesnt take

Answer 5

I mean the equation formater

Answer 6

you can write things like (n+1)! as n! (n+1)

Answer 7

and (n+3)! as (n+2)*(n!)

Answer 8

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

Answer 9

9n^2. sorry im a noob

Answer 10

no problem

Answer 11

let me see what I am left with...

Answer 12

You will be left with a 9n^2 and (n+2) in the denominator

Answer 13

but remember, you are taking the limit as n goes to infinity

Answer 14

what i dont get is doesnt the (n+3)! over (n+2)! cancel out as just (n+3) in the numerator

Answer 15

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