Question

does the series converge or diverge?

n!/(n+2)!

Answer 1

An+1=(n+1)!/(n+3)!
An=n!/(n+2)!

Answer 2

when you cross multiply you get...

(n+1)!/(n+3)!/(n+2)!/n!

Answer 3

(n+1)! = (n+1)(n!)
(n+3)! = (n+3)(n+2)!

Answer 4

then to get them to cancel out you have to use n(n-1)! on (n+3)!

Answer 5

same thing we used in last question!

Answer 6

so then you end up getting (n+1)/(n+3)

Answer 7

which diverges

Answer 8

i don't think ratio test will work here,
you'll get limit = 1

Answer 9

how did u get 1?

Answer 10

lim n-> infinity (n+1)/(n+3) = ...

Answer 11

Long run behavior shows that it will be lim=1

Answer 12

oh crap...so then what test should i use?

Answer 13

n!/(n+2)! = 1/[(n+2)(n+1)]

Answer 14

so we are not using ratio test, right ?

Answer 15

Correct, no ratio test needed.

Answer 16

1/[(n+2)(n+1)] tends to 0, as n----> infinity.

Answer 17

as the series is basically 1/n^2

Answer 18

how do u know its 0?

Answer 19

1/n^2 is basically 1/very large number...as the denominator gets very large, 1/that very large number tends to 0.

Answer 20

i see it now! thank you!

Answer 21

welcome.