Question

Series from 1 to infinity. What test should I use? Ratio test didn't work out for me.
(n!)^2
---------
(2n)!

Answer 1

i think ratio test should work, maybe not

Answer 2

it worked for me.

Answer 3

getting 1/4
@Nanoman what have you tried ? where you stuck ?

Answer 4

basically you use \((n+1)! = (n+1)n! \\ (2(n+1))! =(2n+2)!= (2n+2)(2n+1)(2n)!\)

Answer 5

I got so far in the ratio test:
(2n+2)(2n+1)!/(n+1)^2

Answer 6

where did you (2n)! go ?

Answer 7

ohh....I see what you did there. You expanded that some more to get 2n

Answer 8

yes, so that i can cancel out (2n)!

Answer 9

The other mistake was that I copied it down wrong on paper.

Answer 10

oh wait, looking at my notes I accounted for that. Just a sec, I probably need to rewrite some stuff.

Answer 11

yeah, (n+1)^2 comes in numerator and (2n+2)(2n+1) comes in denominator....

Answer 12

First step is ((n+1)!)^2 /(n!)^2 * (2n)!/(2n+2)! right?

Answer 13

yes.

Answer 14

Okay, now I have (n+1)^2 / (2n+2)(2n+1)

Answer 15

now can you find limit to n -> infinity ?

Answer 16

i would suggest to expand both numerator and denominator and then multiply and divide by n^2.

Answer 17

okay

Answer 18

Wait, couldn't I just use l'hopital's rule now?

Answer 19

ofcourse you can use, if you are allowed to.
but we can solve that without L'Hopitals also, just so you know...

Answer 20

so, are you also getting 1/4 ?

Answer 21

Yeah, just did.

Answer 22

great, which means the series converges....

Answer 23

Yep, which is all the homework question asks for. Thanks!

Answer 24

welcome ^_^