Question

Prove that this series diverges
Summation {n=0)^infinity (2n)!/(n!)^2

Do you use the root or ratio test?

Answer 1

ratio test works nicely

Answer 2

i am pretty sure it does not
does it help to know that
\[\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\]?

Answer 3

oooh "diverges" of course
the terms do not even go to zero

Answer 4

does it matter if the index is n=0 and not n=1?

Answer 5

makes no difference
each term is \(\binom{2n}{n}\) so the terms get larger, not smaller

Answer 6

I used the ratio test and got 0.. which im guessing is incorrect since i lost a point on my homework..

Answer 7

yeah it is incorrect

Answer 8

but the lim as n goest o inf. of 2n/n is indeterminable.. so you use lhopitals rule which gives you 2/1.. sooo?

Answer 9

because 2>1 you knowit diverges?

Answer 10

that's not quite what you get when you apply the ratio test, but even so the limit winds up being 2, yes

Answer 11

when you apply the ratio test, you end up getting 1.. at least thats why i get when using the simplified (2n/n) that satellite just gave..

Answer 12

\[{(2n+1)!\over(n+1)!} \cdot{n!\over (2n)!}=?\]

Answer 13

shouldn't it be (2n+2)!/(n+1)!?

Answer 14

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

Answer 15

(n+1)!/n! = n+1

Answer 16

2!/2! = 1 is what i get.. i factored out the 2! on both sides.. is that okay to do?

Answer 17

no, you can't factor like that
look at what I wrote above, you have to be able to cancel factors in factorial division

(n+1)!/n! = n+1

think about why

Answer 18

becasue (n+1)! = n!(n+1)

Answer 19

exactly, so\[{(2n+1)!\over(n+1)!} \cdot{n!\over (2n)!}=\frac{n!}{(n+1)!}\cdot{(2n+1)!\over(2n)!}=? \]

Answer 20

ok i think i got it..
the first term (n!/(n+1)!) = 1/(n+1)
the second term (2(n+1))!/2n! is also equal to 1 because you make (2(n+1))! = (2n)!(n+1)/(2n)! which gives you (n+1) in the numerator

so combining the two gives you 1?

Answer 21

you can't pull the 2 out like that\[{(2n+1)!\over(n+1)!} \cdot{n!\over (2n)!}=\frac{n!}{(n+1)!}\cdot{(2n+1)!\over(2n)!}=\frac1{(n+1)}\cdot\frac{(2n+1)}1=\frac{2n+1}{n+1}\]

Answer 22

if that confuses you, let 2n=x in the above

Answer 23

\[{(x+1)!\over(n+1)!} \cdot{n!\over (x)!}=\frac{n!}{(n+1)!}\cdot{(x+1)!\over(x)!}=\frac1{(n+1)}\cdot\frac{(x+1)}1=\frac{x+1}{n+1}\]

Answer 24

same rule, you just got confused because of the 2 :P

Answer 25

if a_n is n!/(2n!) then a_(n+1) should be (n+1)!/(2(n+1))!
this is probably where i'm getting most confused

Answer 26

you are confusing (2n)!, the successor of which is (2n+1)!, with 2(n!), the successor of which is 2[(n+1)!]

Answer 27

i thought if the parenthesis was around the entire term 2n then the factorial applies to both

Answer 28

yes, it applies to both:

imagine n=2, then
(2n)!=4!=24
and when n=3
(2n)!=6!=720

but if it were the other way, when n=2
2(n!)=2(2!)=4
and when n=3
2(3!)=2(6)=12

see the difference?

Answer 29

so when you are working out the root test.. and you have to add 1 to n.. you're really adding 1 to 2n since 2 is just a constant.. i think i understand better

Answer 30

ratio test**

Answer 31

thats why you told me to make 2n=x .. yay i understand it. ok great!

Answer 32

Oh now I understand your point...actually, when you put it that way I realize I should be more careful; you can't just replace 2n with x for the reason you stated :/
however what satellite said is true, the factor of 2 will carry either way, though I owe it to you to do the root test properly. let me try that, sorry...

Answer 33

yeah you can just distribute the 2, it's no big change...

Answer 34

haha, ok ill wai until you're done to think more about this :D

Answer 35

\[{(2(n+1))!\over(n+1)!} \cdot{n!\over (2n)!}={(2n+2)!\over(n+1)!} \cdot{n!\over (2n)!}=\\\frac1{(n+1)}\cdot\frac{(2n+2)(2n+1)}1=\frac{(2n+2)(2n+1)}{n+1}\]sorry, typos :P

Answer 36

does that make sense?

Answer 37

give me one second, im going tot ry to workit out the way you did

Answer 38

where does the (2n+2)(2n+1) come from?

Answer 39

oh nvm.. thats whats left after you divide (2n+2)!/2n! correct?

Answer 40

right

Answer 41

so the limit is infinity and it diverges.. that only took an hour haha thanks!!

Answer 42

sorry I misunderstood your doubt earlier
welcome!