Does the series converge?

series of (2n)!/(n!)^2 from n=1 to infinity

Question

Does the series converge?

series of (2n)!/(n!)^2 from n=1 to infinity

megannicole51 · Answer

i got 1 so does that mean it diverges?

megannicole51 · Answer

actually doesnt that mean i need to do another test bc getting 1 wont tell me....i did the ratio test!

megannicole51 · Answer

@agent0smith

agent0smith · Answer

1 means it's inconclusive, doesn't tell you anything. Ugh...this looks familiar but idk what test to do...

megannicole51 · Answer

ill look in my book and see what i can find!

agent0smith · Answer

Wolfram alpha says try the limit test, it doesn't converge

megannicole51 · Answer

what would my bn be for the limit test?

megannicole51 · Answer

would it be 2n/n?

agent0smith · Answer

Not sure :/

anonymous · Answer

i don't think the terms go to zero

megannicole51 · Answer

i figured it out:) thanks guys! i have one with a square root that I'm stuck on ill repost!! help if you can:)

anonymous · Answer

try it with a few numbers, small ones like $5$
you would get $\frac{10!}{(5!)^2}=\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}=\binom{10}{5}$

megannicole51 · Answer

yeah thats what i did!:) thanks!!!

anonymous · Answer

yw

anonymous · Answer

in fact, i am almost certain that 
$$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$$