Series of (n!)^2/(2n)!, does it converge?

Learn how to use the ratio test to show if the series of (n!)^2/(2n)! converge or not.
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[@Nobodyman181]

Now, with !!(Doublefactorial)

[@shimotakanaki]

What's the value of this infinite serie ?

[@lagomoof]

WolframAlpha gives variations of (9 + 2π√3)/27. I have no idea how one would go about proving this.

[@General12th]

That's the sad reality of math. Sometimes, all you can do is prove a series _does_ converge, not _what_ it converges to.

[@xinpingdonohoe3978]

Okay, it's a real pain to prove, but I got that a similar Taylor series has a closed form.
Σ[n=1,∞] x^(2n)/nCr(2n,n)
=x/(4-x²) (x + (4arcsin(x/2)/√(4-x²)))
Evaluate this at x=1 to let 1/nCr(2n,n) simplify to n!²/(2n)!
It gives (9+2π√3)/27

[@Jan_Heckmann]

I have a Idea, and I hope you confirm or disprove it:
I rewrite an to: (n!*n!)/(n!*П from n+1 to 2n),
n! cancels out: n!/П from n+1 to 2n
Now I invert that, because I assume, that if 1/an dinverges to ∞ or -∞ an should convert to 0. All of it is based on this Assumption and on the Assumption, that an converges to 0:
lim n➝∞ (П from n+1 to 2n/n!) . Every Factor of the Numerator is larger than every Factor in the Denominator. More specificly, every Factor is exactly n bigger than a Factor below. I can rewrite the Limit as follows:
lim n➝∞((n/1)*((n+1)/2)*((n+2)/3)*....*((2n-2)/(n-2))*((2n-1)/(n-1))*(2n/n)) . This defenetly dinverges to Infinity. Just look at the first Factor: n/1=n. This Factor alone approaches Infinity. All other Factors are bigger than 0. If the latter step is wrong, maybe the Fact, that there are more and more Factors>2 when n increases might also play a Role, and the Rest is still correct. So 1/an diverges to Infinity, and so an Converges to 0. And WolframAlpha also said if converges to 0.
Any Critisism? 🙂

[@illumexhisoka6181]

I think you can somehow use the beta function
I think math 505 made a video about this

[@carlosgamer2818]

only 5 comments

[@holyshit922]

I am playing with finite sums
I have problem for you
Show that \sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor}(-1)^k*{2n-2k \choose n} \cdot {n \choose k} = \sum\limits_{k=0}^{n} {n \choose k}
without using Legendre polynomial
Who wants to record video about it