Verified answer

The answer is π. I have discovered a truly marvelous proof of this, which however the margin is not large enough to contain.

EDIT: Congrats to kb for answering first.

Here's my approach by Stirling's approximation

http://en.wikipedia.org/wiki/Stirling's_approximat...

lim(n>∞) 4/(4n+1) * Π(k=1 to n) (2k/(2k-1))²

Π(k=1 to n) (2k/(2k-1)) = (2 * 4 * 6 * ... 2n)/(1 * 3 * ... (2n-1))

. . . . . . . . . . . . . . . . = (2 * 4 * 6 * ... 2n)^2 /(2n)!

. . . . . . . . . . . . . . . . = (2^n * n!)^2 /(2n)!

⇒ Π(k=1 to n) (2k/(2k-1))² = 2^(4n) * (n!)^4 / ( (2n)! )^2

⇒ 4/(4n+1) * Π(k=1 to n) (2k/(2k-1))² = 4/(4n+1) * 2^(4n) * (n!)^4 / ( (2n)! )^2

Substitute n! = √(2πn) * (n/e)^n, (2n)! = √(4πn) * (2n/e)^(2n), and simplify the above, you're left with

4/(4n+1) * Π(k=1 to n) (2k/(2k-1))² = π * 4n/(4n + 1) → π

EDIT2:

(fixed some typo above)

On the other hand, we can use Catalan Number too.

http://en.wikipedia.org/wiki/Catalan_number#Proper...

C_n = 1/(n+1) * nCr(2n, n) <==> (n!)^2 /(2n)! = 1/((n+1) * C_n)

Catalan Number asymptotically grows to C_n ~ 4^n /(n^(3/2) * √π)

From above,

4/(4n+1) * Π(k=1 to n) (2k/(2k-1))² = 4/(4n+1) * 2^(4n) * (n!)^4 / ( (2n)! )^2

. . . . . . . . . . . . . . . . . . . . . . . . . = 4/(4n + 1) * 16^n * [ (n!)^2 / (2n)! ]^2

. . . . . . . . . . . . . . . . . . . . . . . . . = 4/(4n + 1) * 16^n * 1/(n + 1)^2 * (C_n)^(-2)

. . . . . . . . . . . . . . . . . . . . . . . . . = 4/(4n + 1) * 16^n * 1/(n + 1)^2 * (n^3 * π)/(16^n)

. . . . . . . . . . . . . . . . . . . . . . . . . = π * 4n^3 /((4n + 1)(n + 1)^2) → π

As starwhite reported, it extremely is a million, because of the fact the 2nd term is the series growth of exp(n). *EDIT* ok i assume it extremely is wise that the respond might desire to be below a million. e^n = ?{ok=0,?} n^ok / ok! which converges for all finite n. even nevertheless we've a partial summation, meaning ?{ok=0,n} n^ok / ok! < e^n And as n will develop, the discrepancy between the LHS and e^n additionally will develop. the situation is to tutor that the blunders is a million/2 * e^n. *EDIT2* listed right here are some techniques I even have so a techniques that i'm merely throwing obtainable. Scythian stumbled on a neat dating employing the ? purposes. yet understanding you, you have some trick up your sleeve that reduces the particular purposes to common ones. you may write ?(n+a million,n)) / ?(n+a million) = [?{n,?} t^n exp(-t) dt] / [?{0,?} t^n exp(-t) dt] the two numerator and denominator seem as though Laplace transforms, with s = a million (which does no longer impression the mixing by the way). you may make a substitution to get [?{0,?} (u+n)^n exp(-u-n) du] / [?{0,?} u^n exp(-u) du] I have no thought the place i flow with this, however the belief is to tutor that this cut back is a million/2 as n --> ?. *EDIT3* great locate Steiner. i will might desire to flow by that with an exceptional enamel comb. great stuff.

I get the feeling this was inspired by Vikram's question. I'm gonna say there's a Pi somewhere in the answer. I doubt I'll be able to say more than that :P

lim(n→∞) (4/(4n+1)) * Π(k=1 to n) (2k/(2k-1))²

= lim(n→∞) (4/(4n+1)) * (2*2*4*4*...*(2n)*(2n)) / (1*1*3*3*...*(2n-1)*(2n-1))

= lim(n→∞) (4(2n+1)/(4n+1)) * (2*2*4*4*...*(2n)*(2n)) / (1*3*3*...*(2n-1)*(2n-1)*(2n+1))

= lim(n→∞) ((8n+4)/(4n+1)) * Π(k=1 to n) (2n)(2n)/((2n-1)(2n+1))

= lim(n→∞) ((8n+4)/(4n+1)) * Π(k=1 to ∞) (2n)(2n)/((2n-1)(2n+1))

= (8/4) * (π/2), via Wallis' Product (for the infinite product)

= π.

Link:

http://en.wikipedia.org/wiki/Wallis_product

Fun problem!