ช่วยกันพิสูจน์หน่อยเร็ว

[prime no.]

พิสูจน์ว่า
1/2 * 3/4 * 5/6 * 7/8 *...* 1997/1998 < 1/44

ช่วยกันหน่อยครับ

(1/2)(3/4)(5/6)...(1997/1998) < (2/3)(4/5)(6/7)...(1998/1999)
[(1/2)(3/4)(5/6)...(1997/1998)]^2 < (1/2)(2/3)(3/4)...(1998/1999) = 1/1999
(1/2)(3/4)(5/6)...(1997/1998) < 1/sqrt(1999) < 1/44

แบบนี้พอได้ไหม
up = 1*3*5*7*9*....* 1997
down = 2*4*6*8*10*...*1998
up*down = 1998 ! (factorial)
down = 2*1 * 2*2 * 2*3 * 2*4 * 2*5 * 2*6 *.....* 2*999
= (2^999) * 999!
up/down = up*down/down^2
= 1998! / { (2^999)* 999! }^2

เสนออีกวิธี

Using
(a) 2/pi \int_0^{pi/2} cos^{2m}(x) dx = [1 3 5 ... (2m-1)]/[2 4 6 ... 2m]
(b) \int_0^1 (1-x^2)^m dx = [2 4 6 ... 2m]/[1 3 5 ... (2m+1)]

Then,
S_m = [1 3 5 ... (2m-1)]/[2 4 6 ... 2m]
....= 2/pi \int_0^{pi/2} cos^{2m}(x) dx
let v=sin(x) then
S_m = 2/pi \int_0^1 (1 - v^2)^{m-1/2} dv
....< 2/pi \int_0^1 (1 - v^2)^{m-1} dv
then
S_m < 2/pi [2 4 6 .. 2(m-1)]/[1 3 5 .. (2m-1)] 2m/2m
S_m < 2/pi 1/S_m 1/2m
S_m < sqrt( 2/pi/2m )

where m = 999 then
S_m < sqrt( 2/pi/1998 ) < 1/44