Wallis公式

Wallis公式：lim(n->+∞) (2^2n *(n!)^2/(2n!))^2 /(2n+1)=π/2

或者是lim(n->+∞) ((2n)!!/(2n-1)!!)^2/(2n+1)=π/2 

(!!是双阶乘，不超过这个整数且具有相同奇偶的正整数相乘，比如5!!=1*3*5,6!!=2*4*6)

下面是证明，最后也有手写的证明

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证明：I(n)=∫(0->π/2) (sinx)^n dx

u=(sinx)^(n-1)                      v'=sinx

u'=(n-1)(sinx)^(n-2) *cosx     v=-cosx

I(n)=-sinx^(n-1) *cosx |(0->π/2) +(n-1)∫(0->π/2) (sinx)^(n-2)(cosx)^2dx

=(n-1)∫(0->π/2) (sinx)^(n-2)(1-(sin^2))dx

=(n-1)∫(0->π/2) (sinx)^(n-2) dx-(n-1)∫(0->π/2) (sinx)^ndx

=(n-1)I(n-2) -(n-1)I(n)

∴ I(n)=(n-1)/n *I(n-2)

∴ I(n)/I(n-2)=(n-1)/n

∴ I(n-2)/I(n)=n/(n-1)

∴ I(2n-1)/I(2n+1)=(2n+1)/(2n)

I(0)=π/2

I(1)=1

∴I(2n)=(2n-1)/(2n) *I(2n-2)=(2n-1)/(2n) *(2n-3)/(2n-2) *I(2n-4)=(2n-1)!!/(2n)!! *π/2

∵当x∈[0,π/2] 0<=sinx<=1

∴(sinx)^(2n+1)<=(sinx)^(2n)<=sin^(2n-1)

∴I(2n+1)<=I(2n)<=I(2n-1)

∴(2n)!!/(2n+1)!! <=(2n-1)!!/(2n)!! *π/2 <=(2n-2)!!/(2n-1)!!

1<=(2n+1)!!/(2n)!! *(2n-1)!!/(2n)!! *π/2 <=(2n+1)!!/2n!! *(2n-2)!!/(2n-1)!!

1<=π/2/(((2n)!!)^2 /((2n-1)!!*(2n+1)!!)))<=(2n+1)/(2n)

lim(n->+∞) (2n+1)/(2n)=2/2=1

由夹逼准则

lim(n->+∞) π/2/(((2n)!!)^2 /((2n-1)!!*(2n+1)!!)))=1

∴lim(n->+∞) ((2n)!!)^2 /((2n-1)!!*(2n+1)!!))=π/2

lim(n->+∞) (2^n n!)^2/((2n-1)!!)^2 /(2n+1)=π/2

lim(n->+∞)(2^n n!)^2 *((2n)!!)^2 /((2n)!)^2 /(2n+1)=π/2

lim(n->+∞) ((2^n n!)^2/(2n)!)^2/(2n+1)=π/2

lim(n->+∞) 2^4n (n!/(2n)!)^2/(2n+1)=π/2

∴lim(n->+∞) 2^4n (n!/(2n)!)^2/(2n+1)=π/2

lim(n->+∞) (2n)!!/(2n-1)!!)^2/(2n+1)=π/2 

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手写证明