数列极限a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (n趋于正无穷大)

数列极限a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (n趋于正无穷大)
数列极限
a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (n趋于正无穷大)

数列极限a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (n趋于正无穷大)
考虑余弦半角公式：cos(θ/2)=√((1+cosθ)/2)
令a[n]=cos(θ/2^n),则a[1]=cos(θ/2)=√2/2
所以θ=π/2
a[n]=cos(π/2^(n+1))
a[1]*a[2]*..*a[n]
=cos(π/4)*cos(π/8)*...*cos(π/2^(n+1))
=2^n*sin(π/2^(n+1))*cos(π/2^(n+1)*cos(π/2^n)*..*cos(π/4)/(2^n*sin(π/2^(n+1)))（反复用正弦倍角公式）
=sin(π/2)/(2^n*sin(π/2^(n+1)))
=1/((π/2)*[sin(π/2^(n+1))/(π/2^(n+1))])
=2/π
上式用到了倍角公式sin2θ=2sinθcosθ,极限公式lim[x->0]sinx/x=1