数列极限a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (
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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)
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) (](/uploads/image/z/17452680-24-0.jpg?t=%E6%95%B0%E5%88%97%E6%9E%81%E9%99%90a1%3D%281%2F2%29%5E%281%2F2%29%2Can%3D%5B%281%2B+an-1%29%2F2%5D%5E%281%2F2%29%2C%E6%B1%82lim+%28a1a2a3.an%29+%28)
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
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
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