作业帮数列极限a1=(1/2)^(1/2),an=[(1+ an-1)/2]^(1/2),求lim (a1a2a3.an) (n趋于正无穷大)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/27 03:45:41
![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)](/uploads/image/z/5603941-37-1.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+++%28n%E8%B6%8B%E4%BA%8E%E6%AD%A3%E6%97%A0%E7%A9%B7%E5%A4%A7%29)
xRAN@J3:EEHt$aY5ѐ0&h nzr)u*nrx=V'9r) FVg2M QVc;] PL ]W($ó.Nyխ} EẸȹ> =p/3SW+4O |WXGufh`[KZ "ipM366gȼ?F*06dCV9f%hF݀k
"@r%b y.f8,yϞ\= L;X3~QII$K?]LCw[K A0< t&MZK7J5Cak%
{O$
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