Question

递推数列求极限

Answer 1

a(n+1)=[1- 1/(n+1)²]an=[n(n+2)/(n+1)²]an
a(n+1)/an=n(n+2)/(n+1)²
an/a(n-1)=(n-1)(n+1)/n²
a(n-1)/a(n-2)=(n-2)n/(n-1)²
…………
a2/a1=1×3/2²
连乘
an/a1=[1×3×2×4×...×(n-1)(n+1)]/(2²×3²×...×n²)
=[1×2×3×...×(n-1)×3×4×...×(n+1)]/(2²×3²×...×n²)
=[1×2×n×(n+1)×(3×4×...×(n-1)²)]/(2×3×...×n)²
=2n(n+1)/(2n)²
=(n+1)/2n
=(1+ 1/n)/2
an=a1·[(1+ 1/n)/2]=(1+ 1/n)/2
n趋向于∞，1/n趋向于0，(1+ 1/n)/2趋向于1/2

liman =1/2
n趋向于∞