Question

已知正项数列｛a｝满足a1=1/2,且a(n+1)=an/(1+an) 1,求正项数列｛a｝的通项公式 2,求和:a1/

Answer 1

n>=3,

a(n)=a(n-1)/(1+a(n-1))=(a(n-2)/(1+a(n-2)))/(1+a(n-2)/(1+a(n-2)))

=a(n-2)/(1+2a(n-2))=(a(n-3)/(1+a(n-3)))/(1+2a(n-3)/(1+a(n-3)))

=a(n-3)/(1+3a(n-3))=.

.

=a2/(1+(n-2)a2)

=a1/(1+(n-1)a1)

=1/(n+1)

n=1,2通项满足通项公式,a(n)=1/(n+1)

a(n)/n=1/(n(n+1))=1/n-1/(n+1)

Sn=1-1/2+1/2-1/3+.+1/n-1/(n+1)

=1-1/(n+1) 无数项相加时,Sn=1