Question

哪位数学达人知道∑1/n

Answer 1

an=Sn-S(n-1)=S(n-1)+2^n

Sn=2S(n-1)+2^n

两边除以2^n

Sn/2^n=S(n-1)/2^(n-1)+1

所以Sn/2^n等差,d=1

S1/2^1=3/2

Sn/2^n=n+1/2

Sn=(2n+1)*2^(n-1)

an=

3,n=1

Sn-S(n-1)=(n+3/2)*2^(n-1),n≥2