Question

求级数∑(1+1/2+…+1/n)/(n+1)(n+2)的和

Answer 1

1. 化简通项un
un=(1+1/2+…+1/n)/[(n+1)*(n+2)]
=[1/(n+1)-1/(n+2)]*(1+1/2+…+1/n)

2. 求前n项部分和Sn
Sn=(1/2-1/3)*1+(1/3-1/4)*(1+1/2)+(1/4-1/5)*(1+1/2+1/3)+...
+[1/n-1/(n+1)]*[1+1/2+…+1/(n-1)]
+[1/(n+1)-1/(n+2)]*(1+1/2+…+1/n)
=1/2+1/3*1/2+1/4*1/3+...+1/(n+1)*1/n-1/(n+2)*(1+1/2+…+1/n)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/n-1/(n+1)]-1/(n+2)*(1+1/2+…+1/n)
=1-1/(n+1)-1/(n+2)*(1+1/2+…+1/n)

3. 求极限lim{n→∞}Sn
利用欧拉常数C=lim{n→∞}[(1+1/2+…+1/n)-lnn]，得
lim{n→∞}[(1+1/2+…+1/n)-lnn]/(n+2)
=lim{n→∞}[(1+1/2+…+1/n)-lnn]*lim{n→∞}1/(n+2)
=C*0
=0
故lim{n→∞}(1+1/2+…+1/n)/(n+2)=lim{n→∞}lnn/(n+2)=0 (洛必达法则)
因此，级数的和
S=lim{n→∞}Sn
=lim{n→∞}[1-1/(n+1)-1/(n+2)*(1+1/2+…+1/n)]
=1-lim{n→∞}1/(n+1)-lim{n→∞}(1+1/2+…+1/n)/(n+2)
=1

追问

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