先看单项的分解式:
an=1/(1+2+3+...+n)=2/n(n+1)=2*[1/n-1/(n+1)]
根据单项的分解式来求和:
Sn=a1+a2+...+an
=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n)
=2/1*2+2/2*3+2/3*4...+2/n(n+1)
=2*[1/1*2+1/2*3+1/3*4...+1/n(n+1)]
=2*[(1/1-1/2)+(1/2-1/3)+...1/n-1/(n+1)]
=2*[1-1/(n+1)]
当n=1时,
s1=2*1/(1+1)=1成立
当n=k时,假设成立
sk=2k/(k+1)
当n=k+1是
s(k+1)=sk+1/(1+2+3+…+(k+1))
=2k/(k+1)+2/((k+1)(k+2))
=2(k/(k+1)+1/(k+1)-1/(k+2))
=2(1-1/(k+2))
=2(k+1)/((k+1)+1)成立