先看單項(xiàng)的分解式：
an=1/（1+2+3+...+n）=2/n(n+1) =2*[1/n-1/(n+1)]
根據(jù)單項(xiàng)的分解式來求和：
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)]
當(dāng)n＝1時(shí),
s1=2*1/(1+1) =1 成立
當(dāng)n＝k時(shí),假設(shè)成立
sk＝2k/(k+1)
當(dāng)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