a(n+1)=an+1/[n(n+1)]
=>a(n+1)-an=1/[n(n+1)]=1/n-1/(n+1)
當(dāng)n≥1時(shí),
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
...
an-a(n-1)=1/(n-1)-1/n
a(n+1)-an=1/n-1/(n+1)
將上述等式兩邊分別相加,則有
(a2-a1)+(a3-a2)+(a4-a3)+...+[an-a(n-1)]+[a(n+1)-an]=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/(n-1)-1/n]+[1/n-1/(n+1)]
=>a(n+1)-a1=1-1/(n+1)
=>a(n+1)=3/2-1/(n+1)
=>an=3/2-1/n