a(n+1)=an+1/n(n+2)a(n+1)-an=1/2[1/n-1/(n+2)]这样a2-a1=1/2(1-1/3)a3-a2=1/2(1/3-1/5).an-a(n-1)=1/2[1/(n-2)-1/n]相加得an-a1=1/2(1-1/n)an=a1+1/2(1-1/n)=2+1/2(1-1/n)=5/2-1/(2n)