已知数an=4n-1,bn=(1/n)(a1+a2+.+an).设bn的前n项和为sn,则lim（1/s1 + 1/s2

a1+a2+...+an=-n+2n*(n+1)=2n^2+n
bn=2n+1
sn=n+n(n+1)=n^2+2n=n(n+2)
1/sn=1/[n(n+2)]=0.5*(1/n-1/(n+2))
极根=lim 0.5*(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-1)-1/(n+1)+1/n-1/(n+2))
=lim 0.5*(1+1/2-1/(n+1)-1/(n+2))
=0.5*3/2=3/4