Question

数列极限

已知数列{an}通项公式an=lg[1+2/(n^2+3n)]，(n=1，2，3，···)，Sn是数列的前n项和，求lim<n→+∞>Sn。

Answer 1

an=lg[1+2/(n^2+3n)]=lg[(n^2+3n+2)/(n^2+3n)]
an=lg[(n+1)(n+2)/n(n+3)]
Sn=lg([2*3*3*4*4*5*5*6*6*7*……n(n+1)*(n+1(n+2)]/[1*4*2*5*3*6*4*7*5*8*……*(n-1(n+2)*n(n+3)])
=lg([3*(n+1)]/(1*(n+3)]
lim<n→+∞>Sn=lg3

Answer 2

an=lg[(n^2+3n+2)/(n^2+3n)]=lg(n+1)+lg(n+2)-lgn-lg(n+3)
Sn=(lg2+lg3-lg1-lg4)+(lg3+lg4-lg2-lg5)+...+(lg(n+1)+lg(n+2)-lgn-lg(n+3))
=lg(n+1)+lg3-lg(n+3)
=lg[(3n+3)/(n+3)]
lim(n→+∞)Sn=lg3

Answer 3

数列极限的概念！