数列an=1\(4 n^2 - 1)的前n项和为Sn,则Sn的极限为?
问题描述:
数列an=1\(4 n^2 - 1)的前n项和为Sn,则Sn的极限为?
0.5
答
an=1/(4n^2-1)=(1/2)[1/(2n-1)-1/(2n+1)]
∴sn=(1/2)[1/1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]=1/2-1/(4n+2)
∴lim sn = 1/2