数列{1/4n^2-1}的前n项和为Sn,则limSn=?
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数列{1/4n^2-1}的前n项和为Sn,则limSn=?
如果是(1/4)n^2-1,那么lim Sn = +∞,因为lim an = +∞ > 0
如果是1/(4 n^2)-1,那么lim Sn = -∞,因为lim an = -1 < 0
如果是1/(4 n^2 - 1),那么lim Sn = 1/2,因为
an = 1/(4 n^2 - 1)=1/(2n-1)(2n+1)=(1/2)[1/(2n-1) - 1/(2n+1)]
Sn = (1/2)[(1-1/3)+(1/3-1/5)+...+1/(2n-1) - 1/(2n+1)] = (1/2)[1-1/(2n+1)]
如果是1/(4 n^2)-1,那么lim Sn = -∞,因为lim an = -1 < 0
如果是1/(4 n^2 - 1),那么lim Sn = 1/2,因为
an = 1/(4 n^2 - 1)=1/(2n-1)(2n+1)=(1/2)[1/(2n-1) - 1/(2n+1)]
Sn = (1/2)[(1-1/3)+(1/3-1/5)+...+1/(2n-1) - 1/(2n+1)] = (1/2)[1-1/(2n+1)]
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