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在无穷数列an的前n项和为s,若an=1/n^2+3n+2,则lim(sn+1/2)
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在无穷数列an的前n项和为s,若an=1/n^2+3n+2,则lim(sn+1/2)
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答案和解析
an = 1/(n^2+3n+2)
=1/[(n+1)(n+2)]
= 1/(n+1) -1/(n+2)
Sn =a1+a2+...+an
= 1/2 - 1/(n+2)
lim(n->∞) ( Sn + 1/2)
=lim(n->∞) (1 -1/(n+2))
=1
=1/[(n+1)(n+2)]
= 1/(n+1) -1/(n+2)
Sn =a1+a2+...+an
= 1/2 - 1/(n+2)
lim(n->∞) ( Sn + 1/2)
=lim(n->∞) (1 -1/(n+2))
=1
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