Question

高数求极限。。麻烦过程详细点

Answer 1

因为：1^2+2^2+……+n^2=n(n+1)(2n+1)/6
令sn=1/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)
因此，利用迫敛性
n(n+1)(2n+1) / 6(n^3+n) < sn < n(n+1)(2n+1) / 6(n^3+1)

lim n(n+1)(2n+1) / 6(n^3+n)
=lim n(n+1)(2n+1)/n^3 / 6(n^3+n)/n^3
=lim (n/n)(1+(1/n))(2+(1/n)) / 6(1+(1/n^2))
=1(1+0)(2+0) / 6(1+0)
=2/6
=1/3

lim n(n+1)(2n+1) / 6(n^3+1)
=lim n(n+1)(2n+1)/n^3 / 6(n^3+1)/n^3
=lim (n/n)(1+(1/n))(2+(1/n)) / 6(1+(1/n^3))
=1(1+0)(2+0) / 6(1+0)
=2/6
=1/3

因此，lim sn=1/3

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