如何用夹逼定理证明lim (1/n²+1/(n+1)²+...+1/(2n)²)=0 n→∞
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如何用夹逼定理证明lim (1/n²+1/(n+1)²+...+1/(2n)²)=0 n→∞
如何用夹逼定理证明lim (1/n²+1/(n+1)²+...+1/(2n)²)=0 n→∞
如何用夹逼定理证明lim (1/n²+1/(n+1)²+...+1/(2n)²)=0 n→∞
令sn=1/n^2+1/(n+1)^2+……+1/(n+n)^2
则,
1/n^2
lim 1/n^2+1/n^2+……+1/n^2(n个相加)
=lim n/n^2
=lim 1/n
=0
因此,根据迫敛性
lim sn=0
有不懂欢迎追问
因为1/4n^2<=1/(n+k)^2<=1/n^2,k=0,1,……,n
上面n+1个不等式相加得
1/4n
根据夹逼定理limSn=0
1/(2n)²<1/(2n-1)²<……<1/(n+1)²<1/n² 推出limn/(2n)²=lim1/(4n)=0 n→∞