3个回答
展开全部
∵(1+1/2+1/3+...+1/n)^(1/n)
≥(1+1/2+1/2^2+..+1/2^n)^(1/n)
=[(1-1/2^n)/(1-1/2)]^(1/n)
=(2-1/2^n)^(1/n)
∴lim(1+1/2+1/3+...+1/n)^(1/n)≥lim(2-1/2^n)^(1/n)=lim2^(1/n)=1
又∵(1+1/2+1/3+...+1/n)^(1/n)
≤(1+1+...+1)^(1/n)
=n^(1/n)
∴lim(1+1/2+1/3+...+1/n)^(1/n)≤limn^(1/n)=1
由夹逼准则
知lim(1+1/2+1/3+...+1/n)^(1/n)=1
希望能帮到你,祝学习进步O(∩_∩)O
≥(1+1/2+1/2^2+..+1/2^n)^(1/n)
=[(1-1/2^n)/(1-1/2)]^(1/n)
=(2-1/2^n)^(1/n)
∴lim(1+1/2+1/3+...+1/n)^(1/n)≥lim(2-1/2^n)^(1/n)=lim2^(1/n)=1
又∵(1+1/2+1/3+...+1/n)^(1/n)
≤(1+1+...+1)^(1/n)
=n^(1/n)
∴lim(1+1/2+1/3+...+1/n)^(1/n)≤limn^(1/n)=1
由夹逼准则
知lim(1+1/2+1/3+...+1/n)^(1/n)=1
希望能帮到你,祝学习进步O(∩_∩)O
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询