高等数学吧 关注：472,275贴子：3,884,139
，他没这水平。
别烦我 试了一下：
Stolz一下变成lim_{n->∞} 1+ ((n+1)^(1/2)-n^(1/2))+...+((n+1)^(1/n)-n^(1/n))+(n+1)^(1/(n+1))
1+(n+1)^(1/(n+1))是趋于2的，所以猜测S(n)=((n+1)^(1/2)-n^(1/2))+...+((n+1)^(1/n)-n^(1/n))趋于0.
对于每一项(n+1)^(1/k)-n^(1/k)有：
0<(n+1)^(1/k)-n^(1/k)=1/k (n+ξ)^(1/k-1)<1/k n^(1/k - 1)< 1/k n^(1/2 - 1) = 1/k n^(-1/2)
于是：
0<S(n)<1/√n Σ{k=2..n} 1/k < 1/√n ∫{1..n} 1/x dx = ln n/√n
再由lim_{n->∞} ln n/√n =0 得证
Stolz一下变成lim_{n->∞} 1+ ((n+1)^(1/2)-n^(1/2))+...+((n+1)^(1/n)-n^(1/n))+(n+1)^(1/(n+1))
1+(n+1)^(1/(n+1))是趋于2的，所以猜测S(n)=((n+1)^(1/2)-n^(1/2))+...+((n+1)^(1/n)-n^(1/n))趋于0.
对于每一项(n+1)^(1/k)-n^(1/k)有：
0<(n+1)^(1/k)-n^(1/k)=1/k (n+ξ)^(1/k-1)<1/k n^(1/k - 1)< 1/k n^(1/2 - 1) = 1/k n^(-1/2)
于是：
0<S(n)<1/√n Σ{k=2..n} 1/k < 1/√n ∫{1..n} 1/x dx = ln n/√n
再由lim_{n->∞} ln n/√n =0 得证
