计算极限 lim(x→∞)ln∑(k=1,x)k^a／lnx（a＞-1）

elim

题：计算 \(\underset{x\to+\infty}{\lim}\dfrac{\ln\underset{1\le\ k\le\ x}{\large\sum}k^a}{\ln x},\;\;(a > -1).\)
解：\(\displaystyle\ln\sum_{1\le\ k\le\ x}k^a-(1+a)\ln \lfloor x\rfloor=\ln\big(\lfloor x\rfloor^{1+a}\sum_{k=1}^{\lfloor x\rfloor}\big({\small\frac{k}{\lfloor x\rfloor}}\big)^a{\small\frac{1}{\lfloor x\rfloor}}\big)\big)-(1+a)\ln\lfloor x\rfloor\)
\(\displaystyle\longrightarrow \ln \int_0^1 x^a dx=\ln\small\frac{1}{1+a}\)
\(\therefore\quad\underset{x\to+\infty}{\lim}\dfrac{\ln\underset{1\le\ k\le\ x}{\large\sum}k^a}{\ln x}=1+a.\)

王守恩

\(\displaystyle\bigg\lceil\frac{\ln\big(\sum_{k=1}^{x^a}k^x\big)}{\ln(x)}\bigg\rceil=a(x+1)-1\)

\(\displaystyle\big\lceil\ \ \ \ \big\rceil\ \ 表示小数部分作\  “1”\)