求 f(a)=lim(n→∞)∑(k=1,n)1／(n+k^a)（a＞0）

elim

题：求 \(\displaystyle\lim_{n\to\infty}\small\sum_{k=1}^n\frac{1}{n+k^a}\;\;(a>0)\)
解：\(0< a< 1{\small\implies}1\leftarrow{\large\frac{n}{n+n^a}}<{\displaystyle\small\sum_{k=1}^n\frac{1}{n+k^a}}<{\large\frac{n}{n+1}}\to 1\)
\(\qquad a>1\displaystyle{\small\implies\sum_{k=1}^n\frac{1}{n+k^a}\le\;}{\small\sum_{k=1}^n\frac{1}{2\sqrt{nk^a}}}\overset{\text{stolz}}{\sim}\,\frac{(n+1)^{-a/2}}{\small\sqrt{n+1}-\sqrt{n}}\to 0\)
\(\qquad\)最后,\(\,a=1,\;\;\displaystyle{\small\sum_{k=1}^n\frac{1}{n+k}=\int_0^1\frac{1}{1+x}dx=}\ln 2.\quad\small\square\)

【Limits,Series, and Fractional Part Integrals】(1.1)

elim

令\(\;f(a)=\displaystyle\lim_{n\to\infty}{\small\sum_{k=1}^n\frac{1}{n+k^a}},\;\;(a>0).\)
则主贴表明\(\;\;f(a)=\begin{cases}1,& 0< a< 1,\\ \ln 2,& a=1,\\ 0,& a>1.\end{cases}\)
\(f((0,\infty)) = \{0,\ln 2,\,1\},\;\;f\,\)的值域只有三个数,

luyuanhong

楼上 elim 的帖子很好！已收藏。

王守恩

Table[\(\displaystyle\lim_{n\to\infty}\sum_{k = 1}^n\frac{1}{n+k^a}\)], {a, -9, 9}
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, Log[2], 0, 0, 0, 0, 0, 0, 0, 0}