求级数的极限 lim(k→∞)∑(n=1,k／2)[√(2n)-√(2n-1)]／√k

awei

求级数极限\(\lim_{k\to\infty}\sum_{n=1}^{\frac{k}{2}}\frac{\sqrt{2n}-\sqrt{2n-1}}{\sqrt{k}}\)
挺有意思的问题

awei

Wolfram Cloud验证和我预想的结果一致的确是1/2，国产的Ai数学的确不好

王守恩

好题！！！

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/1}\frac{\sqrt{2n+0}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{1}{\sqrt{2}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/2}\frac{\sqrt{2n+0}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{1}{\sqrt{4}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/3}\frac{\sqrt{2n+0}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{1}{\sqrt{6}}\)
......

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/1}\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{2}{\sqrt{2}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/2}\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{2}{\sqrt{4}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/3}\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{2}{\sqrt{6}}\)
......

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/1}\frac{\sqrt{2n+2}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{3}{\sqrt{2}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/2}\frac{\sqrt{2n+2}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{3}{\sqrt{4}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/3}\frac{\sqrt{2n+2}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{3}{\sqrt{6}}\)
......

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt{2n+b}-\sqrt{2n-1}}{\sqrt{k}}\bigg]=\frac{b+1}{\sqrt{2a}}\)

awei

王守恩 发表于 2024-6-29 17:35
好题！！！

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/1}\frac{\sqrt{2n+0}-\sqrt{2n-1}}{\ ...

谢谢老师回帖，自己发现的一种奇怪积分方法，还正在琢磨，和黎曼积分，勒贝格积分不一样，

王守恩

往前走！

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt[2]{2*n+b}-\sqrt[2]{2*n}}{\sqrt[2]{k}}\bigg]=\frac{b}{\sqrt[2]{2^{1}*a}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt[3]{3*n+b}-\sqrt[3]{3*n}}{\sqrt[3]{k}}\bigg]=\frac{b}{\sqrt[3]{3^{2}*a}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt[4]{4*n+b}-\sqrt[4]{4*n}}{\sqrt[4]{k}}\bigg]=\frac{b}{\sqrt[4]{4^{3}*a}}\)

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt[5]{5*n+b}-\sqrt[5]{5*n}}{\sqrt[5]{k}}\bigg]=\frac{b}{\sqrt[5]{5^{4}*a}}\)
......

\(\displaystyle\lim_{k\to\infty}\bigg[\sum_{n=1}^{k/a}\frac{\sqrt[c]{c*n+b}-\sqrt[c]{c*n}}{\sqrt[c]{k}}\bigg]=\frac{b}{\sqrt[c]{c^{c-1}*a}}\)

luyuanhong

awei

谢谢陆老师回帖，这个问题plus版应该是，只要x不为0，极限恒等于1/2
\(x\ne0，\lim_{k\to\infty}\sum_{n=1}^{\frac{k}{2}}\frac{\left( 2n\right)^x-\left( 2n-1\right)^x}{k^x}=\frac{1}{2}\)
负数就发散，正数就收敛，没想到负数

luyuanhong

awei

luyuanhong 发表于 2024-7-1 08:49

谢谢陆老师解答，