正项级数的根值判别法

笔记下载:正项级数的根值判别法

根值判别法(柯西判别法)

1,定理:\(\displaystyle\sum a_n\) 正项级数,\(\displaystyle\lim_{n\to\infty} \sqrt[n]{a_n}=L\)

\(\displaystyle\left(\begin{array} {l}L<1\\L>1\\L=1\end{array}\right)\quad\Rightarrow\quad \sum a_n \left(\begin{array}{l}\text{收敛}\\ \text{发散}\\ \text{不知道}\end{array}\right)\)

2,\(\displaystyle \left(b_n\right)^n\),\(\displaystyle n^n\) 适用

例1,判断级数 \(\displaystyle\sum_{n=1}^{\infty} \left(\frac{2n+3}{3n+2}\right)^n\) 的敛散性。

解答:\(\displaystyle\lim_{n\to\infty} \sqrt[n]{a_n}\)

 \(\displaystyle\quad\ =\lim_{n\to\infty} \sqrt[n]{\left(\frac{2n+3}{3n+2}\right)^n}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{2n+3}{3n+2}\)

\(\displaystyle\quad\ =\frac{2}{3}<1\)

\(\displaystyle\Rightarrow\quad\) 级数收敛

例2,判断级数 \(\displaystyle\sum \frac{n^n}{5^{2n+3}}\) 的敛散性。

解答:\(\displaystyle \frac{n^n}{5^{2n+3}}=\frac{n^n}{5^{2n}\cdot 5^3}=\frac{1}{125}\cdot \frac{n^n}{(5^2)^n}=\frac{1}{125}\cdot \frac{n^n}{25^n}\)

\(\displaystyle\Rightarrow\quad \sum_{n=1}^{\infty} \frac{n^n}{5^{2n+3}}\)

 \(\displaystyle\quad\ =\frac{1}{125} \sum\left(\frac{n}{25}\right)^n\)

\(\displaystyle\Rightarrow\quad \lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty} \frac{n}{25}=\infty\)

\(\displaystyle\Rightarrow\quad\) 级数发散

注:可使用另一种方法:\(\displaystyle\lim_{n\to\infty} a_n=+\infty\ne 0, \quad\Rightarrow\quad\)级数发散。