比较判别法的极限形式

笔记下载:比较判别法的极限形式

1,定理(比较判别法的极限形式):

设 \(\displaystyle\sum a_n\),\(\displaystyle\sum b_n\) 都是正项级数

若 \(\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n}=l\)

\(\displaystyle\begin{cases} (1)\ 0<l<+\infty &\Rightarrow\quad\sum a_n\text{与}\sum b_n\text{同敛散}\\ (2)\ l=0 &\Rightarrow\quad\text{若}\sum b_n\text{收敛,则}\sum a_n\text{收敛}\\ (3)\ l=+\infty &\Rightarrow\quad\text{若}\sum b_n\text{发散,则}\sum a_n\text{发散}\end{cases}\)

证明:

(1)存在\(\displaystyle N>0\),使得当 \(\displaystyle n>N\) 时,\(\displaystyle |\frac{a_n}{b_n}-l|<\epsilon\) 时任何 \(\displaystyle \epsilon>0\) 成立

取 \(\displaystyle \epsilon=\frac{l}{2}\)

\(\displaystyle\Rightarrow\quad|\frac{a_n}{b_n}-l|<\frac{l}{2}\)

\(\displaystyle\Rightarrow\quad \frac{1}{2}l<\frac{a_n}{b_n}<\frac{3}{2}l\)

\(\displaystyle\Rightarrow\quad\frac{1}{2}lb_n<a_n<\frac{3}{2}lb_n\)

由级数的性质,\(\displaystyle\quad\Rightarrow\quad\sum b_n\) 收敛,则 \(\displaystyle\frac{1}{2}lb_n\),\(\displaystyle\frac{3}{2}lb_n\) 收敛

\(\displaystyle\qquad\qquad\qquad\quad\ \Rightarrow\quad\sum b_n\) 发散,则 \(\displaystyle\frac{1}{2}lb_n\),\(\displaystyle\frac{3}{2}lb_n\) 发散

\(\displaystyle\Rightarrow\quad\) 由比较判别法,\(\displaystyle a_n\) 与 \(\displaystyle b_n\) 同敛散

(2)\(\displaystyle \sum b_n\) 收敛,\(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=0\)

\(\displaystyle\Rightarrow\quad a_n<b_n\quad (n>N)\)

\(\displaystyle\Rightarrow\quad\) 比较判别法, \(\displaystyle\sum a_n\) 收敛

(3)\(\displaystyle \sum b_n\) 发散,\(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=+\infty\)

\(\displaystyle\Rightarrow\quad a_n>b_n\quad (n>N)\)

\(\displaystyle\Rightarrow\quad\) 比较判别法, \(\displaystyle\sum a_n\) 发散

2,如何应用?找级数的主要部分

例1,判别级数 \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n^2-5n-1}\) 的敛散性。

解答:\(\displaystyle\lim_{n\to\infty} \frac{\frac{1}{2n^2-5n-1}}{\frac{1}{2n^2}}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{2n^2}{2n^2-5n-1}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{2}{2-\frac{5}{n}-\frac{1}{n^2}}\)

\(\displaystyle\quad\ =1\)

所以,\(\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n^2-5n-1}\) 收敛

例2,判别级数 \(\displaystyle\sum_{n=1}^{\infty}\frac{2n^2+3n}{\sqrt{5+n^7}}\) 的敛散性。

解答:级数主要部分,分子是 \(\displaystyle 2n^2\),分子是 \(\displaystyle \sqrt n^7=n^{\frac{7}{2}}\)

\(\displaystyle\Rightarrow\quad \frac{2n^2}{n^{\frac{7}{2}}}=2\cdot \frac{1}{n^\frac{3}{2}}\), 所以

\(\displaystyle\lim_{n\to\infty} \frac{\frac{2n^2+3n}{\sqrt{5+n^7}}}{\frac{1}{n^{\frac{3}{2}}}}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{2n^{\frac{7}{2}}+3n^{\frac{5}{2}}}{\sqrt{5+n^7}}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{2+3\cdot\frac{1}{n}}{\sqrt{\frac{5}{n^7}+1}}\)

\(\displaystyle\quad\ =2\)

因为 \(\displaystyle\sum \frac{1}{n^\frac{3}{4}}\) 收敛,

所以,由比较判别法的极限形式,得 \(\displaystyle\sum_{n=1}^{\infty}\frac{2n^2+3n}{\sqrt{5+n^7}}\) 收敛

例3,判别级数 \(\displaystyle\sum \frac{3n^3-2n^2}{n^4+n^2+1}\) 的敛散性。

解答:\(\displaystyle\lim_{n\to\infty} \frac{\frac{3n^3-2n^2}{n^4+n^2+1}}{\frac{1}{n}}\)

\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{3n^4-2n^3}{n^4+n^2+1}=3\)

因为 \(\displaystyle \sum \frac{1}{n}\) 发散,

所以,由比较判别法的极限形式,得 \(\displaystyle\sum \frac{3n^3-2n^2}{n^4+n^2+1}\) 发散。

3,适用范围:\(\displaystyle n\) 的代数式