正项级数的比较判别法

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比较判别法:\(\displaystyle \sum a_n, a_n\ge0\),正项级数

1,定理1:单调有界数列必有极限

2,定理2:正项级数收敛的充分必要条件是 \(\displaystyle s_n\) 有上界

证明:\(\displaystyle s_n\)单调递增,有下界 \(\displaystyle a_1\)

所以,若 \(\displaystyle s_n\) 有上界\(\displaystyle\quad\Rightarrow\quad s_n\) 单调递增有界 \(\displaystyle\quad\Rightarrow\quad s_n\) 有极限

3,定理3(比较判别法):

若 \(\displaystyle a_n\) 和 \(\displaystyle b_n\) 均 \(\displaystyle \ge 0\),且 \(\displaystyle a_n<b_n\quad\left(n=1,\ 2,\ \cdots\right)\)

(1)大收则小收:若 \(\displaystyle\sum b_n\) 收敛\(\displaystyle\quad\Rightarrow\quad\sum a_n\) 收敛

(2)小散则大散:若 \(\displaystyle\sum a_n\) 发散\(\displaystyle\quad\Rightarrow\quad\sum b_n\) 发散

证明:

(1)\(\displaystyle t_n=\sum_{k=1}^n b_k\),\(\displaystyle t=\sum_{n=1}^\infty b_n\),\(\displaystyle s_n=\sum_{k=1}^n a_n\)

\(\displaystyle\Rightarrow\quad t_n\le t\),而 \(\displaystyle s_n\le t_n\quad\Rightarrow\quad s_n \le t\quad\Rightarrow\quad s_n\) 有上界\(\displaystyle\quad\Rightarrow\quad\sum a_n\) 收敛

(2)\(\displaystyle\sum a_n\) 发散\(\displaystyle\quad\Rightarrow\quad s_n\rightarrow+\infty\quad\Rightarrow\quad t_n\rightarrow+\infty\) (因为:\(\displaystyle t_n\ge s_n\))

4,常用的比较级数

(1)几何级数

\(\displaystyle\sum_{n=0}^{\infty} ar^n\):\(\displaystyle\quad \begin{cases}\text{发散,}\quad &|r|\ge 1\\ \text{收敛,}\quad &|r|< 1\end{cases}\)

(2)\(\displaystyle p-\)级数

\(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}\):\(\displaystyle\quad \begin{cases} p>1\quad &\text{收敛}\\ p\le 1\quad &\text{发散}\end{cases}\)

5,方法:将级数放缩,变成\(\displaystyle p-\)级数或者几何级数

(要证明发散,缩小发散级数;要证明收敛,放大成一个收敛级数)

例1,判定级数 \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n^2+5n+1}\) 是否收敛?

解答:\(\displaystyle \frac{1}{2n^2+5n+1}<\frac{1}{2n^2}\)

因为 \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n^2}=\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}\) 收敛

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

例2,判定级数 \(\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n}\) 是否收敛?

解答:\(\displaystyle \frac{\ln n}{n}>\frac{1}{n}\qquad n\ge 3\)

因为 \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}\) 发散,所以\(\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n}\) 发散

例3,判定级数\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n+1}\) 是否收敛?

解答:\(\displaystyle \frac{1}{2^n+1} < \frac{1}{2^n}=\left(\frac{1}{2}\right)^n\)

因为 \(\displaystyle \sum\left(\frac{1}{2}\right)^n\) 收敛,所以 \(\displaystyle \sum\frac{1}{2^n+1}\) 收敛

注:预估收敛还是发散,要观察主要部分。