交错级数的判别法

笔记下载:交错级数的判别法

1,交错级数:

\(\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n=a_1-a_2+a_3-a_4+\cdots,\ a_n\ge 0\)

2,定理(莱布尼茨判别法):

交错级数 \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n,\ a_n\ge 0\),

满足:\(\left.\begin{array}{l}\displaystyle \lim_{n\to\infty} a_n=0\\ \displaystyle a_{n+1}\le a_n\end{array}\right\}\quad\Rightarrow\quad \displaystyle\sum_{n=1}^{\infty}(-1)^na_n\) 收敛

证明:

\(\displaystyle s_2=a_1-a_2\ge o\)

\(\displaystyle\ s_4=a_1-a_2+a_3-a_4\ge 0,\ \ s_4\ge s_2\quad\)(因: \(\displaystyle a_3-a_4\ge0\))

\(\displaystyle \cdots\)

\(\displaystyle\ s_{2n}\ge s_{2(n-1)}\)

\(\displaystyle\Rightarrow\quad s_{2n}\) 单调增加,\(\displaystyle\ s_{2n}\ge 0\)

又因:\(\displaystyle s_{2n}=a_1-a_2+a_3-a_4+\cdots+a_{2n-1}-a_{2n}\)

 \(\displaystyle\qquad\qquad=a_1-(a_2-a_3)-(a_4-a_5)-\cdots-\left(a_{2n-2}-a_{2n-1}\right)-\left(a_{2n}\right)\)

 \(\displaystyle\qquad\qquad\le a_1\)

\(\displaystyle\Rightarrow\quad s_{2n}\) 有上界\(\displaystyle a_1\)

\(\displaystyle\Rightarrow\quad \{s_{2n}\}\) 有极限 \(\displaystyle s\)(单调有界数列)

\(\displaystyle\quad\lim_{n\to\infty} s_{2n+1}\)

\(\displaystyle=\lim_{n\to\infty} \left(s_{2n}+(-1)^{2n}a_{2n+1}\right)\)

\(\displaystyle=\lim_{n\to\infty} s_{2n}+\lim_{n\to\infty} a_{2n+1}\)

\(\displaystyle=s\)

所以,\(\displaystyle s_{2n}\) 和 \(\displaystyle s_{2n+1}\) 都有同一极限\(\displaystyle s\),得 \(\displaystyle s_n\) 有极限 \(\displaystyle s\)

例1,证明 \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) 收敛

证明:\(\left.\begin{array}{l}\displaystyle \lim_{n\to\infty} \frac{1}{n}=0\\ \displaystyle \frac{1}{n+1}<\frac{1}{n}\end{array}\right\}\quad\Rightarrow\quad\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) 收敛

例2,证明 \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt n}\) 收敛

例3,判断级数 \(\displaystyle\sum (-1)^{n+1}\frac{n^2}{1+n^3}\) 是否收敛

解答:

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

如何证明 \(a_{n+1}\le a_n?\quad\) 令 \(\displaystyle f(x)=\frac{x^2}{1+x^3}\)

\(\displaystyle f'(x)=\frac{2x(1+x^3)-3x^2\cdot x^2}{(1+x^3)^2}\)

\(\displaystyle\qquad=\frac{2x+2x^4-3x^4}{(x^3+1)^2}\)

\(\displaystyle\qquad=\frac{2x-x^4}{(x^3+1)^2}\)

\(\displaystyle\qquad=\frac{x(2-x^3)}{(x^3+1)^2}\)

当\(\displaystyle x^3<2\qquad f(x)\) 单调增加

当\(\displaystyle x^3>2\qquad f(x)\) 单调减少

所以,只要 \(\displaystyle n\ge 2\qquad\frac{n^2}{n^3+1}\) 单调递减

\(\displaystyle\Rightarrow\quad\frac{(n+1)^2}{(n+1)^3+1}\le \frac{n^2}{n^3+1}\)

\(\displaystyle\Rightarrow\quad a_{n+1}\le a_n\qquad(n\ge 2)\)

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

3,判断 \(\displaystyle a_{n+1}\le a_n\) 的方法:

(1)直接计算(直接比较)

(2)\(\displaystyle a_n=f(n)\),用\(\displaystyle f'(x)\) 来判断

4,交错级数的误差:\(\displaystyle |s-s_n|\le a_{n+1}\)

因 \(\displaystyle s=\sum_{n=1}^{\infty} (-1)^{n+1} a_n\),则 \(\displaystyle s_n=\sum_{k=1}^n (-1)^{k+1} a_k\),所以:

\(\displaystyle\quad |s-s_n|\)

\(\displaystyle=|\sum_{k={n+1}}^{\infty} (-1)^{k+1} a_k|\)

\(\displaystyle=a_{n+1}-a_{n+2}+a_{n+3}-\cdots\)

\(\displaystyle=a_{n+1}-(a_{n+2}-a_{n+3})-(\cdots)+\cdots\)

\(\displaystyle\le a_{n+1}\)