级数发散的充分条件与收敛级数的性质

笔记下载:级数发散的充分条件与收敛级数的性质

1,定理(级数收敛的必要条件):

\(\displaystyle\sum_{n=1}^\infty a_n\) 收敛\(\displaystyle\quad\Rightarrow\quad\lim_{n\to\infty} a_n=0\)

证明:\(\displaystyle a_n=s_n-s_{n-1},\quad\sum_{n=1}^\infty a_n\) 收敛

\(\displaystyle\Rightarrow\ \lim_{n\to\infty} a_n=\lim_{n\to\infty}\left(s_n-s_{n-1}\right)=\lim_{n\to\infty}s_n-\lim_{n\to\infty}s_{n-1}=s-s=0\)

2,定理(级数发散的充分条件):

若 \(\displaystyle\lim_{n\to\infty} a_n\ne 0\quad\Rightarrow\quad\sum_{n=1}^\infty a_n\) 发散

例1,\(\displaystyle\sum_{n=1}^\infty \frac{1n^2}{n^2+n-2}\) 发散

证明:\(\displaystyle\lim_{n\to\infty} a_n=\lim_{n\to\infty}\frac{2n^2}{n^2+n-2}=\lim_{n\to\infty} \frac{2}{1+\frac{1}{n}-\frac{2}{n^2}}=2\ne o\)

例2,\(\displaystyle\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^n\) 发散

证明:因为 \(\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e\ne 0\)

3,定理(收敛级数的性质):

(1)\(\displaystyle\sum {u_n\pm v_n}=\sum u_n\pm\sum v_n\)

\(\displaystyle\qquad\quad\)若 \(\displaystyle\sum u_n\) 和 \(\displaystyle\sum v_n\) 都收敛,则\(\displaystyle\sum {u_n\pm v_n}\) 收敛

(2)\(\displaystyle\sum ku_n=k\sum u_n\)

\(\displaystyle\qquad\quad\)当\(\displaystyle k\ne 0\) 时,\(\displaystyle\sum ku_n\) 与\(\displaystyle\sum u_n\) 同敛散

(3)去掉有限项或加上有限项,不改变敛散性

证明:(1)设 \(\displaystyle\delta_n=\sum_{k=1}^n u_k\),\(\displaystyle\sigma_n=\sum_{k=1}^n v_k\),\(\displaystyle s_n=\sum_{k=1}^n {u_k\pm v_k}\)

\(\displaystyle\Rightarrow\quad s_n=\delta_n+\sigma_n\)

\(\displaystyle\Rightarrow\quad \lim_{n\to\infty} s_n=\lim_{n\to\infty} (\delta_n+\sigma_n)=\delta+\sigma\)

(2)、(3)类似

例3,求级数 \(\displaystyle\sum_{n=1}^\infty \left(\frac{3}{4^n}-\frac{2}{5^{n-1}}\right)\)

解答:\(\displaystyle\sum_{n=1}^\infty \left(\frac{3}{4^n}-\frac{2}{5^{n-1}}\right)\)

\(\displaystyle\quad\ =\sum_{n=1}^\infty \frac{3}{4^n}-\sum_{n=1}^\infty\frac{2}{5^{n-1}}\)

\(\displaystyle\quad\ =\frac{3}{4}\sum_{n=1}^\infty \frac{1}{4^{n-1}}-2\sum_{n=1}^\infty \left(\frac{1}{5}\right)^{n-1}\)

\(\displaystyle\quad\ =\frac{3}{4}\cdot\frac{1}{1-\frac{1}{4}}-2\cdot\frac{1}{1-\frac{1}{5}}\)

\(\displaystyle\quad\ =1-\frac{5}{2}\)

\(\displaystyle\quad\ ==-\frac{3}{2}\)