1，无穷级数：
无限个数之和 \(\displaystyle\ a_1+a_2+ \cdots +a_n+ \cdots=\sum_{n=1}^{\infty} a_n\)
称为（常数项）无穷级数。
2，\(\displaystyle s_1=a_1, \ s_2=a_1+a_2, \ s_3=a_1+a_2+a_3,\ \cdots, \ s_n=\sum_{i=1}^n a_i,\ \cdots\)
\(\displaystyle\quad\ \ s_1,\ s_2,\ \cdots, s_n,\ \cdots=\{s_n\}_{n=1}^\infty\)
\(\displaystyle\quad\ \left(s_n\right)_{n=1}^\infty:\ \sum_{n=1}^\infty a_n\ \)的部分和数列
3，\(\displaystyle \sum_{n=1}^\infty a_n\)
\(\displaystyle\begin {cases}\text{收敛}&\displaystyle\Leftrightarrow\lim_{n\to \infty}s_n=s\\ \text{发散}&\displaystyle \Leftrightarrow\lim_{n\to \infty}s_n \text{不存在}\end{cases}\)
例1，求\(\displaystyle 1-1+1-1+ \cdots\ +(-1)^{n-1}+\ \cdots\ =\sum_{n=1}^\infty (-1)^{n-1}\qquad\) 发散
解答：\(\displaystyle \lim_{n\to\infty}s_{2n-1}=1\ne \lim_{n\to\infty}s_{2n}=0\)
例2，求\(\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)}\quad\)收敛，且\(\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)}=1\)
证明：\(\displaystyle s_n=\sum_{i=1}^n \frac{1}{i(i+1)}=\sum_{i=1}^n \left(\frac{1}{i}-\frac{1}{i+1}\right)\)
\(\displaystyle\frac{1}{i(i+1)}=\frac{A}{i}+\frac{B}{i+1}\quad\Rightarrow\quad A(i+1)+Bi=1\)
\(\displaystyle i=0\ \ \ \quad\Rightarrow\quad A=1\)
\(\displaystyle i=-1\quad\Rightarrow\quad B=-1\)
\(\displaystyle s_n=\sum_{i=1}^n \frac{1}{i(i+1)}\)
\(\displaystyle\quad=\sum_{i=1}^n\left(\frac{1}{i}-\frac{1}{i+1}\right)\)
\(\displaystyle\quad=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots\ +\frac{1}{n}-\frac{1}{n+1}\)
\(\displaystyle\quad=1-\frac{1}{n+1}\)
\(\displaystyle\lim_{n\to\infty} s_n=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1\quad\Rightarrow\quad\sum_{n=1}^\infty \frac{1}{n(n+1)}=1\)
例3，几何级数
\(\displaystyle\sum_{n=0}^\infty ar^n=a+ar+ar^2+\cdots+ar^{n-1}+ar^n+\cdots\)
解答：\(\displaystyle s_0=a,\)
\(\displaystyle\quad\quad\ \ \ s_1=a+ar,\)
\(\displaystyle\quad\quad\ \ \ \cdots\)
\(\displaystyle\quad\quad\ \ \ s_n=a+ar+\cdots+ar^n=a(1+r+r^2+\cdots\ +r^n)=a\cdot\frac{1-r^{n+1}}{1-r}\)
（1）\(\displaystyle |r|>1\quad\Rightarrow\quad\lim_{n\to\infty} s_n=\lim_{n\to\infty} a\cdot \frac{1-r^{n+1}}{1-r}=\pm \infty\)
（2）\(\displaystyle |r|=1\)
\(\displaystyle\quad r=1\ \)时，\(\displaystyle s_n=\sum_{k=0}^n a=(n+1)a\)，\(\displaystyle \lim_{n\to\infty} s_n=\pm\infty\)
\(\displaystyle\quad r=-1\)时，\(\displaystyle s_n=\sum_{k=0}^n a(-1)^n\)，\(\displaystyle \lim_{n\to\infty} s_n\ \)不存在
（3）\(\displaystyle |r|<1\quad\Rightarrow\quad\lim_{n\to\infty} s_n=\lim_{n\to\infty} a\cdot\frac{1-r^{n+1}}{1-r}=\frac{a}{1-r}\)
\(\displaystyle \Rightarrow\quad\)结论：几何级数\(\displaystyle\sum_{n=0}^\infty ar^n\)
\(\begin {cases} |r|\ge 1&\displaystyle\quad\Rightarrow\quad\text{发散}\\|r|<1&\displaystyle\quad\Rightarrow\quad\text{收敛于}\ \frac{a}{1-r}\end{cases}\)
例4，调和级数\(\displaystyle\ \sum_{n=1}^\infty \frac{1}{n}\ \)发散
证明：\(\displaystyle s_1=1,\ s_2=1+\frac{1}{2}=\frac{3}{2},\ \)
\(\displaystyle\qquad\quad s_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\displaystyle\qquad\qquad=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)>\frac{3}{2}+\frac{1}{2}=2\)
\(\displaystyle\qquad\quad s_8=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}\)
\(\displaystyle\qquad\qquad=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)>1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=1+\frac{3}{2}\)
\(\displaystyle\qquad\quad s_{16}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{16}>1+\frac{3}{2}+\left(\frac{1}{9}+\cdots +\frac{1}{16}\right)>1+\frac{3}{2}+\frac{1}{2}=1+\frac{4}{2}\)
\(\displaystyle\qquad\quad s_{2n}>1+\frac{n}{2}\)
\(\displaystyle\Rightarrow\quad\lim_{n\to\infty} s_{2n}=+\infty\qquad\{s_{2n}\} \) 发散， \(\displaystyle \Rightarrow\quad\{s_n\}_{n=1}^\infty\)发散