无穷级数

达朗贝尔判别法

lim⁡n→∞un+1un=ρ\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \rho
  1. ρ<1\rho \lt 1 收敛
  2. ρ>1\rho \gt 1 发散

柯西判别法

lim⁡n→∞unn=ρ\lim_{n \to \infty} \sqrt[n]{u_n} = \rho
  1. ρ<1\rho \lt 1 收敛
  2. ρ>1\rho \gt 1 发散

阿贝尔定理求收敛域

lim⁡n→∞∣un+1∣∣un∣=ρ\lim_{n \to \infty} \frac{|u_{n+1}|}{|u_n|} = \rho

或者

lim⁡n→∞unn=ρ\lim_{n \to \infty} \sqrt[n]{u_{n}} = \rho

解出 x 得值域，并判断端点是否收敛

ex=∑m=0∞xnn!,−∞<x<+∞e^x=\sum^\infty_{m=0}\frac{x^n}{n!}, -\infty \lt x \lt +\infty 11+x=∑n=0∞(−1)nxn,−1<x<1\frac{1}{1+x} = \sum^\infty_{n=0}(-1)^nx^n,-1 \lt x \lt 1 11−x=∑n=0∞xn,−1<x<1\frac{1}{1-x} = \sum^\infty_{n=0}x^n, -1 \lt x \lt 1 ln⁡(1+x)=∑n=1∞(−1)n−1xnn,−1<x≤1\ln(1+x) = \sum^{\infty}_{n=1}(-1)^{n-1}\frac{x^n}{n}, -1 < x \le 1 sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!,−∞<x<+∞\sin x = \sum^\infty_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}, -\infty \lt x \lt +\infty cos⁡x=∑x=0∞(−1)nx2n(2n)!,−∞<x<+∞\cos x = \sum^\infty_{x=0}(-1)^n\frac{x^{2n}}{(2n)!}, -\infty \lt x \lt +\infty