Công thức tổng dải số

$\sum _{k=0}^{n}c=nc$ where $c$ is some constant.

$\sum _{k=0}^{n}k=\frac{n(n+1)}{2}$

$\sum _{k=0}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

$\sum _{k=0}^n{k}^3=\frac{n^2(n+1{)}^2}{4}$

$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots =e^x$

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}{x}^n=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots =\ln (1+x)\text{ for }|x|<1$

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n)!}{x}^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =\cos (x)\text{ for all }x\in ℂ$