فهرست سری‌های ریاضی

• ${\displaystyle \sum _{m=1}^{n}m=\frac{n(n+1)}{2}}$
• ${\displaystyle \sum _{m=1}^n{m}^2=\frac{n(n+1)(2n+1)}{6}=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}$
• ${\displaystyle \sum _{m=1}^n{m}^3={\left[\frac{n(n+1)}{2}\right]}^2=\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}={\left(\sum _{m=1}^{n}m\right)}^2}$
• ${\displaystyle \sum _{m=1}^n{m}^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}=\frac{6n^5+15n^4+10n^3-n}{30}}$
• ${\displaystyle \sum _{m=0}^n{m}^s=\frac{(n+1{)}^{s+1}}{s+1}+\sum _{k=1}^s\frac{B_k}{s-k+1}(\genfrac{}{}{0ex}{}{s}{k})(n+1{)}^{s-k+1}}$

• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^2}=\frac{{\pi }^2}{6}}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^4}=\frac{{\pi }^4}{90}}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^{2n}}=(-1{)}^{n+1}\frac{B_{2n}(2\pi {)}^{2n}}{2(2n)!}}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{m}^{-s}=\prod _{p\text{ prime}}\frac{1}{1-p^{-s}}=\zeta (s)}$

${\displaystyle \sum _{m=0}^n{x}^m=\frac{1-x^{n+1}}{1-x}=1+\frac{1}{r}\left(1-\frac{1}{(1+r{)}^n}\right)}$

${\displaystyle \sum _{m=0}^{\mathrm{\infty }}{x}^{2m}=\frac{1}{1-x^2}}$

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}m{x}^m=\frac{x}{(1-x{)}^2}}$

${\displaystyle \sum _{m=1}^{n}m{x}^m=x\frac{1-x^n}{(1-x{)}^2}-\frac{n{x}^{n+1}}{1-x}}$

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{m}^2{x}^m=\frac{x(1+x)}{(1-x{)}^3}}$

${\displaystyle \sum _{m=1}^n{m}^2{x}^m=\frac{x(1+x-(n+1{)}^2{x}^n+(2n^2+2n-1)x^{n+1}-n^2{x}^{n+2})}{(1-x{)}^3}}$

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{m}^3{x}^m=\frac{x(1+4x+x^2)}{(1-x{)}^4}}$

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{m}^4{x}^m=\frac{x(1+x)(1+10x+x^2)}{(1-x{)}^5}}$

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{m}^k{x}^m={\mathrm{Li}}_{-k}(x),}$

• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{x^m}{m}=\ln \left(\frac{1}{1-x}\right)}$
  برای
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{2m+1}{x}^{2m+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots =\arctan (x)}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{x^{2m+1}}{2m+1}=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}(x)}$
  برای

• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{x^m}{m!}=e^x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{m!}{x}^m=\frac{1}{e^x}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}m\frac{x^m}{m!}=x{e}^x}$
  (توزیع پواسون)
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}{m}^2\frac{x^m}{m!}=(x+x^2)e^x}$
  (مشتق دوم توزیع پواسون)
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}{m}^3\frac{x^m}{m!}=(x+3x^2+x^3)e^x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}{m}^4\frac{x^m}{m!}=(x+7x^2+6x^3+x^4)e^x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}{m}^n\frac{x^m}{m!}=x\frac{d}{dx}\sum _{m=0}^{\mathrm{\infty }}{m}^{n-1}\frac{x^m}{m!}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{(2m+1)!}{x}^{2m+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots =\sin x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m}{(2m)!}{x}^{2m}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots =\cos x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{x^{2m+1}}{(2m+1)!}=\sinh x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{x^{2m}}{(2m)!}=\cosh x}$

• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(4m)!}{{16}^m\sqrt{2}(2m)!(2m+1)!}{x}^m=\sqrt{\frac{1-\sqrt{1-x}}{x}}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{4^m(m){!}^2}{(m+1)(2m+1)!}{x}^{2m}={\left(\frac{\arcsin x}{x}\right)}^2}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{\prod _{n=0}^{m-1}(4n^2+1)}{(2m)!}{x}^{2m}+\sum _{m=0}^{\mathrm{\infty }}\frac{4^m\prod _{n=1}^m(\frac{1}{2}-n+n^2)}{(2m+1)!}{x}^{2m+1}=e^{\arcsin x}}$

• ${\displaystyle (1+x{)}^{\alpha }=\sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{\alpha }{m})x^m}$

${\displaystyle (\genfrac{}{}{0ex}{}{\alpha }{n})=\prod _{k=1}^n\frac{\alpha -k+1}{k}=\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}$

• ${\displaystyle \sqrt{1+x}=\sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^m(2m)!}{(1-2m)m{!}^2{4}^m}{x}^m}$
  برای

• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m+n}{m})x^m=\frac{1}{(1-x{)}^{n+1}}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{1}{m+1}(\genfrac{}{}{0ex}{}{2m}{m})x^m=\frac{1}{2x}(1-\sqrt{1-4x})}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2m}{m})x^m=\frac{1}{\sqrt{1-4x}}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2m+n}{m})x^m=\frac{1}{\sqrt{1-4x}}{\left(\frac{1-\sqrt{1-4x}}{2x}\right)}^n}$

• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{B_m}{m!}{x}^m=\frac{x}{e^x-1}}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-4{)}^m{B}_{2m}}{(2m)!}{x}^{2m}=x\cot x}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{(-1{)}^{m-1}{2}^{2m}(2^{2m}-1)B_{2m}}{(2m)!}{x}^{2m-1}=\tan x}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{(-1{)}^{m-1}(4^m-2)B_{2m}}{(2m)!}{x}^{2m}=\frac{x}{\sin x}}$

• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{H}_m{x}^m=\frac{\log \frac{1}{1-x}}{1-x}}$
• ${\displaystyle \sum _{m=2}^{\mathrm{\infty }}\frac{H_{2m-1}}{m}{x}^m=\frac{1}{2}{\left(\log \frac{1}{1-x}\right)}^2}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{(-1{)}^{m-1}{H}_{2m}}{2m+1}{x}^{2m+1}=\frac{1}{2}\arctan x\log (1+x^2)}$
• ${\displaystyle \sum _{m=0}^{\mathrm{\infty }}\frac{\sum _{n=0}^{2m}\frac{(-1{)}^n}{2n+1}}{4m+2}{x}^{4m+2}=\frac{1}{4}\arctan x\log \frac{1+x}{1-x}}$

• ${\displaystyle \sum _{m=0}^n(\genfrac{}{}{0ex}{}{n}{m})=2^n}$
• ${\displaystyle \sum _{m=0}^n(\genfrac{}{}{0ex}{}{n}{m})a^{(n-m)}{b}^m=(a+b{)}^n}$
• ${\displaystyle \sum _{m=0}^n(-1{)}^i(\genfrac{}{}{0ex}{}{n}{m})=0}$
• ${\displaystyle \sum _{m=0}^n(\genfrac{}{}{0ex}{}{m}{k})=(\genfrac{}{}{0ex}{}{n+1}{k+1})}$
• ${\displaystyle \sum _{m=0}^n(\genfrac{}{}{0ex}{}{k+m}{m})=(\genfrac{}{}{0ex}{}{k+n+1}{n})}$
• ${\displaystyle \sum _{m=0}^r(\genfrac{}{}{0ex}{}{r}{m})(\genfrac{}{}{0ex}{}{s}{n-m})=(\genfrac{}{}{0ex}{}{r+s}{n})}$

• ${\displaystyle \sum _{m=1}^n\sin \left(\frac{m\pi }{n}\right)=0}$
• ${\displaystyle \sum _{m=1}^n\cos \left(\frac{m\pi }{n}\right)=0}$

• ${\displaystyle \sum _{m=b+1}^{\mathrm{\infty }}\frac{b}{m^2-b^2}=\frac{1}{2}{H}_{2b}}$
• ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}\frac{y}{m^2+y^2}=-\frac{1}{2y}+\frac{\pi }{2}\coth (\pi y)}$