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

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

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

${\displaystyle \sum _{m=0}^{n}x^{m}={\frac {1-x^{n+1}}{1-x}}=1+{\frac {1}{r}}\left(1-{\frac {1}{(1+r)^{n}}}\right)}$ که ${\displaystyle r>0}$ و شکست در تجزیه (پاسخ نامعتبر MathML همراه SVG یا PNG جایگزین (توصیه شده برای مرورگرهای مدرن و ابزارهای کمکی) ("Math extension cannot connect to Restbase.") از سرور "http://localhost:6011/fa.teknopedia.teknokrat.ac.id/v1/":): {\displaystyle x=\frac{1}{1+r}.\,\!}

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

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

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

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

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

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

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

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

• ${\displaystyle \sum _{m=1}^{\infty }{\frac {x^{m}}{m}}=\ln \left({\frac {1}{1-x}}\right)}$ برای ${\displaystyle |x|<1\!}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{2m+1}}x^{2m+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)\,\!}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m+1}}{2m+1}}=\mathrm {arctanh} (x)}$ برای ${\displaystyle |x|<1\,\!}$

• ${\displaystyle \sum _{m=0}^{\infty }{\frac {x^{m}}{m!}}=e^{x}}$
• شکست در تجزیه (پاسخ نامعتبر MathML همراه SVG یا PNG جایگزین (توصیه شده برای مرورگرهای مدرن و ابزارهای کمکی) ("Math extension cannot connect to Restbase.") از سرور "http://localhost:6011/fa.teknopedia.teknokrat.ac.id/v1/":): {\displaystyle \sum^{\infty}_{m=0} \frac{(-1)^m}{m!} x^{m} = \frac{1}{e^{x}} }
• شکست در تجزیه (پاسخ نامعتبر MathML همراه SVG یا PNG جایگزین (توصیه شده برای مرورگرهای مدرن و ابزارهای کمکی) ("Math extension cannot connect to Restbase.") از سرور "http://localhost:6011/fa.teknopedia.teknokrat.ac.id/v1/":): {\displaystyle \sum^{\infty}_{m=0} m \frac{x^m}{m!} = x e^x} (توزیع پواسون)
• ${\displaystyle \sum _{m=0}^{\infty }m^{2}{\frac {x^{m}}{m!}}=(x+x^{2})e^{x}}$ (مشتق دوم توزیع پواسون)
• ${\displaystyle \sum _{m=0}^{\infty }m^{3}{\frac {x^{m}}{m!}}=(x+3x^{2}+x^{3})e^{x}}$
• ${\displaystyle \sum _{m=0}^{\infty }m^{4}{\frac {x^{m}}{m!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}}$
• شکست در تجزیه (پاسخ نامعتبر MathML همراه SVG یا PNG جایگزین (توصیه شده برای مرورگرهای مدرن و ابزارهای کمکی) ("Math extension cannot connect to Restbase.") از سرور "http://localhost:6011/fa.teknopedia.teknokrat.ac.id/v1/":): {\displaystyle \sum^{\infty}_{m=0} m^n \frac{x^m}{m!} = x \frac{d}{dx}\sum^{\infty}_{m=0} m^{n-1} \frac{x^m}{m!}}
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m+1)!}}x^{2m+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m)!}}x^{2m}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m+1}}{(2m+1)!}}=\sinh x}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m}}{(2m)!}}=\cosh x}$

• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(2m)!}{4^{m}(m!)^{2}(2m+1)}}x^{2m+1}=\arcsin x\quad {\mbox{ for }}|x|<1\!}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}(2m)!}{4^{m}(m!)^{2}(2m+1)}}x^{2m+1}=\mathrm {arcsinh} (x)\quad {\mbox{ for }}|x|<1\!}$
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(4m)!}{16^{m}{\sqrt {2}}(2m)!(2m+1)!}}x^{m}={\sqrt {\frac {1-{\sqrt {1-x}}}{x}}}}$ ^[۱]
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {4^{m}(m)!^{2}}{(m+1)(2m+1)!}}x^{2m}=\left({\frac {\arcsin {x}}{x}}\right)^{2}}$ ^[۱]
• شکست در تجزیه (پاسخ نامعتبر MathML همراه SVG یا PNG جایگزین (توصیه شده برای مرورگرهای مدرن و ابزارهای کمکی) ("Math extension cannot connect to Restbase.") از سرور "http://localhost:6011/fa.teknopedia.teknokrat.ac.id/v1/":): {\displaystyle \sum^{\infty}_{m=0} \frac{\prod_{n=0}^{m-1}(4n^2+1)}{(2m)!} x^{2m} + \sum^{\infty}_{m=0} \frac{4^m \prod_{n=1}^{m}(\frac{1}{2}-n+n^2)}{(2m+1)!} x^{2m+1} = e^{\arcsin{x}} } ^[۱]

• ${\displaystyle (1+x)^{-1}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }(-x)^{m}&|x|<1\\\displaystyle \sum _{m=1}^{\infty }-(x)^{-m}&|x|>1\\\end{cases}}}$

• ${\displaystyle (a+x)^{n}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{n-m}x^{m}&|x|\!<\!|a|\\\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{m}x^{n-m}&|x|\!>\!|a|\\\end{cases}}}$

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

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

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

• ^[۲] ${\displaystyle \sum _{m=0}^{\infty }{m+n \choose m}x^{m}={\frac {1}{(1-x)^{n+1}}}}$
• ^[۲] ${\displaystyle \sum _{m=0}^{\infty }{\frac {1}{m+1}}{2m \choose m}x^{m}={\frac {1}{2x}}(1-{\sqrt {1-4x}})}$
• ^[۲] ${\displaystyle \sum _{m=0}^{\infty }{2m \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}}$
• ^[۲] ${\displaystyle \sum _{m=0}^{\infty }{2m+n \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}\left({\frac {1-{\sqrt {1-4x}}}{2x}}\right)^{n}}$

• ${\displaystyle \sum _{m=0}^{\infty }{\frac {B_{m}}{m!}}x^{m}={\frac {x}{e^{x}-1}}}$ ^[۱]
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-4)^{m}B_{2m}}{(2m)!}}x^{2m}=x\cot {x}}$ ^[۱]
• ${\displaystyle \sum _{m=1}^{\infty }{\frac {(-1)^{m-1}2^{2m}(2^{2m}-1)B_{2m}}{(2m)!}}x^{2m-1}=\tan {x}}$ ^[۱]
• ${\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m-1}(4^{m}-2)B_{2m}}{(2m)!}}x^{2m}={\frac {x}{\sin {x}}}}$ ^[۱]