តារាងនៃស៊េរីគណិតវិទ្យា

ផលបូកនៃស្វ័យគុណ

$\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}$
$\sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}$
$\sum _{i=1}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left[\sum _{i=1}^{n}i\right]^{2}$
$\sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}$
$\sum _{i=0}^{n}i^{s}={\frac {(n+1)^{s+1}}{s+1}}+\sum _{k=1}^{s}{\frac {B_{k}}{s-k+1}}{s \choose k}(n+1)^{s-k+1}$

ដែល

B_{k}

$\sum _{i=1}^{\infty }i^{-s}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}=\zeta (s)$

ដែល

\zeta (s)

ផលបូកអនន្ត (ចំពោះ $\|x\|\leq 1,x\neq 1$ )		ផលបូកមិនអនន្ត
$\sum _{i=0}^{\infty }x^{i}={\frac {1}{1-x}}$		$\sum _{i=0}^{n}x^{i}={\frac {1-x^{n+1}}{1-x}}$
$\sum _{i=1}^{\infty }ix^{i}={\frac {x}{(1-x)^{2}}}$		$\sum _{i=1}^{n}ix^{i}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}$
$\sum _{i=1}^{\infty }i^{2}x^{i}={\frac {x(1+x)}{(1-x)^{3}}}$		$\sum _{i=1}^{n}i^{2}x^{i}=$ ${\frac {x(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}{(1-x)^{3}}}$
$\sum _{i=1}^{\infty }i^{3}x^{i}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}$
$\sum _{i=1}^{\infty }i^{4}x^{i}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}$
$\sum _{i=1}^{\infty }i^{k}x^{i}=x{\frac {d}{dx}}\left(\sum _{i=1}^{\infty }i^{\left(k-1\right)}x^{i}\right)=Li_{-k}(x)$ ដែល $Li_{s}(x)$ ជាពហុលោការីត(Polylogarithm)នៃ $x$ .

$\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}=\log _{e}\left({\frac {1}{1-x}}\right)\!$ ចំពោះ $|x|\leq 1,\,x\not =-1$

$\sum _{i=0}^{\infty }{\frac {(-1)^{i}}{2i+1}}x^{2i+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)$

$\sum _{i=0}^{\infty }{\frac {x^{2i+1}}{2i+1}}=\mathrm {arctanh} (x)\!$ ចំពោះ $|x|<1\!$

ស៊េរីជាច្រើនដែលកើតឡើងពីទ្រឹស្តីបទតាយលើ(Taylor's theorem)មានមេគុណដែលមានហ្វាក់តូរ្យែល ។

$\sum _{i=0}^{\infty }i{\frac {x^{i}}{i!}}=xe^{x}$ (c.f. មធ្យមនៃរបាយព័យសុង(Poisson distribution))
$\sum _{i=0}^{\infty }i^{2}{\frac {x^{i}}{i!}}=(x+x^{2})e^{x}$ (c.f. ម៉ូម៉ង់ទី២នៃរបាយព័យសុង)
$\sum _{i=0}^{\infty }i^{3}{\frac {x^{i}}{i!}}=(x+3x^{2}+x^{3})e^{x}$
$\sum _{i=0}^{\infty }i^{4}{\frac {x^{i}}{i!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}$

$\sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}x^{2i+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x$

$\sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}x^{2i}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x$

$\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}=\arcsin x\!$ ចំពោះ $|x|<1\!$

$\sum _{i=0}^{\infty }{\frac {(-1)^{i}(2i)!}{4^{i}(i!)^{2}(2i+1)}}x^{2i+1}=\mathrm {arsinh} (x)\!$ ចំពោះ $|x|<1\!$

ស៊េរីទ្វេធា (រួមទាំងរឹសការេនៃ $\alpha =1/2$ ហើយស៊េរីធរណីមាត្រអនន្តចំពោះ $\alpha =-1$ ):

${\sqrt {1+x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)n!^{2}4^{n}}}x^{n}\!$ ចំពោះ $|x|<1\!$

ទំរង់ទូទៅ:

$(1+x)^{\alpha }=\sum _{n=0}^{\infty }{\alpha \choose n}x^{n}\!$ ចំពោះ $|x|<1\!$ ហើយចំពោះគ្រប់ចំនួនកុំផ្លិច $\alpha \!$

ដោយសំរួលមេគុណទ្វេធា

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

^[១] $\sum _{i=0}^{\infty }{i+n \choose i}x^{i}={\frac {1}{(1-x)^{n+1}}}$
^[១] $\sum _{i=0}^{\infty }{\frac {1}{i+1}}{2i \choose i}x^{i}={\frac {1}{2x}}({\sqrt {1-4x}})$
^[១] $\sum _{i=0}^{\infty }{2i \choose i}x^{i}={\frac {1}{\sqrt {1-4x}}}$
^[១] $\sum _{i=0}^{\infty }{2i+n \choose i}x^{i}={\frac {1}{\sqrt {1-4x}}}\left({\frac {1-{\sqrt {1-4x}}}{2x}}\right)^{n}$