Spisak matematičkih redova

Ovaj spisak matematičkih redova sadrži formule za konačne i beskonačne sume. Može se koristiti zajedno sa drugim alatima za izračunavanje suma.

Sume stepeni

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

gdje je

B_{k}

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

gdje je

\zeta (s)

Beskonačna suma (za $\|x\|<1$ )	Konačna suma
$\sum _{i=0}^{\infty }x^{i}={\frac {1}{1-x}}\,\!$	$\sum _{i=0}^{n}x^{i}={\frac {1-x^{n+1}}{1-x}}=1+{\frac {1}{r}}(1-{\frac {1}{(1+r)^{n}}})$ gdje je $r>0$ i $x={\frac {1}{1+r}}.\,\!$
$\sum _{i=0}^{\infty }x^{2i}={\frac {1}{1-x^{2}}}\,\!$
$\sum _{i=1}^{\infty }ix^{i}={\frac {x}{(1-x)^{2}}}\,\!$	$\sum _{i=1}^{n}ix^{i}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}\,\!$
$\sum _{i=1}^{\infty }i^{2}x^{i}={\frac {x(1+x)}{(1-x)^{3}}}\,\!$	$\sum _{i=1}^{n}i^{2}x^{i}={\frac {x(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}{(1-x)^{3}}}\,\!$
$\sum _{i=1}^{\infty }i^{3}x^{i}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}\,\!$
$\sum _{i=1}^{\infty }i^{4}x^{i}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}\,\!$
$\sum _{i=1}^{\infty }i^{k}x^{i}=\operatorname {Li} _{-k}(x),\,\!$ gdje je Li_s(x) polilogaritam od x.

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

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

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

Mnogi potencijalni redovi, koji se dobijaju iz Taylorovog teorema imaju koeficijent koji sadrži faktorijel.

$\sum _{i=0}^{\infty }i{\frac {x^{i}}{i!}}=xe^{x}$ (sredina Poissonove distribucije)
$\sum _{i=0}^{\infty }i^{2}{\frac {x^{i}}{i!}}=(x+x^{2})e^{x}$ (Drugi moment Poissonove distribucije)
$\sum _{i=0}^{\infty }i^{3}{\frac {x^{i}}{i!}}=(x+3x^{2}+x^{3})e^{x}$
$\sum _{i=0}^{\infty }i^{4}{\frac {x^{i}}{i!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}$

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

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

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

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

Binomni red (uključujući kvadratni korjen za $\alpha =1/2$ i beskonačni geometrijski red za $\alpha =-1$ ):

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

Opći oblik:

$(1+x)^{\alpha }=\sum _{n=0}^{\infty }{\alpha \choose n}x^{n}\quad {\mbox{ za sve }}|x|<1{\mbox{ i sve kompleksne brojeve }}\alpha \!$

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

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