ស្វ័យគុណទ្វេធា

Carl von Linné, Alexander Roslin, 1775. Oil painting in the portrait collection at Gripsholm Castle. (Photo credit: Wikipedia)

**ភាគបែងធម្មតា

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

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

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

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

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

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

$\sqrt{1+x} = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}x^n \!$ 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] $\sum_{i=0}^\infty {2i + n \choose i} x^i = \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n$

ផលបូក នៃ ស៊ីនុស និងកូស៊ីនុស មានក្នុងស៊េរីហ្វូរៀរ(Fourier series) ។

$\sum_{n=b+1}^{\infty} \frac{b}{n^2 - b^2} = \sum_{n=1}^{2b} \frac{1}{2n}$

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

ដែល $B_k$ ជាចំនួនប៊ែរនូយី(Bernoulli number)ទីk ។

$\sum_{i=1}^\infty i^{-s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = \zeta(s)$

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