List of series

This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums.

Sums of powers

[math]\displaystyle{ \sum_{i=1}^n i = \frac{n(n+1)}{2}\,\! }[/math]
See also triangle number. This is one of the most useful series: many applications can be found throughout mathematics.
[math]\displaystyle{ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \,\! }[/math]
[math]\displaystyle{ \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\,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^{n} i^{4} = \frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}=\frac{6 n^5+15 n^4+10 n^3-n}{30}\,\! }[/math]
[math]\displaystyle{ \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}\,\! }[/math]
Where [math]\displaystyle{ B_k\, }[/math] is the [math]\displaystyle{ k\, }[/math]th Bernoulli number, [math]\displaystyle{ B_1\, }[/math] is negative and [math]\displaystyle{ s\choose k }[/math] is the binomial coefficient (choose function).
[math]\displaystyle{ \sum_{i=1}^\infty i^{-s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = \zeta(s)\,\! }[/math]

Where [math]\displaystyle{ \zeta(s)\, }[/math] is the Riemann zeta function.

Infinite sum (for [math]\displaystyle{ \|x\| \lt 1 }[/math])	Finite sum
[math]\displaystyle{ \sum_{i=0}^\infty x^i= \frac{1}{1-x}\,\! }[/math]	[math]\displaystyle{ \sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x} = 1+\frac{1}{r}\left(1-\frac{1}{(1+r)^n}\right) }[/math] where [math]\displaystyle{ r\gt 0 }[/math] and [math]\displaystyle{ x=\frac{1}{1+r}.\,\! }[/math]
[math]\displaystyle{ \sum_{i=0}^\infty x^{2i}= \frac{1}{1-x^2}\,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^\infty i x^i = \frac{x}{(1-x)^2}\,\! }[/math]	[math]\displaystyle{ \sum_{i=1}^n i x^i = x\frac{1-x^n}{(1-x)^2} - \frac{n x^{n+1}}{1-x}\,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3}\,\! }[/math]	[math]\displaystyle{ \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} \,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^{\infty} i^3 x^i =\frac{x(1+4x+x^2)}{(1-x)^4}\,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^{\infty} i^4 x^i =\frac{x(1+x)(1+10x+x^2)}{(1-x)^5}\,\! }[/math]
[math]\displaystyle{ \sum_{i=1}^{\infty} i^k x^i = \operatorname{Li}_{-k}(x),\,\! }[/math] where Li_s(x) is the polylogarithm of x.

[math]\displaystyle{ \sum^{\infty}_{n=1} \frac{x^n}n = \log_e\left(\frac{1}{1-x}\right) \quad\mbox{ for } |x| \lt 1 \! }[/math]

[math]\displaystyle{ \sum^{\infty}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots = \arctan(x)\,\! }[/math]

[math]\displaystyle{ \sum^{\infty}_{n=0} \frac{x^{2n+1}}{2n+1} = \mathrm{arctanh} (x) \quad\mbox{ for } |x| \lt 1\,\! }[/math]

[math]\displaystyle{ \sum^{\infty}_{n=1} \frac{1}{n^2} = \frac{\pi^2}{6}\,\! }[/math]
[math]\displaystyle{ \sum^{\infty}_{n=1} \frac{1}{n^4} = \frac{\pi^4}{90}\,\! }[/math]

[math]\displaystyle{ \sum^{\infty}_{n=1} \frac{y}{n^2+y^2} = -\frac{1}{2y}+\frac{\pi}{2}\coth(\pi y) }[/math]

Many power series which arise from Taylor's theorem have a coefficient containing a factorial.

[math]\displaystyle{ \sum^{\infty}_{i=0} i \frac{x^i}{i!} = x e^x }[/math] (c.f. mean of Poisson distribution)
[math]\displaystyle{ \sum^{\infty}_{i=0} i^2 \frac{x^i}{i!} = (x + x^2) e^x }[/math] (c.f. second moment of Poisson distribution)
[math]\displaystyle{ \sum^{\infty}_{i=0} i^3 \frac{x^i}{i!} = (x + 3x^2 + x^3) e^x }[/math]
[math]\displaystyle{ \sum^{\infty}_{i=0} i^4 \frac{x^i}{i!} = (x + 7x^2 + 6x^3 + x^4) e^x }[/math]

[math]\displaystyle{ \sum^{\infty}_{i=0} \frac{(-1)^i}{(2i+1)!} x^{2i+1}= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sin x }[/math]

[math]\displaystyle{ \sum^{\infty}_{i=0} \frac{(-1)^i}{(2i)!} x^{2i} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x }[/math]

[math]\displaystyle{ \sum^{\infty}_{i=0} \frac{x^{2i+1}}{(2i+1)!} = \sinh x }[/math]

[math]\displaystyle{ \sum^{\infty}_{i=0} \frac{x^{2i}}{(2i)!} = \cosh x }[/math]

[math]\displaystyle{ \sum^{\infin}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1} = \arcsin x\quad\mbox{ for } |x| \lt 1\! }[/math]

[math]\displaystyle{ \sum^{\infty}_{i=0} \frac{(-1)^i (2i)!}{4^i (i!)^2 (2i+1)} x^{2i+1} = \mathrm{arcsinh}(x) \quad\mbox{ for } |x| \lt 1\! }[/math]

[math]\displaystyle{ (1+x)^{-1} = \begin{cases} \displaystyle \sum_{i=0}^\infty (-x)^i & |x|\lt 1 \\ \displaystyle \sum_{i=1}^\infty -(x)^{-i} & |x|\gt 1 \\ \end{cases} }[/math]

[math]\displaystyle{ (a+x)^n = \begin{cases} \displaystyle \sum_{i=0}^\infty \binom{n}{i} a^{n-i} x^i & |x| \! \lt \! |a| \\ \displaystyle \sum_{i=0}^\infty \binom{n}{i} a^i x^{n-i} & |x| \! \gt \! |a| \\ \end{cases} }[/math]

[math]\displaystyle{ (1+x)^\alpha = \sum_{i=0}^\infty {\alpha \choose i} x^i\quad\mbox{ for all } |x| \lt 1 \mbox{ and all complex } \alpha\! }[/math]

[math]\displaystyle{ {\alpha\choose n} = \prod_{k=1}^n \frac{\alpha-k+1}k = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\! }[/math]

[math]\displaystyle{ \sqrt{1+x} = \sum_{i=0}^\infty \frac{(-1)^i(2i)!}{(1-2i)i!^24^i}x^i \quad\mbox{ for } |x|\lt 1\! }[/math]

Miscellaneous:

^[1] [math]\displaystyle{ \sum_{i=0}^\infty {i+n \choose i} x^i = \frac{1}{(1-x)^{n+1}} }[/math]
^[1] [math]\displaystyle{ \sum_{i=0}^\infty \frac{1}{i+1}{2i \choose i} x^i = \frac{1}{2x}(1-\sqrt{1-4x}) }[/math]
^[1] [math]\displaystyle{ \sum_{i=0}^\infty {2i \choose i} x^i = \frac{1}{\sqrt{1-4x}} }[/math]
^[1] [math]\displaystyle{ \sum_{i=0}^\infty {2i + n \choose i} x^i = \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n }[/math]

[math]\displaystyle{ \sum_{i=0}^n {n \choose i} = 2^n }[/math]
[math]\displaystyle{ \sum_{i=0}^n {n \choose i}a^{(n-i)} b^i = (a + b)^n }[/math]
[math]\displaystyle{ \sum_{i=0}^n (-1)^i{n \choose i} = 0 }[/math]
[math]\displaystyle{ \sum_{i=0}^n {i \choose k} = { n+1 \choose k+1 } }[/math]
[math]\displaystyle{ \sum_{i=0}^n {k+i \choose i} = { k + n + 1 \choose n } }[/math]
[math]\displaystyle{ \sum_{i=0}^r {r \choose i}{s \choose n-i} = {r + s \choose n} }[/math]

[math]\displaystyle{ \sum_{n=b+1}^{\infty} \frac{b}{n^2 - b^2} = \sum_{n=1}^{2b} \frac{1}{2n} }[/math]