This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums. This article is an orphan, as no other articles link to it. Please introduce links to this page from related articles; try the Find link tool for suggestions. (May 2012) Sums of powers ∑ i = 1 n i = n ( n + 1 ) 2 {\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}\,\!} See also triangle number. This is one of the most useful series: many applications can be found throughout mathematics. ∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = n 3 3 + n 2 2 + n 6 {\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}}\,\!} ∑ i = 1 n i 3 = [ n ( n + 1 ) 2 ] 2 = n 4 4 + n 3 2 + n 2 4 = ( ∑ i = 1 n i ) 2 {\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}\,\!} ∑ i = 1 n i 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 = 6 n 5 + 15 n 4 + 10 n 3 − n 30 {\displaystyle \sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {6n^{5}+15n^{4}+10n^{3}-n}{30}}\,\!} ∑ i = 0 n i s = ( n + 1 ) s + 1 s + 1 + ∑ k = 1 s B k s − k + 1 ( s k ) ( n + 1 ) s − k + 1 {\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}\,\!} Where B k {\displaystyle B_{k}\,} is the k {\displaystyle k\,} th Bernoulli number, B 1 {\displaystyle B_{1}\,} is negative and ( s k ) {\displaystyle s \choose k} is the binomial coefficient (choose function). ∑ i = 1 ∞ i − s = ∏ p prime 1 1 − p − s = ζ ( s ) {\displaystyle \sum _{i=1}^{\infty }i^{-s}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}=\zeta (s)\,\!} Where ζ ( s ) {\displaystyle \zeta (s)\,} is the Riemann zeta function. 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Infinite sum (for | x | < 1 {\displaystyle |x|<1} )Finite sum ∑ i = 0 ∞ x i = 1 1 − x {\displaystyle \sum _{i=0}^{\infty }x^{i}={\frac {1}{1-x}}\,\!} ∑ i = 0 n x i = 1 − x n + 1 1 − x = 1 + 1 r ( 1 − 1 ( 1 + r ) n ) {\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)} where r > 0 {\displaystyle r>0} and x = 1 1 + r . {\displaystyle x={\frac {1}{1+r}}.\,\!} ∑ i = 0 ∞ x 2 i = 1 1 − x 2 {\displaystyle \sum _{i=0}^{\infty }x^{2i}={\frac {1}{1-x^{2}}}\,\!} ∑ i = 1 ∞ i x i = x ( 1 − x ) 2 {\displaystyle \sum _{i=1}^{\infty }ix^{i}={\frac {x}{(1-x)^{2}}}\,\!} ∑ i = 1 n i x i = x 1 − x n ( 1 − x ) 2 − n x n + 1 1 − x {\displaystyle \sum _{i=1}^{n}ix^{i}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}\,\!} ∑ i = 1 ∞ i 2 x i = x ( 1 + x ) ( 1 − x ) 3 {\displaystyle \sum _{i=1}^{\infty }i^{2}x^{i}={\frac {x(1+x)}{(1-x)^{3}}}\,\!} ∑ i = 1 n i 2 x i = x ( 1 + x − ( n + 1 ) 2 x n + ( 2 n 2 + 2 n − 1 ) x n + 1 − n 2 x n + 2 ) ( 1 − x ) 3 {\displaystyle \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}}}\,\!} ∑ i = 1 ∞ i 3 x i = x ( 1 + 4 x + x 2 ) ( 1 − x ) 4 {\displaystyle \sum _{i=1}^{\infty }i^{3}x^{i}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}\,\!} ∑ i = 1 ∞ i 4 x i = x ( 1 + x ) ( 1 + 10 x + x 2 ) ( 1 − x ) 5 {\displaystyle \sum _{i=1}^{\infty }i^{4}x^{i}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}\,\!} ∑ i = 1 ∞ i k x i = Li − k ( x ) , {\displaystyle \sum _{i=1}^{\infty }i^{k}x^{i}=\operatorname {Li} _{-k}(x),\,\!} where Lis(x) is the polylogarithm of x. Close Simple denominators ∑ n = 1 ∞ x n n = log e ( 1 1 − x ) for | x | < 1 {\displaystyle \sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=\log _{e}\left({\frac {1}{1-x}}\right)\quad {\mbox{ for }}|x|<1\!} ∑ n = 0 ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 = x − x 3 3 + x 5 5 − ⋯ = arctan ( x ) {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)\,\!} ∑ n = 0 ∞ x 2 n + 1 2 n + 1 = a r c t a n h ( x ) for | x | < 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {x^{2n+1}}{2n+1}}=\mathrm {arctanh} (x)\quad {\mbox{ for }}|x|<1\,\!} ∑ n = 1 ∞ 1 n 2 = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}\,\!} ∑ n = 1 ∞ 1 n 4 = π 4 90 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{4}}}={\frac {\pi ^{4}}{90}}\,\!} ∑ n = 1 ∞ y n 2 + y 2 = − 1 2 y + π 2 coth ( π y ) {\displaystyle \sum _{n=1}^{\infty }{\frac {y}{n^{2}+y^{2}}}=-{\frac {1}{2y}}+{\frac {\pi }{2}}\coth(\pi y)} Factorial denominators Many power series which arise from Taylor's theorem have a coefficient containing a factorial. ∑ i = 0 ∞ x i i ! = e x {\displaystyle \sum _{i=0}^{\infty }{\frac {x^{i}}{i!}}=e^{x}} ∑ i = 0 ∞ i x i i ! = x e x {\displaystyle \sum _{i=0}^{\infty }i{\frac {x^{i}}{i!}}=xe^{x}} (c.f. mean of Poisson distribution) ∑ i = 0 ∞ i 2 x i i ! = ( x + x 2 ) e x {\displaystyle \sum _{i=0}^{\infty }i^{2}{\frac {x^{i}}{i!}}=(x+x^{2})e^{x}} (c.f. second moment of Poisson distribution) ∑ i = 0 ∞ i 3 x i i ! = ( x + 3 x 2 + x 3 ) e x {\displaystyle \sum _{i=0}^{\infty }i^{3}{\frac {x^{i}}{i!}}=(x+3x^{2}+x^{3})e^{x}} ∑ i = 0 ∞ i 4 x i i ! = ( x + 7 x 2 + 6 x 3 + x 4 ) e x {\displaystyle \sum _{i=0}^{\infty }i^{4}{\frac {x^{i}}{i!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}} ∑ i = 0 ∞ ( − 1 ) i ( 2 i + 1 ) ! x 2 i + 1 = x − x 3 3 ! + x 5 5 ! − ⋯ = sin x {\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i+1)!}}x^{2i+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x} ∑ i = 0 ∞ ( − 1 ) i ( 2 i ) ! x 2 i = 1 − x 2 2 ! + x 4 4 ! − ⋯ = cos x {\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}}{(2i)!}}x^{2i}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x} ∑ i = 0 ∞ x 2 i + 1 ( 2 i + 1 ) ! = sinh x {\displaystyle \sum _{i=0}^{\infty }{\frac {x^{2i+1}}{(2i+1)!}}=\sinh x} ∑ i = 0 ∞ x 2 i ( 2 i ) ! = cosh x {\displaystyle \sum _{i=0}^{\infty }{\frac {x^{2i}}{(2i)!}}=\cosh x} Modified-factorial denominators ∑ n = 0 ∞ ( 2 n ) ! 4 n ( n ! ) 2 ( 2 n + 1 ) x 2 n + 1 = arcsin x for | x | < 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}=\arcsin x\quad {\mbox{ for }}|x|<1\!} ∑ i = 0 ∞ ( − 1 ) i ( 2 i ) ! 4 i ( i ! ) 2 ( 2 i + 1 ) x 2 i + 1 = a r c s i n h ( x ) for | x | < 1 {\displaystyle \sum _{i=0}^{\infty }{\frac {(-1)^{i}(2i)!}{4^{i}(i!)^{2}(2i+1)}}x^{2i+1}=\mathrm {arcsinh} (x)\quad {\mbox{ for }}|x|<1\!} Binomial series Geometric series: ( 1 + x ) − 1 = { ∑ i = 0 ∞ ( − x ) i | x | < 1 ∑ i = 1 ∞ − ( x ) − i | x | > 1 {\displaystyle (1+x)^{-1}={\begin{cases}\displaystyle \sum _{i=0}^{\infty }(-x)^{i}&|x|<1\\\displaystyle \sum _{i=1}^{\infty }-(x)^{-i}&|x|>1\\\end{cases}}} Binomial Theorem: ( a + x ) n = { ∑ i = 0 ∞ ( n i ) a n − i x i | x | < | a | ∑ i = 0 ∞ ( n i ) a i x n − i | x | > | a | {\displaystyle (a+x)^{n}={\begin{cases}\displaystyle \sum _{i=0}^{\infty }{\binom {n}{i}}a^{n-i}x^{i}&|x|\!<\!|a|\\\displaystyle \sum _{i=0}^{\infty }{\binom {n}{i}}a^{i}x^{n-i}&|x|\!>\!|a|\\\end{cases}}} ( 1 + x ) α = ∑ i = 0 ∞ ( α i ) x i for all | x | < 1 and all complex α {\displaystyle (1+x)^{\alpha }=\sum _{i=0}^{\infty }{\alpha \choose i}x^{i}\quad {\mbox{ for all }}|x|<1{\mbox{ and all complex }}\alpha \!} with generalized binomial coefficients ( α n ) = ∏ k = 1 n α − k + 1 k = α ( α − 1 ) ⋯ ( α − n + 1 ) n ! {\displaystyle {\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}\!} Square root: 1 + x = ∑ i = 0 ∞ ( − 1 ) i ( 2 i ) ! ( 1 − 2 i ) i ! 2 4 i x i for | x | < 1 {\displaystyle {\sqrt {1+x}}=\sum _{i=0}^{\infty }{\frac {(-1)^{i}(2i)!}{(1-2i)i!^{2}4^{i}}}x^{i}\quad {\mbox{ for }}|x|<1\!} Miscellaneous: [1] ∑ i = 0 ∞ ( i + n i ) x i = 1 ( 1 − x ) n + 1 {\displaystyle \sum _{i=0}^{\infty }{i+n \choose i}x^{i}={\frac {1}{(1-x)^{n+1}}}} [1] ∑ i = 0 ∞ 1 i + 1 ( 2 i i ) x i = 1 2 x ( 1 − 1 − 4 x ) {\displaystyle \sum _{i=0}^{\infty }{\frac {1}{i+1}}{2i \choose i}x^{i}={\frac {1}{2x}}(1-{\sqrt {1-4x}})} [1] ∑ i = 0 ∞ ( 2 i i ) x i = 1 1 − 4 x {\displaystyle \sum _{i=0}^{\infty }{2i \choose i}x^{i}={\frac {1}{\sqrt {1-4x}}}} [1] ∑ i = 0 ∞ ( 2 i + n i ) x i = 1 1 − 4 x ( 1 − 1 − 4 x 2 x ) n {\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}} Remove adsBinomial coefficients ∑ i = 0 n ( n i ) = 2 n {\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n}} ∑ i = 0 n ( n i ) a ( n − i ) b i = ( a + b ) n {\displaystyle \sum _{i=0}^{n}{n \choose i}a^{(n-i)}b^{i}=(a+b)^{n}} ∑ i = 0 n ( − 1 ) i ( n i ) = 0 {\displaystyle \sum _{i=0}^{n}(-1)^{i}{n \choose i}=0} ∑ i = 0 n ( i k ) = ( n + 1 k + 1 ) {\displaystyle \sum _{i=0}^{n}{i \choose k}={n+1 \choose k+1}} ∑ i = 0 n ( k + i i ) = ( k + n + 1 n ) {\displaystyle \sum _{i=0}^{n}{k+i \choose i}={k+n+1 \choose n}} ∑ i = 0 r ( r i ) ( s n − i ) = ( r + s n ) {\displaystyle \sum _{i=0}^{r}{r \choose i}{s \choose n-i}={r+s \choose n}} Remove adsTrigonometric functions Sums of sines and cosines arise in Fourier series. ∑ i = 1 n sin ( i π n ) = 0 {\displaystyle \sum _{i=1}^{n}\sin \left({\frac {i\pi }{n}}\right)=0} ∑ i = 1 n cos ( i π n ) = 0 {\displaystyle \sum _{i=1}^{n}\cos \left({\frac {i\pi }{n}}\right)=0} Remove adsUnclassified ∑ n = b + 1 ∞ b n 2 − b 2 = ∑ n = 1 2 b 1 2 n {\displaystyle \sum _{n=b+1}^{\infty }{\frac {b}{n^{2}-b^{2}}}=\sum _{n=1}^{2b}{\frac {1}{2n}}} Remove adsRelated pages Series (mathematics) List of integrals Summation Taylor series Binomial theorem Gregory's series NotesLoading content...ReferencesLoading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. 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