List Of Mathematical Series - Wikipedia

See also: Series (mathematics) § Examples of numerical series, and Summation

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

• Here, 0 0 {\displaystyle 0^{0}} is taken to have the value 1 {\displaystyle 1}
• { x } {\displaystyle \{x\}} denotes the fractional part of x {\displaystyle x}
• B n ( x ) {\displaystyle B_{n}(x)} is a Bernoulli polynomial.
• B n {\displaystyle B_{n}} is a Bernoulli number, and here, B 1 = − 1 2 . {\displaystyle B_{1}=-{\frac {1}{2}}.}
• E n {\displaystyle E_{n}} is an Euler number.
• ζ ( s ) {\displaystyle \zeta (s)} is the Riemann zeta function.
• Γ ( z ) {\displaystyle \Gamma (z)} is the gamma function.
• ψ n ( z ) {\displaystyle \psi _{n}(z)} is a polygamma function.
• Li s ⁡ ( z ) {\displaystyle \operatorname {Li} _{s}(z)} is a polylogarithm.
• ( n k ) {\displaystyle n \choose k} is binomial coefficient
• exp ⁡ ( x ) {\displaystyle \exp(x)} denotes exponential of x {\displaystyle x}

Sums of powers

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See Faulhaber's formula.

• ∑ k = 0 m k n − 1 = B n ( m + 1 ) − B n n {\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}}

The first few values are:

• ∑ k = 1 m k = m ( m + 1 ) 2 {\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}}
• ∑ k = 1 m k 2 = m ( m + 1 ) ( 2 m + 1 ) 6 = m 3 3 + m 2 2 + m 6 {\displaystyle \sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}}
• ∑ k = 1 m k 3 = [ m ( m + 1 ) 2 ] 2 = m 4 4 + m 3 2 + m 2 4 {\displaystyle \sum _{k=1}^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}}

See zeta constants.

• ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}

The first few values are:

• ζ ( 2 ) = ∑ k = 1 ∞ 1 k 2 = π 2 6 {\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} (the Basel problem)
• ζ ( 4 ) = ∑ k = 1 ∞ 1 k 4 = π 4 90 {\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}}
• ζ ( 6 ) = ∑ k = 1 ∞ 1 k 6 = π 6 945 {\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}}

Power series

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Low-order polylogarithms

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Finite sums:

• ∑ k = m n z k = z m − z n + 1 1 − z {\displaystyle \sum _{k=m}^{n}z^{k}={\frac {z^{m}-z^{n+1}}{1-z}}} (geometric series)
• ∑ k = 0 n z k = 1 − z n + 1 1 − z {\displaystyle \sum _{k=0}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}}
• ∑ k = 1 n z k = 1 − z n + 1 1 − z − 1 = z − z n + 1 1 − z {\displaystyle \sum _{k=1}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}-1={\frac {z-z^{n+1}}{1-z}}}
• ∑ k = 1 n k z k = z 1 − ( n + 1 ) z n + n z n + 1 ( 1 − z ) 2 {\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}}
• ∑ k = 1 n k 2 z k = z 1 + z − ( n + 1 ) 2 z n + ( 2 n 2 + 2 n − 1 ) z n + 1 − n 2 z n + 2 ( 1 − z ) 3 {\displaystyle \sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}}
• ∑ k = 0 n k m z k = ( z d d z ) m 1 − z n + 1 1 − z {\displaystyle \sum _{k=0}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}}

Infinite sums, valid for | z | < 1 {\displaystyle |z|<1} (see polylogarithm):

• Li n ⁡ ( z ) = ∑ k = 1 ∞ z k k n {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}

The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form:

• d d z Li n ⁡ ( z ) = Li n − 1 ⁡ ( z ) z {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}}
• Li 1 ⁡ ( z ) = ∑ k = 1 ∞ z k k = − ln ⁡ ( 1 − z ) {\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)}
• Li 0 ⁡ ( z ) = ∑ k = 1 ∞ z k = z 1 − z {\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}}
• Li − 1 ⁡ ( z ) = ∑ k = 1 ∞ k z k = z ( 1 − z ) 2 {\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}}
• Li − 2 ⁡ ( z ) = ∑ k = 1 ∞ k 2 z k = z ( 1 + z ) ( 1 − z ) 3 {\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}}
• Li − 3 ⁡ ( z ) = ∑ k = 1 ∞ k 3 z k = z ( 1 + 4 z + z 2 ) ( 1 − z ) 4 {\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}}
• Li − 4 ⁡ ( z ) = ∑ k = 1 ∞ k 4 z k = z ( 1 + z ) ( 1 + 10 z + z 2 ) ( 1 − z ) 5 {\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}}

Exponential function

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• ∑ k = 0 ∞ z k k ! = e z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}}
• ∑ k = 0 ∞ k z k k ! = z e z {\displaystyle \sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} (cf. mean of Poisson distribution)
• ∑ k = 0 ∞ k 2 z k k ! = ( z + z 2 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} (cf. second moment of Poisson distribution)
• ∑ k = 0 ∞ k 3 z k k ! = ( z + 3 z 2 + z 3 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}}
• ∑ k = 0 ∞ k 4 z k k ! = ( z + 7 z 2 + 6 z 3 + z 4 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}}
• ∑ k = 0 ∞ k n z k k ! = z d d z ∑ k = 0 ∞ k n − 1 z k k ! = e z T n ( z ) {\displaystyle \sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)}

where T n ( z ) {\displaystyle T_{n}(z)} is the Touchard polynomials.

Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship

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• ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) ! = sin ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z}
• ∑ k = 0 ∞ z 2 k + 1 ( 2 k + 1 ) ! = sinh ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z}
• ∑ k = 0 ∞ ( − 1 ) k z 2 k ( 2 k ) ! = cos ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z}
• ∑ k = 0 ∞ z 2 k ( 2 k ) ! = cosh ⁡ z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z}
• ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tan ⁡ z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tan z,|z|<{\frac {\pi }{2}}}
• ∑ k = 1 ∞ ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tanh ⁡ z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}}
• ∑ k = 0 ∞ ( − 1 ) k 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = cot ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi }
• ∑ k = 0 ∞ 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = coth ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi }
• ∑ k = 0 ∞ ( − 1 ) k − 1 ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csc ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi }
• ∑ k = 0 ∞ − ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csch ⁡ z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi }
• ∑ k = 0 ∞ ( − 1 ) k E 2 k z 2 k ( 2 k ) ! = sech ⁡ z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}}
• ∑ k = 0 ∞ E 2 k z 2 k ( 2 k ) ! = sec ⁡ z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}}
• ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k ( 2 k ) ! = ver ⁡ z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} (versine)
• ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k 2 ( 2 k ) ! = hav ⁡ z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z} [1] (haversine)
• ∑ k = 0 ∞ ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsin ⁡ z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1}
• ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsinh ⁡ z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1}
• ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 2 k + 1 = arctan ⁡ z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1}
• ∑ k = 0 ∞ z 2 k + 1 2 k + 1 = arctanh ⁡ z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1}
• ln ⁡ 2 + ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 k ) ! z 2 k 2 2 k + 1 k ( k ! ) 2 = ln ⁡ ( 1 + 1 + z 2 ) , | z | ≤ 1 {\displaystyle \ln 2+\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left(1+{\sqrt {1+z^{2}}}\right),|z|\leq 1}
• ∑ k = 2 ∞ ( k ⋅ arctanh ⁡ ( 1 k ) − 1 ) = 3 − ln ⁡ ( 4 π ) 2 {\displaystyle \sum _{k=2}^{\infty }\left(k\cdot \operatorname {arctanh} \left({\frac {1}{k}}\right)-1\right)={\frac {3-\ln(4\pi )}{2}}}

Modified-factorial denominators

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• ∑ k = 0 ∞ ( 4 k ) ! 2 4 k 2 ( 2 k ) ! ( 2 k + 1 ) ! z k = 1 − 1 − z z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1} [2]
• ∑ k = 0 ∞ 2 2 k ( k ! ) 2 ( k + 1 ) ( 2 k + 1 ) ! z 2 k + 2 = ( arcsin ⁡ z ) 2 , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1} [2]
• ∑ n = 0 ∞ ∏ k = 0 n − 1 ( 4 k 2 + α 2 ) ( 2 n ) ! z 2 n + ∑ n = 0 ∞ α ∏ k = 0 n − 1 [ ( 2 k + 1 ) 2 + α 2 ] ( 2 n + 1 ) ! z 2 n + 1 = e α arcsin ⁡ z , | z | ≤ 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(4k^{2}+\alpha ^{2})}{(2n)!}}z^{2n}+\sum _{n=0}^{\infty }{\frac {\alpha \prod _{k=0}^{n-1}[(2k+1)^{2}+\alpha ^{2}]}{(2n+1)!}}z^{2n+1}=e^{\alpha \arcsin {z}},|z|\leq 1}

Binomial coefficients

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• ( 1 + z ) α = ∑ k = 0 ∞ ( α k ) z k , | z | < 1 {\displaystyle (1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1} (see Binomial theorem § Newton's generalized binomial theorem)
• [3] ∑ k = 0 ∞ ( α + k − 1 k ) z k = 1 ( 1 − z ) α , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1}
• [3] ∑ k = 0 ∞ 1 k + 1 ( 2 k k ) z k = 1 − 1 − 4 z 2 z , | z | ≤ 1 4 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}} , generating function of the Catalan numbers
• [3] ∑ k = 0 ∞ ( 2 k k ) z k = 1 1 − 4 z , | z | < 1 4 {\displaystyle \sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}} , generating function of the Central binomial coefficients
• [3] ∑ k = 0 ∞ ( 2 k + α k ) z k = 1 1 − 4 z ( 1 − 1 − 4 z 2 z ) α , | z | < 1 4 {\displaystyle \sum _{k=0}^{\infty }{2k+\alpha \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}}\left({\frac {1-{\sqrt {1-4z}}}{2z}}\right)^{\alpha },|z|<{\frac {1}{4}}}
• ∑ k = 0 ∞ 1 ( N + k n ) = N ( n − 1 ) ( N n ) , N ≥ n {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{N+k \choose n}}={\frac {N}{(n-1){N \choose n}}},\quad N\geq n}

Harmonic numbers

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(See harmonic numbers, themselves defined H n = ∑ j = 1 n 1 j {\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}} , and H ( x ) {\displaystyle H(x)} generalized to the real numbers)

• ∑ k = 1 ∞ H k z k = − ln ⁡ ( 1 − z ) 1 − z , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1}
• ∑ k = 1 ∞ H k k + 1 z k + 1 = 1 2 [ ln ⁡ ( 1 − z ) ] 2 , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k+1}}z^{k+1}={\frac {1}{2}}\left[\ln(1-z)\right]^{2},\qquad |z|<1}
• ∑ k = 1 ∞ ( − 1 ) k − 1 H 2 k 2 k + 1 z 2 k + 1 = 1 2 arctan ⁡ z log ⁡ ( 1 + z 2 ) , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}H_{2k}}{2k+1}}z^{2k+1}={\frac {1}{2}}\arctan {z}\log {(1+z^{2})},\qquad |z|<1} [2]
• ∑ n = 0 ∞ ∑ k = 0 2 n ( − 1 ) k 2 k + 1 z 4 n + 2 4 n + 2 = 1 4 arctan ⁡ z log ⁡ 1 + z 1 − z , | z | < 1 {\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{2n}{\frac {(-1)^{k}}{2k+1}}{\frac {z^{4n+2}}{4n+2}}={\frac {1}{4}}\arctan {z}\log {\frac {1+z}{1-z}},\qquad |z|<1} [2]
• ∑ n = 0 ∞ x 2 n 2 ( n + x ) = x π 2 6 − H ( x ) {\displaystyle \sum _{n=0}^{\infty }{\frac {x^{2}}{n^{2}(n+x)}}=x{\frac {\pi ^{2}}{6}}-H(x)}

Binomial coefficients

[edit] Main article: Binomial coefficient

• ∑ k = 0 n ( n k ) = 2 n {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}
• ∑ k = 0 n ( n k ) 2 = ( 2 n n ) {\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}}
• ∑ k = 0 n ( − 1 ) k ( n k ) = 0 ,  where  n ≥ 1 {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ where }}n\geq 1}
• ∑ k = 0 n ( k m ) = ( n + 1 m + 1 ) {\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}}
• ∑ k = 0 n ( m + k − 1 k ) = ( n + m n ) {\displaystyle \sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}} (see Multiset)
• ∑ k = 0 n ( α k ) ( β n − k ) = ( α + β n ) , where   α + β ≥ n {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n},{\text{where}}\ \alpha +\beta \geq n} (see Vandermonde identity)
• ∑ A   ∈   P ( E ) 1 = 2 n , where  E  is a finite set, and card( E ) = n {\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}1=2^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}
• ∑ { ( A ,   B )   ∈   ( P ( E ) ) 2 A   ⊂   B 1 = 3 n , where  E  is a finite set, and card( E ) = n {\displaystyle \sum _{\begin{cases}(A,\ B)\ \in \ ({\mathcal {P}}(E))^{2}\\A\ \subset \ B\end{cases}}1=3^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}
• ∑ A   ∈   P ( E ) c a r d ( A ) = n 2 n − 1 , where  E  is a finite set, and card( E ) = n {\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}card(A)=n2^{n-1}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}

Trigonometric functions

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Sums of sines and cosines arise in Fourier series.

• ∑ k = 1 ∞ cos ⁡ ( k θ ) k = − 1 2 ln ⁡ ( 2 − 2 cos ⁡ θ ) = − ln ⁡ ( 2 sin ⁡ θ 2 ) , 0 < θ < 2 π {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta )=-\ln \left(2\sin {\frac {\theta }{2}}\right),0<\theta <2\pi }
• ∑ k = 1 ∞ sin ⁡ ( k θ ) k = π − θ 2 , 0 < θ < 2 π {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi }
• ∑ k = 1 ∞ ( − 1 ) k − 1 k cos ⁡ ( k θ ) = 1 2 ln ⁡ ( 2 + 2 cos ⁡ θ ) = ln ⁡ ( 2 cos ⁡ θ 2 ) , 0 ≤ θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\cos(k\theta )={\frac {1}{2}}\ln(2+2\cos \theta )=\ln \left(2\cos {\frac {\theta }{2}}\right),0\leq \theta <\pi }
• ∑ k = 1 ∞ ( − 1 ) k − 1 k sin ⁡ ( k θ ) = θ 2 , − π 2 ≤ θ ≤ π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\sin(k\theta )={\frac {\theta }{2}},-{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}}
• ∑ k = 1 ∞ cos ⁡ ( 2 k θ ) 2 k = − 1 2 ln ⁡ ( 2 sin ⁡ θ ) , 0 < θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(2k\theta )}{2k}}=-{\frac {1}{2}}\ln(2\sin \theta ),0<\theta <\pi }
• ∑ k = 1 ∞ sin ⁡ ( 2 k θ ) 2 k = π − 2 θ 4 , 0 < θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2k\theta )}{2k}}={\frac {\pi -2\theta }{4}},0<\theta <\pi }
• ∑ k = 0 ∞ cos ⁡ [ ( 2 k + 1 ) θ ] 2 k + 1 = 1 2 ln ⁡ ( cot ⁡ θ 2 ) , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\cot {\frac {\theta }{2}}\right),0<\theta <\pi }
• ∑ k = 0 ∞ sin ⁡ [ ( 2 k + 1 ) θ ] 2 k + 1 = π 4 , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi } ,[4]
• ∑ k = 1 ∞ sin ⁡ ( 2 π k x ) k = π ( 1 2 − { x } ) ,   x ∈ R {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}=\pi \left({\dfrac {1}{2}}-\{x\}\right),\ x\in \mathbb {R} }
• ∑ k = 1 ∞ sin ⁡ ( 2 π k x ) k 2 n − 1 = ( − 1 ) n ( 2 π ) 2 n − 1 2 ( 2 n − 1 ) ! B 2 n − 1 ( { x } ) ,   x ∈ R ,   n ∈ N {\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\sin \left(2\pi kx\right)}{k^{2n-1}}}=(-1)^{n}{\frac {(2\pi )^{2n-1}}{2(2n-1)!}}B_{2n-1}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} }
• ∑ k = 1 ∞ cos ⁡ ( 2 π k x ) k 2 n = ( − 1 ) n − 1 ( 2 π ) 2 n 2 ( 2 n ) ! B 2 n ( { x } ) ,   x ∈ R ,   n ∈ N {\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\cos \left(2\pi kx\right)}{k^{2n}}}=(-1)^{n-1}{\frac {(2\pi )^{2n}}{2(2n)!}}B_{2n}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} }
• B n ( x ) = − n ! 2 n − 1 π n ∑ k = 1 ∞ 1 k n cos ⁡ ( 2 π k x − π n 2 ) , 0 < x < 1 {\displaystyle B_{n}(x)=-{\frac {n!}{2^{n-1}\pi ^{n}}}\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}\cos \left(2\pi kx-{\frac {\pi n}{2}}\right),0<x<1} [5]
• ∑ k = 0 n sin ⁡ ( θ + k α ) = sin ⁡ ( n + 1 ) α 2 sin ⁡ ( θ + n α 2 ) sin ⁡ α 2 {\displaystyle \sum _{k=0}^{n}\sin(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\sin(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}
• ∑ k = 0 n cos ⁡ ( θ + k α ) = sin ⁡ ( n + 1 ) α 2 cos ⁡ ( θ + n α 2 ) sin ⁡ α 2 {\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}
• ∑ k = 1 n − 1 sin ⁡ π k n = cot ⁡ π 2 n {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}}
• ∑ k = 1 n − 1 sin ⁡ 2 π k n = 0 {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0}
• ∑ k = 0 n − 1 csc 2 ⁡ ( θ + π k n ) = n 2 csc 2 ⁡ ( n θ ) {\displaystyle \sum _{k=0}^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )} [6]
• ∑ k = 1 n − 1 csc 2 ⁡ π k n = n 2 − 1 3 {\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}}
• ∑ k = 1 n − 1 csc 4 ⁡ π k n = n 4 + 10 n 2 − 11 45 {\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}}

Roots of unity

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A n {\displaystyle n} 'th root of unity is a solution to the equation z n = 1 {\displaystyle z^{n}=1} and they can be written like:

z m = exp ⁡ ( i ⋅ 2 π m n ) , m ∈ { 0 , ⋯ , n − 1 } {\displaystyle z_{m}=\exp \left(i\cdot {\frac {2\pi m}{n}}\right),\quad m\in \{0,\cdots ,n-1\}}

The following summation identities hold:

∑ m ≠ 0 n − 1 1 1 − z m = n − 1 2 {\displaystyle \sum _{m\neq 0}^{n-1}{\frac {1}{1-z_{m}}}={\frac {n-1}{2}}} ∑ m ≠ 0 n − 1 1 ( 1 − z m ) 2 = ( n − 1 ) ( 5 − n ) 12 {\displaystyle \sum _{m\neq 0}^{n-1}{\frac {1}{(1-z_{m})^{2}}}={\frac {(n-1)(5-n)}{12}}}

Let k {\displaystyle k} be an integer 0 < k < n {\displaystyle 0<k<n} then we also got:

∑ m = 0 n − 1 z m k = 0 {\displaystyle \sum _{m=0}^{n-1}z_{m}^{k}=0} ∑ m = 0 n − 1 ( x − z m ) k = n ⋅ x k {\displaystyle \sum _{m=0}^{n-1}(x-z_{m})^{k}=n\cdot x^{k}} ∑ m ≠ 0 n − 1 ( z m ) k 1 − z m = 2 k − n − 1 2 {\displaystyle \sum _{m\neq 0}^{n-1}{\frac {(z_{m})^{k}}{1-z_{m}}}={\frac {2k-n-1}{2}}} ∑ m ≠ 0 n − 1 ( z m ) k ( 1 − z m ) 2 = 6 k ( n + 2 − k ) − ( n + 1 ) ( n + 5 ) 12 {\displaystyle \sum _{m\neq 0}^{n-1}{\frac {(z_{m})^{k}}{(1-z_{m})^{2}}}={\frac {6k(n+2-k)-(n+1)(n+5)}{12}}}

Rational functions

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• ∑ n = a + 1 ∞ a n 2 − a 2 = 1 2 H 2 a {\displaystyle \sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}} [7]
• ∑ n = 0 ∞ 1 n 2 + a 2 = 1 + a π coth ⁡ ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}}
• ∑ n = 0 ∞ ( − 1 ) n n 2 + a 2 = 1 + a π csch ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n^{2}+a^{2}}}={\frac {1+a\pi \;{\text{csch}}(a\pi )}{2a^{2}}}}
• ∑ n = 0 ∞ ( 2 n + 1 ) ( − 1 ) n ( 2 n + 1 ) 2 + a 2 = π 4 sech ( a π 2 ) {\displaystyle \sum _{n=0}^{\infty }{\frac {(2n+1)(-1)^{n}}{(2n+1)^{2}+a^{2}}}={\frac {\pi }{4}}{\text{sech}}\left({\frac {a\pi }{2}}\right)}
• ∑ n = 0 ∞ 1 n 4 + 4 a 4 = 1 8 a 4 + π ( sinh ⁡ ( 2 π a ) + sin ⁡ ( 2 π a ) ) 8 a 3 ( cosh ⁡ ( 2 π a ) − cos ⁡ ( 2 π a ) ) {\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1}{8a^{4}}}+{\dfrac {\pi (\sinh(2\pi a)+\sin(2\pi a))}{8a^{3}(\cosh(2\pi a)-\cos(2\pi a))}}}
• An infinite series of any rational function of n {\displaystyle n} can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition,[8] as explained here. This fact can also be applied to finite series of rational functions, allowing the result to be computed in constant time even when the series contains a large number of terms.

Exponential function

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• 1 p ∑ n = 0 p − 1 exp ⁡ ( 2 π i n 2 q p ) = e π i / 4 2 q ∑ n = 0 2 q − 1 exp ⁡ ( − π i n 2 p 2 q ) {\displaystyle \displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)} (see the Landsberg–Schaar relation)
• ∑ n = − ∞ ∞ e − π n 2 = π 4 Γ ( 3 4 ) {\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}}

Numeric series

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These numeric series can be found by plugging in numbers from the series listed above.

Alternating harmonic series

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• ∑ k = 1 ∞ ( − 1 ) k + 1 k = 1 1 − 1 2 + 1 3 − 1 4 + ⋯ = ln ⁡ 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots =\ln 2}
• ∑ k = 1 ∞ ( − 1 ) k + 1 2 k − 1 = 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2k-1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}}

Alternating arithmetic series

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Let S ( a , b ) {\displaystyle S(a,b)} be defined as:

S ( a , b ) = ∑ k = 0 ∞ ( − 1 ) k a k + b {\displaystyle S(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+b}}}

where a , b > 0 {\displaystyle a,b>0} are positive whole numbers. Then if g c d ( a , b ) = c {\displaystyle gcd(a,b)=c} we can write a = c α {\displaystyle a=c\alpha } and b = c β {\displaystyle b=c\beta } , where g c d ( α , β ) = 1 {\displaystyle gcd(\alpha ,\beta )=1} , and get:

S ( a , b ) = S ( c α , c β ) = ∑ k = 0 ∞ ( − 1 ) k c d k + c e = 1 c ∑ k = 0 ∞ ( − 1 ) k α k + β = S ( α , β ) c {\displaystyle S(a,b)=S(c\alpha ,c\beta )=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{cdk+ce}}={\frac {1}{c}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\alpha k+\beta }}={\frac {S(\alpha ,\beta )}{c}}}

Now if b > a {\displaystyle b>a} we can, per Euclid's division lemma, write b = c a + d {\displaystyle b=ca+d} where a > d > 0 {\displaystyle a>d>0} and then

S ( a , b ) = S ( a , c a + d ) = ∑ k = 0 ∞ ( − 1 ) k a k + c a + d = ∑ k = 0 ∞ ( − 1 ) k a ( k + c ) + d = ∑ k = c ∞ ( − 1 ) k − c a k + d {\displaystyle S(a,b)=S(a,ca+d)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+ca+d}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{a(k+c)+d}}=\sum _{k=c}^{\infty }{\frac {(-1)^{k-c}}{ak+d}}}

where we now can add the remaining rows back and subtract them to give us:

S ( a , b ) = ( − 1 ) c ( ∑ k = 0 ∞ ( − 1 ) k a k + d − ∑ k = 0 c − 1 ( − 1 ) k a k + d ) = ( − 1 ) c ( S ( a , d ) − ∑ k = 0 c − 1 ( − 1 ) k a k + d ) {\displaystyle S(a,b)=(-1)^{c}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+d}}-\sum _{k=0}^{c-1}{\frac {(-1)^{k}}{ak+d}}\right)=(-1)^{c}\left(S(a,d)-\sum _{k=0}^{c-1}{\frac {(-1)^{k}}{ak+d}}\right)}

what that means is that all the infinite choices of a {\displaystyle a} and b {\displaystyle b} can essentially be boiled down to the cases where g c d ( a , b ) = 1 {\displaystyle gcd(a,b)=1} and a > b > 0 {\displaystyle a>b>0} . If we assume those two things we can then write:

S ( a , b ) = 1 a ( π 2 sin ⁡ ( π b a ) − 2 ∑ m = 0 ⌊ a 2 ⌋ cos ⁡ ( π ( 2 m + 1 ) b a ) ln ⁡ ( sin ⁡ ( π 2 m + 1 2 a ) ) ) {\displaystyle S(a,b)={\frac {1}{a}}\left({\frac {\pi }{2\sin \left({\frac {\pi b}{a}}\right)}}-2\sum _{m=0}^{\lfloor {\frac {a}{2}}\rfloor }\cos \left(\pi {\frac {(2m+1)b}{a}}\right)\ln \left(\sin \left(\pi {\frac {2m+1}{2a}}\right)\right)\right)}

and in the case of using a negative sign instead:

S − ( a , b ) = ∑ k = 0 ∞ ( − 1 ) k a k − b {\displaystyle S_{-}(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak-b}}}

the same two rules apply from above apply and then we can do the following for the case with a > b > 0 {\displaystyle a>b>0} (since a > a − b > 0 {\displaystyle a>a-b>0} ):

S − ( a , b ) = ∑ k = 0 ∞ ( − 1 ) k a k − b = − 1 b + ∑ k = 0 ∞ ( − 1 ) k + 1 a ( k + 1 ) − b = − 1 b − ∑ k = 0 ∞ ( − 1 ) k a k + ( a − b ) = − S ( a , a − b ) − 1 b {\displaystyle S_{-}(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak-b}}=-{\frac {1}{b}}+\sum _{k=0}^{\infty }{\frac {(-1)^{k+1}}{a(k+1)-b}}=-{\frac {1}{b}}-\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+(a-b)}}=-S(a,a-b)-{\frac {1}{b}}}

Let us test out the formula: S ( 3 , 2 ) = 1 3 ( π 2 sin ⁡ ( 2 π 3 ) − 2 ( cos ⁡ ( 2 π 3 ) ln ⁡ ( sin ⁡ ( π 6 ) ) + cos ⁡ ( 2 π ) ln ⁡ ( sin ⁡ ( π 2 ) ) ) ) = π 3 3 − ln ⁡ ( 2 ) 3 {\displaystyle S(3,2)={\frac {1}{3}}\left({\frac {\pi }{2\sin \left({\frac {2\pi }{3}}\right)}}-2\left(\cos \left({\frac {2\pi }{3}}\right)\ln \left(\sin \left({\frac {\pi }{6}}\right)\right)+\cos \left(2\pi \right)\ln \left(\sin \left({\frac {\pi }{2}}\right)\right)\right)\right)={\frac {\pi }{3{\sqrt {3}}}}-{\frac {\ln(2)}{3}}}

Sum of reciprocal of factorials

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• ∑ k = 0 ∞ 1 k ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + ⋯ = e {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+\cdots =e}
• ∑ k = 0 ∞ 1 ( 2 k ) ! = 1 0 ! + 1 2 ! + 1 4 ! + 1 6 ! + 1 8 ! + ⋯ = 1 2 ( e + 1 e ) = cosh ⁡ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k)!}}={\frac {1}{0!}}+{\frac {1}{2!}}+{\frac {1}{4!}}+{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots ={\frac {1}{2}}\left(e+{\frac {1}{e}}\right)=\cosh 1}
• ∑ k = 0 ∞ 1 ( 3 k ) ! = 1 0 ! + 1 3 ! + 1 6 ! + 1 9 ! + 1 12 ! + ⋯ = 1 3 ( e + 2 e cos ⁡ 3 2 ) {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(3k)!}}={\frac {1}{0!}}+{\frac {1}{3!}}+{\frac {1}{6!}}+{\frac {1}{9!}}+{\frac {1}{12!}}+\cdots ={\frac {1}{3}}\left(e+{\frac {2}{\sqrt {e}}}\cos {\frac {\sqrt {3}}{2}}\right)}
• ∑ k = 0 ∞ 1 ( 4 k ) ! = 1 0 ! + 1 4 ! + 1 8 ! + 1 12 ! + 1 16 ! + ⋯ = 1 2 ( cos ⁡ 1 + cosh ⁡ 1 ) {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(4k)!}}={\frac {1}{0!}}+{\frac {1}{4!}}+{\frac {1}{8!}}+{\frac {1}{12!}}+{\frac {1}{16!}}+\cdots ={\frac {1}{2}}\left(\cos 1+\cosh 1\right)}

Trigonometry and π

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• ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! = 1 1 ! − 1 3 ! + 1 5 ! − 1 7 ! + 1 9 ! + ⋯ = sin ⁡ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}={\frac {1}{1!}}-{\frac {1}{3!}}+{\frac {1}{5!}}-{\frac {1}{7!}}+{\frac {1}{9!}}+\cdots =\sin 1}
• ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! = 1 0 ! − 1 2 ! + 1 4 ! − 1 6 ! + 1 8 ! + ⋯ = cos ⁡ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}={\frac {1}{0!}}-{\frac {1}{2!}}+{\frac {1}{4!}}-{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots =\cos 1}
• ∑ k = 1 ∞ 1 k 2 + 1 = 1 2 + 1 5 + 1 10 + 1 17 + ⋯ = 1 2 ( π coth ⁡ π − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}+1}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \coth \pi -1)}
• ∑ k = 1 ∞ ( − 1 ) k k 2 + 1 = − 1 2 + 1 5 − 1 10 + 1 17 + ⋯ = 1 2 ( π csch ⁡ π − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{2}+1}}=-{\frac {1}{2}}+{\frac {1}{5}}-{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \operatorname {csch} \pi -1)}
• 3 + 4 2 × 3 × 4 − 4 4 × 5 × 6 + 4 6 × 7 × 8 − 4 8 × 9 × 10 + ⋯ = π {\displaystyle 3+{\frac {4}{2\times 3\times 4}}-{\frac {4}{4\times 5\times 6}}+{\frac {4}{6\times 7\times 8}}-{\frac {4}{8\times 9\times 10}}+\cdots =\pi }

Reciprocal of tetrahedral numbers

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• ∑ k = 1 ∞ 1 T e k = 1 1 + 1 4 + 1 10 + 1 20 + 1 35 + ⋯ = 3 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{Te_{k}}}={\frac {1}{1}}+{\frac {1}{4}}+{\frac {1}{10}}+{\frac {1}{20}}+{\frac {1}{35}}+\cdots ={\frac {3}{2}}}

Where T e n = ∑ k = 1 n T k {\displaystyle Te_{n}=\sum _{k=1}^{n}T_{k}}

Exponential and logarithms

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• ∑ k = 0 ∞ 1 ( 2 k + 1 ) ( 2 k + 2 ) = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + 1 7 × 8 + 1 9 × 10 + ⋯ = ln ⁡ 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)(2k+2)}}={\frac {1}{1\times 2}}+{\frac {1}{3\times 4}}+{\frac {1}{5\times 6}}+{\frac {1}{7\times 8}}+{\frac {1}{9\times 10}}+\cdots =\ln 2}
• ∑ k = 1 ∞ 1 2 k k = 1 2 + 1 8 + 1 24 + 1 64 + 1 160 + ⋯ = ln ⁡ 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}k}}={\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{24}}+{\frac {1}{64}}+{\frac {1}{160}}+\cdots =\ln 2}
• ∑ k = 1 ∞ ( − 1 ) k + 1 2 k k + ∑ k = 1 ∞ ( − 1 ) k + 1 3 k k = ( 1 2 + 1 3 ) − ( 1 8 + 1 18 ) + ( 1 24 + 1 81 ) − ( 1 64 + 1 324 ) + ⋯ = ln ⁡ 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2^{k}k}}+\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{3^{k}k}}={\Bigg (}{\frac {1}{2}}+{\frac {1}{3}}{\Bigg )}-{\Bigg (}{\frac {1}{8}}+{\frac {1}{18}}{\Bigg )}+{\Bigg (}{\frac {1}{24}}+{\frac {1}{81}}{\Bigg )}-{\Bigg (}{\frac {1}{64}}+{\frac {1}{324}}{\Bigg )}+\cdots =\ln 2}
• ∑ k = 1 ∞ 1 3 k k + ∑ k = 1 ∞ 1 4 k k = ( 1 3 + 1 4 ) + ( 1 18 + 1 32 ) + ( 1 81 + 1 192 ) + ( 1 324 + 1 1024 ) + ⋯ = ln ⁡ 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{3^{k}k}}+\sum _{k=1}^{\infty }{\frac {1}{4^{k}k}}={\Bigg (}{\frac {1}{3}}+{\frac {1}{4}}{\Bigg )}+{\Bigg (}{\frac {1}{18}}+{\frac {1}{32}}{\Bigg )}+{\Bigg (}{\frac {1}{81}}+{\frac {1}{192}}{\Bigg )}+{\Bigg (}{\frac {1}{324}}+{\frac {1}{1024}}{\Bigg )}+\cdots =\ln 2}
• ∑ k = 1 ∞ 1 n k k = ln ⁡ ( n n − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{n^{k}k}}=\ln \left({\frac {n}{n-1}}\right)} , that is ∀ n > 1 {\displaystyle \forall n>1}

See also

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• Series (mathematics)
• List of integrals
• Summation § Identities
• Taylor series
• Binomial theorem
• Gregory's series
• On-Line Encyclopedia of Integer Sequences

Notes

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1. ^ Weisstein, Eric W. "Haversine". MathWorld. Wolfram Research, Inc. Archived from the original on 2005-03-10. Retrieved 2015-11-06.
2. ^ a b c d Wilf, Herbert R. (1994). generatingfunctionology (PDF). Academic Press, Inc.
3. ^ a b c d "Theoretical computer science cheat sheet" (PDF).
4. ^ Calculate the Fourier expansion of the function f ( x ) = π 4 {\displaystyle f(x)={\frac {\pi }{4}}} on the interval 0 < x < π {\displaystyle 0<x<\pi } :
   • π 4 = ∑ n = 0 ∞ c n sin ⁡ [ n x ] + d n cos ⁡ [ n x ] {\displaystyle {\frac {\pi }{4}}=\sum _{n=0}^{\infty }c_{n}\sin[nx]+d_{n}\cos[nx]}
   ⇒ { c n = { 1 n ( n  odd ) 0 ( n  even ) d n = 0 ( ∀ n ) {\displaystyle \Rightarrow {\begin{cases}c_{n}={\begin{cases}{\frac {1}{n}}\quad (n{\text{ odd}})\\0\quad (n{\text{ even}})\end{cases}}\\d_{n}=0\quad (\forall n)\end{cases}}}
5. ^ "Bernoulli polynomials: Series representations (subsection 06/02)". Wolfram Research. Retrieved 2 June 2011.
6. ^ Hofbauer, Josef. "A simple proof of 1 + 1/22 + 1/32 + ··· = π2/6 and related identities" (PDF). Retrieved 2 June 2011.
7. ^ Sondow, Jonathan; Weisstein, Eric W. "Riemann Zeta Function (eq. 52)". MathWorld—A Wolfram Web Resource.
8. ^ Abramowitz, Milton; Stegun, Irene (1964). "6.4 Polygamma functions". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Courier Corporation. p. 260. ISBN 0-486-61272-4. {{cite book}}: ISBN / Date incompatibility (help)

References

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• Many books with a list of integrals also have a list of series.