Here, 0 0 } is taken to have the value 1 { x } } denotes the fractional part of x B n ( x ) (x)} is a Bernoulli polynomial.
Di sini, 0 0 } dianggap memiliki nilai 1 B n ( x ) (x)} adalah polinomial Bernoulli. B n } adalah bilangan Bernoulli, dan di sini, B 1− 1 2-{\frac {1}{2}}} , E n } adalah bilangan Euler.
ζ ( s ) is the Riemann zeta function.
ζ ( s ) adalah fungsi zeta Riemann.
( n k ) is binomial coefficient exp ⁡ ( x ) denotes exponential of x See Faulhaber's formula.
( n k ) {k}}} adalah koefisien binomial exp ⁡ ( x ) melambangkan eksponensial dari x Lihat rumus Faulhaber ∑ k0 m k n − 1B n ( m + 1 )
∑ k0 m k n − 1B n ( m + 1 ) − B n n ^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} The first few values are: ∑ k1 m km ( m + 1 ) 2 ^{m}k={\frac {m(m+1)}{2}}} ∑ k1 m k 2m ( m + 1 ) ( 2 m + 1 ) 6m 3 3 + m 2 2 + m 6 ^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}} ∑ k1 m k 3[ m ( m + 1 ) 2 ] 2m 4 4 + m 3 2 + m 2 4 ^{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.
− B n n ^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} Beberapa nilai pertamanya adalahː ∑ k1 m km ( m + 1 ) 2 ^{m}k={\frac {m(m+1)}{2}}} ∑ k1 m k 2m ( m + 1 ) ( 2 m + 1 ) 6m 3 3 + m 2 2 + m 6 ^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}} ∑ k1 m k 3[ m ( m + 1 ) 2 ] 2m 4 4 + m 3 2 + m 2 4 ^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}} Lihat konstanta zeta.
∑ k1 ∞ 1 k 2 n( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! ^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} The first few values are: ζ ( 2 )
∑ k1 ∞ 1 k 2 n( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! ^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} Beberapa nilai pertamanya adalahː ζ ( 2 )∑ k1 ∞ 1 k 2π 2 6 ^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} (Masalah Basel) ζ ( 4 )
∑ k1 ∞ 1 k 2π 2 6 ^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} (the Basel problem) ζ ( 4 )
∑ k1 ∞ 1 k 6π 6 945 ^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}}
∑ k1 ∞ z k k n _{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:
∑ k1 ∞ z k k n _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}} Berikut ini adalah sebuah sifat yang berguna untuk menghitung polilogaritma urutan bilangan bulat rendah secara rekursif dalam bentuk tertutup:
Li n − 1 ⁡ ( z ) z }{\mathrm {d} z}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}} Li 1 ⁡ ( z )∑ k1 ∞ z k k− ln ⁡ ( 1 − z ) _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)} Li 0 ⁡ ( z )
Li n − 1 ⁡ ( z ) z {dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}} Li 1 ⁡ ( z )∑ k1 ∞ z k k− ln ⁡ ( 1 − z ) _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)} Li 0 ⁡ ( z )
z e z ^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} (cf. mean of Poisson distribution) ∑ k0 ∞ k 2 z k k !
z e z ^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} (bandingkan rata-rata distribusi Poisson) ∑ k0 ∞ k 2 z k k !
( z + z 2 ) e z ^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} (cf. second moment of Poisson distribution) ∑ k0 ∞ k 3 z k k !
( z + z 2 ) e z ^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} (bandingkan momen kedua distribusi Poisson) ∑ k0 ∞ k 3 z k k !
e z T n ( z ) ^{\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 ) (z)} is the Touchard polynomials.
e z T n ( z ) ^{\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)} dengan T n ( z ) (z)} adalah polinomial Touchard.
sec ⁡ z , | z | < π 2 ^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}} ∑ k1 ∞ ( − 1 ) k − 1 z 2 k ( 2 k ) !ver ⁡ z ^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} (versine) ∑ k1 ∞ ( − 1 ) k − 1 z 2 k 2 ( 2 k ) !
sec ⁡ z , | z | < π 2 ^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}} ∑ k1 ∞ ( − 1 ) k − 1 z 2 k ( 2 k ) !ver ⁡ z ^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} (versinus) ∑ k1 ∞ ( − 1 ) k − 1 z 2 k 2 ( 2 k ) !
hav ⁡ z ^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z} (haversine) ∑ k0 ∞ ( 2 k ) !
hav ⁡ z ^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z} (haversinus) ∑ k0 ∞ ( 2 k ) !
z 2 k 2 2 k + 1 k ( k ! ) 2ln ⁡ ( 1 + 1 + z 2 ) , | z | ≤ 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} ∑ k2 ∞ ( k ⋅ arctanh ⁡ ( 1 k )
z 2 k 2 2 k + 1 k ( k ! ) 2ln ⁡ ( 1 + 1 + z 2 ) , | z | ≤ 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} ∑ k0 ∞ ( 4 k ) ! 2 4 k 2 ( 2 k ) ! ( 2 k + 1 ) !
∑ k0 ∞ ( 4 k ) ! 2 4 k 2 ( 2 k ) ! ( 2 k + 1 ) ! z k1 − 1 − z z , | z | < 1 ^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1} ∑ k0 ∞ 2 2 k ( k ! ) 2 ( k + 1 ) ( 2 k + 1 ) !
z k1 − 1 − z z , | z | < 1 ^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1} ∑ k0 ∞ 2 2 k ( k ! )
z 2 k + 2( arcsin ⁡ z ) 2 , | z | ≤ 1 ^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1} ∑ n0 ∞ ∏ k0 n − 1 ( 4 k 2 + α 2 ) ( 2 n ) ! z 2 n + ∑ n0 ∞ α ∏ k0 n − 1 [ ( 2 k + 1 ) 2 + α 2 ] ( 2 n + 1 ) !
z 2 k + 2( arcsin ⁡ z ) 2 , | z | ≤ 1 ^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1} ∑ n0 ∞ ∏ k0 n − 1 ( 4 k 2 + α 2 ) ( 2 n ) !
ln ⁡ ( 2 cos ⁡ θ 2 ) , 0 ≤ θ < π ^{\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 } ∑ k1 ∞ ( − 1 ) k − 1 k sin ⁡ ( k θ )θ 2 , − π 2 ≤ θ ≤ π 2 ^{\infty }{\frac {(-1)^{k-1}}{k}}\sin(k\theta )={\frac {\theta }{2}},-{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}} ∑ k1 ∞ cos ⁡ ( 2 k θ ) 2 k− 1 2 ln ⁡ ( 2 sin ⁡ θ ) , 0 < θ < π ^{\infty }{\frac {\cos(2k\theta )}{2k}}=-{\frac {1}{2}}\ln(2\sin \theta ),0<\theta <\pi } ∑ k1 ∞ sin ⁡ ( 2 k θ ) 2 kπ − 2 θ 4 , 0 < θ < π ^{\infty }{\frac {\sin(2k\theta )}{2k}}={\frac {\pi -2\theta }{4}},0<\theta <\pi } ∑ k0 ∞ cos ⁡ [ ( 2 k + 1 ) θ ] 2 k + 11 2 ln ⁡ ( cot ⁡ θ 2 ) , 0 < θ < π ^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\cot {\frac {\theta }{2}}\right),0<\theta <\pi } ∑ k0 ∞ sin ⁡ [ ( 2 k + 1 ) θ ] 2 k + 1π 4 , 0 < θ < π ^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi } , ∑ k1 ∞ sin ⁡ ( 2 π k x ) kπ ( 1 2 − { x } ) , x ∈ R ^{\infty }{\frac {\sin(2\pi kx)}{k}}=\pi \left({\dfrac {1}{2}}-\{x\}\right),\ x\in \mathbb {R} } ∑ k1 ∞ sin ⁡ ( 2 π k x )
kπ − θ 2 , 0 < θ < 2 π ^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi } ∑ k1 ∞ cos ⁡ ( k θ ) k− 1 2 ln ⁡ ( 2 − 2 cos ⁡ θ ) , θ ∈ R ^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} } ∑ k0 ∞ sin ⁡ [ ( 2 k + 1 ) θ ] 2 k + 1π 4 , 0 < θ < π ^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi } , B n ( x )− n ! 2 n − 1 π n ∑ k1 ∞ 1 k n cos ⁡ ( 2 π k x − π n 2 ) , 0 < x < 1 (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} ∑ k0 n sin ⁡ ( θ + k α )sin ⁡ ( n + 1 ) α 2 sin ⁡ ( θ + n α 2 ) sin ⁡ α 2 ^{n}\sin(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\sin(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}} ∑ k0 n cos ⁡ ( θ + k α )sin ⁡ ( n + 1 ) α 2 cos ⁡ ( θ + n α 2 ) sin ⁡ α 2 ^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}} ∑ k1 n − 1 sin ⁡ π k ncot ⁡ π 2 n ^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}} ∑ k1 n − 1 sin ⁡ 2 π k n0 ^{n-1}\sin {\frac {2\pi k}{n}}=0} ∑ k0 n − 1 csc 2 ⁡ ( θ + π k n )n 2 csc 2 ⁡ ( n θ ) ^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )} ∑ k1 n − 1 csc 2 ⁡ π k nn 2 − 1 3 ^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}} ∑ k1 n − 1 csc 4 ⁡ π k nn 4 + 10 n 2 − 11 45 ^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}} ∑ na + 1 ∞ a n 2 − a 21 2 H 2 a ^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}} ∑ n0 ∞ 1 n 2 + a 21 + a π coth ⁡ ( a π ) 2 a 2 ^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}} ∑ n0 ∞ 1 n 4 + 4 a 41 8 a 4 + π ( sinh ⁡ ( 2 π a ) + sin ⁡ ( 2 π a ) ) 8 a 3 ( cosh ⁡ ( 2 π a ) − cos ⁡ ( 2 π a ) ) ^{\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))}}} Suatu deret tak terhingga dari setiap fungsi rasional n dapat direduksi menjadi suatu deret terhingga dari fungsi poligamma, dengan menggunakan dekomposisi pecahan parsial.
k 1 − z m2 k − n − 1 2 ^{n-1}{\frac {(z_{m})^{k}}{1-z_{m}}}={\frac {2k-n-1}{2}}} ∑ m ≠ 0 n − 1 ( z m ) k ( 1 − z m ) 26 k ( n + 2 − k )
m 1 − z n + 1 1 − z ^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}} Penjumlahan tak terhingga, sah untuk |
1 p ∑ n0 p − 1 exp ⁡ ( 2 π i n 2 q p )e π i / 4 2 q ∑ n0 2 q − 1 exp ⁡ ( − π i n 2 p 2 q ) {\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 2π 4 Γ ( 3 4 ) ^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}} These numeric series can be found by plugging in numbers from the series listed above.
e π i / 4 2 q ∑ n0 2 q − 1 exp ⁡ ( − π i n 2 p 2 q ) {\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)} (lihat relasi Landsberg–Schaar) ∑ n− ∞ ∞ e − π n 2π 4 Γ ( 3 4 ) ^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}} Banyak buku-buku dengan sebuah daftar integral juga memiliki sebuah daftar deret.
k + 1 k1 1 − 1 2 + 1 3 − 1 4 + ⋯ln ⁡ 2 ^{\infty }{\frac {(-1)^{k+1}}{k}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots\ln 2} ∑ k1 ∞ ( − 1 ) k + 1 2 k − 11 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯π 4 ^{\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}}}
∑ k1 ∞ 1 k 2 n( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! ^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} Beberapa nilai pertamanya adalahː ζ ( 2 )∑ k1 ∞ 1 k 2π 2 6 ^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} (Masalah Basel) ζ ( 4 )
∑ k0 ∞ ( − 1 ) k c d k + c e1 c ∑ k0 ∞ ( − 1 ) k α k + βS ( α , β ) c ^{\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 we can, per Euclid's division lemma, write bc a + d where a > d > 0 and then S ( a , b )S ( a , c a + d )
0 , ^{n}(-1)^{k}{n \choose k}=0,} dengan n ≥ 1 ∑ k0 n ( k m )( n + 1 m + 1 ) ^{n}{k \choose m}={n+1 \choose m+1}} ∑ k0 n ( m + k − 1 k )( n + m n ) ^{n}{m+k-1 \choose k}={n+m \choose n}} (lihat multihimpunan) ∑ k0 n ( α k ) ( β n − k )( α + β n ) ^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}} (lihat identitas Vandermonde) Penjumlahan fungsi sinus dan kosinus muncul dalam deret Fourier.
∑ k0 ∞ ( − 1 ) k a k + c a + d∑ k0 ∞ ( − 1 )
ln ⁡ 2 + ∑ k1 ∞ ( − 1 ) k − 1 ( 2 k ) !
k a ( k + c ) + d∑ kc ∞ ( − 1 ) k − c a k + d ^{\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 ( ∑ k0 ∞ ( − 1 )
− B n n ^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} Beberapa nilai pertamanya adalahː ∑ k1 m km ( m + 1 ) 2 ^{m}k={\frac {m(m+1)}{2}}} ∑ k1 m k 2m ( m + 1 ) ( 2 m + 1 ) 6m 3 3 + m 2 2 + m 6 ^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}} ∑ k1 m k 3[ m ( m + 1 ) 2 ] 2m 4 4 + m 3 2 + m 2 4 ^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}} Lihat konstanta zeta.
( − 1 ) c ( S ( a , d ) − ∑ k0 c − 1 ( − 1 ) k a k + d ) \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 and b can essentially be boiled down to the cases where g c d ( a , b )
− B n n ^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} Beberapa nilai pertamanya adalahː ∑ k1 m km ( m + 1 ) 2 ^{m}k={\frac {m(m+1)}{2}}} ∑ k1 m k 2m ( m + 1 ) ( 2 m + 1 ) 6m 3 3 + m 2 2 + m 6 ^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}} ∑ k1 m k 3[ m ( m + 1 ) 2 ] 2m 4 4 + m 3 2 + m 2 4 ^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}} Lihat konstanta zeta.
k ( 2 k + 1 ) !
2 ( k + 1 ) ( 2 k + 1 ) !
+ 1 9 ! + ⋯sin ⁡ 1 ^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}={\frac {1}{1!}}-{\frac {1}{3!}}+{\frac {1}{5!}}-{\frac {1}{7!}}+{\frac {1}{9!}}+\cdots\sin 1} ∑ k0 ∞ ( − 1 ) k ( 2 k ) !
sin ⁡ z ^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z} ∑ k0 ∞ z 2 k + 1 ( 2 k + 1 ) !