Цей список містить формули для рядів та сум
∑ k = 0 n k s = ( n + 1 ) s + 1 s + 1 + ∑ k = 1 s B k s − k + 1 ( s k ) ( n + 1 ) s − k + 1 {\displaystyle \sum _{k=0}^{n}k^{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}} формула Фольхабера [en] Значення при n = 1 , 2 , … , 12 : {\displaystyle n=1,2,\dots ,12:}
∑ k = 1 n k = n ( n + 1 ) 2 = ( n + 1 ) 3 − n 3 − 1 6 {\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}={\frac {(n+1)^{3}-n^{3}-1}{6}}\,\!} ∑ k = p q k = p + ( p + 1 ) + ( p + 2 ) + ( p + 3 ) + … + ( q − 1 ) + q = ( p + q ) ( q − p + 1 ) 2 {\displaystyle \sum _{k=p}^{q}k=p+(p+1)+(p+2)+(p+3)+\ldots +(q-1)+q={\frac {(p+q)(q-p+1)}{2}}} ∑ k = 1 n 2 k = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + … + ( 2 n − 2 ) + 2 n = n ( n + 1 ) {\displaystyle \sum _{k=1}^{n}2k=2+4+6+8+10+12+14+16+\ldots +(2n-2)+2n=n(n+1)} ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = n 3 3 + n 2 2 + n 6 {\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}} ∑ k = 1 n k 3 = ( n ( n + 1 ) 2 ) 2 = n 4 4 + n 3 2 + n 2 4 = ( ∑ k = 1 n k ) 2 {\displaystyle \sum _{k=1}^{n}k^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left(\sum _{k=1}^{n}k\right)^{2}} сума перших n {\displaystyle n} [en] ∑ k = 1 n k 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 {\displaystyle \sum _{k=1}^{n}k^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}} ∑ k = 1 n k 5 = n 2 ( n + 1 ) 2 ( 2 n 2 + 2 n − 1 ) 12 {\displaystyle \sum _{k=1}^{n}k^{5}={\frac {n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}}} ∑ k = 1 n k 6 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 4 + 6 n 3 − 3 n + 1 ) 42 {\displaystyle \sum _{k=1}^{n}k^{6}={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}} ∑ k = 1 n k 7 = n 2 ( n + 1 ) 2 ( 3 n 4 + 6 n 3 − n 2 − 4 n + 2 ) 24 {\displaystyle \sum _{k=1}^{n}k^{7}={\frac {n^{2}(n+1)^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)}{24}}} ∑ k = 1 n k 8 = n ( n + 1 ) ( 2 n + 1 ) ( 5 n 6 + 15 n 5 + 5 n 4 − 15 n 3 − n 2 + 9 n − 3 ) 24 {\displaystyle \sum _{k=1}^{n}k^{8}={\frac {n(n+1)(2n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3)}{24}}} ∑ k = 1 n k 9 = n 2 ( n + 1 ) 2 ( n 2 + n − 1 ) ( 2 n 4 + 4 n 3 − n 2 − 3 n + 3 ) 20 {\displaystyle \sum _{k=1}^{n}k^{9}={\frac {n^{2}(n+1)^{2}(n^{2}+n-1)(2n^{4}+4n^{3}-n^{2}-3n+3)}{20}}} ∑ k = 1 n k 10 = n ( n + 1 ) ( 2 n + 1 ) ( n 2 + n − 1 ) ( 3 n 6 + 9 n 5 + 2 n 4 − 11 n 3 + 3 n 2 + 10 n − 5 ) 66 {\displaystyle \sum _{k=1}^{n}k^{10}={\frac {n(n+1)(2n+1)(n^{2}+n-1)(3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5)}{66}}} ∑ k = 1 n k 11 = n 2 ( n + 1 ) 2 ( 2 n 8 + 8 n 7 + 4 n 6 − 16 n 5 − 5 n 4 + 26 n 3 − 3 n 2 − 20 n + 10 ) 24 {\displaystyle \sum _{k=1}^{n}k^{11}={\frac {n^{2}(n+1)^{2}(2n^{8}+8n^{7}+4n^{6}-16n^{5}-5n^{4}+26n^{3}-3n^{2}-20n+10)}{24}}} ∑ k = 1 n k 12 = n ( n + 1 ) ( 2 n + 1 ) ( 105 n 10 + 525 n 9 + 525 n 8 − 1050 n 7 − 1190 n 6 + 2310 n 5 + 1420 n 4 − 3285 n 3 − 287 n 2 + 2073 n − 691 ) 2730 {\displaystyle \sum _{k=1}^{n}k^{12}={\frac {n(n+1)(2n+1)(105n^{10}+525n^{9}+525n^{8}-1050n^{7}-1190n^{6}+2310n^{5}+1420n^{4}-3285n^{3}-287n^{2}+2073n-691)}{2730}}} Суми степенів непарних чисел:
∑ k = 1 n ( 2 k − 1 ) = 1 + 3 + 5 + 7 + 9 + … + ( 2 n − 3 ) + ( 2 n − 1 ) = n 2 {\displaystyle \sum _{k=1}^{n}(2k-1)=1+3+5+7+9+\ldots +(2n-3)+(2n-1)=n^{2}} ∑ k = 1 n ( 2 k − 1 ) 2 = 1 2 + 3 2 + 5 2 + 7 2 + 9 2 + … + ( 2 n − 3 ) 2 + ( 2 n − 1 ) 2 = n ( 4 n 2 − 1 ) 3 {\displaystyle \sum _{k=1}^{n}(2k-1)^{2}=1^{2}+3^{2}+5^{2}+7^{2}+9^{2}+\ldots +(2n-3)^{2}+(2n-1)^{2}={\frac {n(4n^{2}-1)}{3}}} ∑ k = 1 n ( 2 k − 1 ) 3 = 1 3 + 3 3 + 5 3 + 7 3 + 9 3 + … + ( 2 n − 3 ) 3 + ( 2 n − 1 ) 3 = n 2 ( 2 n 2 − 1 ) {\displaystyle \sum _{k=1}^{n}(2k-1)^{3}=1^{3}+3^{3}+5^{3}+7^{3}+9^{3}+\ldots +(2n-3)^{3}+(2n-1)^{3}=n^{2}(2n^{2}-1)} Окремі значення дзета-функції Рімана [en]
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)!}}} Значення при n = 1 , 2 , 3 : {\displaystyle n=1,2,3:}
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}}} ряд обернених квадратів 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}}}
∑ k = 1 ∞ ( − 1 ) k + 1 k 2 s = ( 1 − 1 2 2 s − 1 ) ∑ k = 1 ∞ 1 k 2 s {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2s}}}=\left(1-{\frac {1}{2^{2s-1}}}\right)\sum _{k=1}^{\infty }{\frac {1}{k^{2s}}}} ∑ k = 0 ∞ 1 ( 2 k + 1 ) 2 = π 2 8 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}} ∑ k = 0 ∞ 1 ( 2 k + 1 ) 4 = π 4 96 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{4}}}={\frac {\pi ^{4}}{96}}} ∑ k = 0 ∞ 1 ( 2 k + 1 ) 6 = π 6 960 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{6}}}={\frac {\pi ^{6}}{960}}} Скінченні суми:
∑ 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}}} геометрична прогресія ∑ 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 = 1 n k m z k = ( z d d z ) m 1 − z n + 1 1 − z {\displaystyle \sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}} Нескінченні суми, виконується при | z | < 1 {\displaystyle |z|<1}
Li n ( z ) = ∑ k = 1 ∞ z k k n {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}} Корисна властивість для рекурсивного обчислення полілогарифмів:
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}}} Полілогарифми малих по модулю цілих порядків:
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}}}} ∑ 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}} ∑ 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}} ∑ 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)} де T n ( z ) {\displaystyle T_{n}(z)} поліном Тушара [en]
[ ред. | ред. код ] ∑ 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 ) ! = sh z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\operatorname {sh} 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 ) ! = ch z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\operatorname {ch} z} ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tg z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {tg} z,|z|<{\frac {\pi }{2}}} ∑ k = 1 ∞ ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = th z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {th} z,|z|<{\frac {\pi }{2}}} ∑ k = 0 ∞ ( − 1 ) k 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = ctg z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {ctg} z,|z|<\pi } ∑ k = 0 ∞ 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = cth z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {cth} 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} синус-верзус ∑ 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] гаверсинус ∑ 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|\leqslant 1} ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arsh z , | z | ⩽ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arsh} {z},|z|\leqslant 1} ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 2 k + 1 = arctg z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\operatorname {arctg} z,|z|<1} ∑ k = 0 ∞ z 2 k + 1 2 k + 1 = arth z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arth} z,|z|<1} ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 k ) ! z 2 k 2 2 k + 1 k ( k ! ) 2 = ln ( 1 + 1 + z 2 2 ) , | z | ⩽ 1 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left({\frac {1+{\sqrt {1+z^{2}}}}{2}}\right),|z|\leqslant 1} ∑ 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|\leqslant 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|\leqslant 1} ( 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} ∑ 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} ∑ 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}}} генератриса чисел Каталана ∑ 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}}} центральних біноміальних коефіцієнтів ∑ 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 = 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 arctg 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}}\operatorname {arctg} {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 arctg 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}}\operatorname {arctg} {z}\log {\frac {1+z}{1-z}},\qquad |z|<1} [ 2] ∑ k = 0 n ( n k ) = 2 n {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}} ∑ k = 0 n ( − 1 ) k ( n k ) = 0 , {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,} n ≠ 0 {\displaystyle n\neq 0} ∑ 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}} ∑ k = 0 n ( α k ) ( β n − k ) = ( α + β n ) {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}} тотожність Вандермонда ∑ 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 ( ctg θ 2 ) , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\operatorname {ctg} {\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 } [ 3] ∑ 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} [ 4] ∑ 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 = ctg π 2 n {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\operatorname {ctg} {\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 )} [ 5] ∑ 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}}} Ряд раціональних функцій від n {\displaystyle n} розкладання на прості дроби .[ 6] сталий час , навіть якщо сума містить велику кількість членів.
∑ 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}} ∑ n = 0 ∞ 1 n 2 + a 2 = 1 + a π cth ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \operatorname {cth} (a\pi )}{2a^{2}}}} ∑ n = 0 ∞ 1 n 4 + 4 a 4 = 1 8 a 4 + π ( sh ( 2 π a ) + sin ( 2 π a ) ) 8 a 3 ( ch ( 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 (\operatorname {sh} (2\pi a)+\sin(2\pi a))}{8a^{3}(\operatorname {ch} (2\pi a)-\cos(2\pi a))}}} 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)} співвідношення Ландсберга–Шара [en] ∑ 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)}}} Ці числові ряди можна знайти за допомогою рядів, наведених вище.
∑ 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}}} ∑ 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 ) = ch 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}}(e+{\frac {1}{e}})=\operatorname {ch} 1} ∑ 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 cth 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 \operatorname {cth} \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 } ∑ k = 1 ∞ 1 T k = 1 1 + 1 3 + 1 6 + 1 10 + 1 15 + ⋯ = 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{T_{k}}}={\frac {1}{1}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{10}}+{\frac {1}{15}}+\cdots =2} де T n = ∑ k = 1 n k {\displaystyle T_{n}=\sum _{k=1}^{n}k} n {\displaystyle n} трикутне число
∑ 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}}} де T e n = ∑ k = 1 n T k {\displaystyle Te_{n}=\sum _{k=1}^{n}T_{k}} n {\displaystyle n} тетраедричне число
∑ 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} ↑ Weisstein, Eric W. Haversine . MathWorld Архів оригіналу за 10 березня 2005. ↑ а б в г Wilf, Herbert R. (1994). generatingfunctionology (PDF) . Academic Press, Inc. ↑ Знайдіть розклад в ряд Фур’є функції f ( x ) = π 4 {\displaystyle f(x)={\frac {\pi }{4}}} 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 - непарне 0 , n - парне d n = 0 , n ∈ N {\displaystyle \Rightarrow {\begin{cases}c_{n}={\begin{cases}{\frac {1}{n}},\quad n{\text{- непарне}}\\0,\quad n{\text{- парне}}\end{cases}}\\d_{n}=0,\quad n\in \mathbb {N} \end{cases}}} ↑ Bernoulli polynomials: Series representations (subsection 06/02) . Wolfram Research .↑ Hofbauer, Josef. A simple proof of 1 + 1/22 + 1/32 + ··· = π 2 /6 and related identities (PDF) . ↑ Abramowitz, Milton; Stegun, Irene (1964). 6.4 Polygamma functions . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables . с. 260 . ISBN 0-486-61272-4 
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