歐拉的另一錯誤

歐拉将兩種伯努利數混為一談了，并且還犯了另一錯誤：

設 k ≠ 1 , k > 0 , a > 0 ， z ≠ 0 k\ne1,k>0,a>0， z\ne0 k​=1,k>0,a>0，z​=0

由于

∫ 0 ∞ 1 ( 2 k − 1 ) ( x + 1 ( 2 k − 1 ) ( k − 1 ) z k − 1 ) k d x = z \int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{1}{(2^k-1)(k-1)z} }\right)^{k} }dx=z ∫0∞​(2k−1)(x+k−1(2k−1)(k−1)z1​

​)k1​dx=z

是以有

∫ 0 ∞ 1 ( 2 k − 1 ) ( x + 1 ( 2 k − 1 ) ( k − 1 ) ζ ( k ) k − 1 ) k d x = ζ ( k ) \int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{1}{(2^k-1)(k-1)\zeta(k)} }\right)^{k} }dx=\zeta(k) ∫0∞​(2k−1)(x+k−1(2k−1)(k−1)ζ(k)1​

​)k1​dx=ζ(k)

又因為

∫ 0 ∞ 1 ( 2 k − 1 ) ( x + a ( 2 k − 1 ) π k k − 1 ) k d x = π k ( k − 1 ) a \int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{a}{(2^k-1)\pi^{k}} }\right)^{k} }dx=\frac{\pi^{k}}{(k-1)a} ∫0∞​(2k−1)(x+k−1(2k−1)πka​

​)k1​dx=(k−1)aπk​

是以有：

∑ x = 1 ∞ 1 x k = ∑ x = 0 ∞ 1 ( 2 k − 1 ) ( x + 1 / 2 ) k \sum_{x=1}^{\infty}{\frac{1}{x^{k}}}=\sum_{x=0}^{\infty}{\frac{1}{(2^k-1)(x+1/2)^{k}}} x=1∑∞​xk1​=x=0∑∞​(2k−1)(x+1/2)k1​

= ∫ 0 ∞ 1 ( 2 k − 1 ) ( x + a ( 2 k − 1 ) π k k − 1 ) k d x = π k ( k − 1 ) a =\int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{a}{(2^k-1)\pi^{k}} }\right)^{k} }dx=\frac{\pi^{k}}{(k-1)a} =∫0∞​(2k−1)(x+k−1(2k−1)πka​

​)k1​dx=(k−1)aπk​

ζ ( k ) = ∫ 0 ∞ 1 ( 2 k − 1 ) ( x + 1 ( 2 k − 1 ) ( k − 1 ) ζ ( k ) k − 1 ) k d x = π k ( k − 1 ) a \zeta(k)=\int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{1}{(2^k-1)(k-1)\zeta(k)} }\right)^{k} }dx=\frac{\pi^{k}}{(k-1)a} ζ(k)=∫0∞​(2k−1)(x+k−1(2k−1)(k−1)ζ(k)1​

​)k1​dx=(k−1)aπk​

a ( 2 k − 1 ) π k = 1 ( 2 k − 1 ) ( k − 1 ) ζ ( k ) \frac{a}{(2^k-1)\pi^{k}}= \frac{1}{(2^k-1)(k-1)\zeta(k)} (2k−1)πka​=(2k−1)(k−1)ζ(k)1​

π k ( k − 1 ) a = 2 k − 1 ∣ B k ∣ π k k ! = ζ ( k ) \frac{\pi^{k}}{(k-1)a}=\frac{2^{k-1}|B_{k}|\pi^k}{k!}=\zeta(k) (k−1)aπk​=k!2k−1∣Bk​∣πk​=ζ(k)

a = k ! ( k − 1 ) 2 k − 1 ∣ B k ∣ = π k ( k − 1 ) ζ ( k ) a=\frac{k!}{(k-1)2^{k-1}|B_{k}|}=\frac{\pi^k}{(k-1)\zeta(k)} a=(k−1)2k−1∣Bk​∣k!​=(k−1)ζ(k)πk​

∣ B k ∣ = k ! ( k − 1 ) 2 k − 1 a = k ! ζ ( k ) 2 k − 1 π k ， k ≠ 1 |B_{k}|=\frac{k!}{(k-1)2^{k-1}a}=\frac{k!\zeta(k)}{2^{k-1}\pi^k}，k\ne1 ∣Bk​∣=(k−1)2k−1ak!​=2k−1πkk!ζ(k)​，k​=1

k = 1 k=1 k=1 時， ζ ( 1 ) = ∞ \zeta(1)=\infty ζ(1)=∞

B 1 = ∞ B_{1}=\infty B1​=∞ 并不是 B 1 = − 1 / 2 B_{1}=-1/2 B1​=−1/2 或者說用在這兒是不成立的。

這些等式對 k ≥ 2 k\geq2 k≥2 都成立，即 k k k 為奇數時也成立！顯然 B k ≠ 0 B_{k}\ne0 Bk​​=0

是以說： B 3 = B 5 = B 7 = ⋯ = 0 B_{3}=B_{5}=B_{7}=\cdots=0 B3​=B5​=B7​=⋯=0 是錯誤的!

伯努利數有兩種

(1): 正幂 x k x^{k} xk 伯努利數 B 3 = B 5 = B 7 = ⋯ = 0 , B 0 = 1 , B 1 = 1 / 2 , B 2 = 1 / 6 , B 4 = − 1 / 30 , ⋯ \\B_{3}=B_{5}=B_{7}=\cdots=0,\\ B_{0}=1,B_{1}=1/2,B_{2}=1/6,B_{4}=-1/30,\cdots B3​=B5​=B7​=⋯=0,B0​=1,B1​=1/2,B2​=1/6,B4​=−1/30,⋯

隻能用在正幂有限項求和：

∑ x = 1 n x k = B k + 1 ( n ) − B k + 1 k + 1 = ∫ 0 n B k ( x ) d x \sum_{x=1}^{n}{x^k}=\frac{B_{k+1}(n)-B_{k+1}}{k+1}=\int_{0}^{n}B_{k}(x)dx x=1∑n​xk=k+1Bk+1​(n)−Bk+1​​=∫0n​Bk​(x)dx

其中 B k ( x ) B_{k}(x) Bk​(x) 為伯努利多項式；

嚴格的說 B 1 = − 1 2 B_{1}=-\frac{1}{2} B1​=−21​ 的一支還是正幂系的，因為出現有 B 2 k B_{2k} B2k​ 的幂級數都是正幂的

且它們的 B 3 = B 5 = B 7 = ⋯ = 0 B_{3}=B_{5}=B_{7}=\cdots=0 B3​=B5​=B7​=⋯=0

x e x − 1 = ∑ k = 0 ∞ B k k ! x k , B 1 = − 1 2 , ∣ x ∣ < 2 π \frac{x}{e^x-1}=\sum_{k=0}^{\infty}{\frac{B_{k}}{k!}}x^k , B_{1}=-\frac{1}{2},|x|<2\pi ex−1x​=k=0∑∞​k!Bk​​xk,B1​=−21​,∣x∣<2π

tan ⁡ x = ∑ k = 1 ∞ 4 k ( 4 k − 1 ) ∣ B 2 k ∣ x 2 k − 1 ( 2 k ) ! , ∣ x ∣ < π 2 \tan x=\sum_{k=1}^{\infty}{4^k(4^k-1)\frac{|B_{2k}|x^{2k-1}}{(2k)!}},|x|<\frac{\pi}{2} tanx=k=1∑∞​4k(4k−1)(2k)!∣B2k​∣x2k−1​,∣x∣<2π​

(2): 負幂 x − k x^{-k} x−k 伯努利數

∣ B k ∣ = k ! ζ ( k ) 2 k − 1 π k ， k ≠ 1 |B_{k}|=\frac{k!\zeta(k)}{2^{k-1}\pi^k}，k\ne1 ∣Bk​∣=2k−1πkk!ζ(k)​，k​=1 注意：奇偶數都成立

B 1 = ∞ , B 2 = 1 / 6 , B 4 = − 1 / 30 , ⋯ B_{1}=\infty,B_{2}=1/6,B_{4}=-1/30,\cdots B1​=∞,B2​=1/6,B4​=−1/30,⋯ 正是歐拉的 ∣ B 2 k ∣ = ( 2 k ) ! ζ ( 2 k ) 2 2 k − 1 π 2 k |B_{2k}|=\frac{(2k)!\zeta(2k)}{2^{2k-1}\pi^{2k}} ∣B2k​∣=22k−1π2k(2k)!ζ(2k)​

∣ B 3 ∣ = 3 ζ ( 3 ) 2 π 3 ， ∣ B 5 ∣ = 15 ζ ( 5 ) 2 π 5 ， ∣ B 7 ∣ = 7 ! ζ ( 7 ) 2 6 π 7 ， ⋯ ≠ 0 , |B_{3}|=\frac{3\zeta(3)}{2\pi^3}，|B_{5}|=\frac{15\zeta(5)}{2\pi^5}，|B_{7}|=\frac{7!\zeta(7)}{2^{6}\pi^7}，\cdots\ne0, ∣B3​∣=2π33ζ(3)​，∣B5​∣=2π515ζ(5)​，∣B7​∣=26π77!ζ(7)​，⋯​=0,

ζ ( k ) = ∫ 0 ∞ 1 ( 2 k − 1 ) ( x + k ! 2 k − 1 ( 2 k − 1 ) ( k − 1 ) ∣ B k ∣ π k k − 1 ) k d x \zeta(k)=\int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{k!}{2^{k-1}(2^k-1)(k-1) |B_{k}|\pi^k}}\right)^{k} }dx ζ(k)=∫0∞​(2k−1)(x+k−12k−1(2k−1)(k−1)∣Bk​∣πkk!​

​)k1​dx

難怪歐拉的眼睛會失明(18世紀沒有電燈沒有電腦，這些計算把他的眼睛傷害了)。

具體看一看：

( 1 ) : ∑ x = 1 ∞ 1 x = ∞ , B 1 = ∞ (1):\sum_{x=1}^{\infty}{\frac{1}{x}}=\infty,B_{1}=\infty (1):x=1∑∞​x1​=∞,B1​=∞

( 2 ) : ∑ x = 1 ∞ 1 x 2 = ∑ x = 0 ∞ 1 3 ( x + 1 / 2 ) 2 = ∫ 0 ∞ 1 3 ( x + 6 3 π 2 ) 2 d x = π 2 6 = π 2 a (2):\sum_{x=1}^{\infty}{\frac{1}{x^2}}=\sum_{x=0}^{\infty}{\frac{1}{3(x+1/2)^2}}=\int_{0}^{\infty}\frac{1}{3\left( x+\frac{6}{3\pi^2}\right)^2}dx=\frac{\pi^2}{6}=\frac{\pi^2}{a} (2):x=1∑∞​x21​=x=0∑∞​3(x+1/2)21​=∫0∞​3(x+3π26​)21​dx=6π2​=aπ2​

a = 6 × 1 = 6 a=6\times1=6 a=6×1=6

∣ B 2 ∣ = 2 ! ( 2 − 1 ) 2 2 − 1 6 = 2 ! ζ ( 2 ) 2 2 − 1 π 2 = 1 6 |B_{2}|=\frac{2!}{(2-1)2^{2-1}6}=\frac{2!\zeta(2)}{2^{2-1}\pi^2}=\frac{1}{6} ∣B2​∣=(2−1)22−162!​=22−1π22!ζ(2)​=61​

( 3 ) : ∑ x = 1 ∞ 1 x 3 = ∑ x = 0 ∞ 1 7 ( x + 1 / 2 ) 3 = ∫ 0 ∞ 1 7 ( x + 12.897 7 π 3 ) 3 d x = π 3 25.79 ⋅ ⋅ = π 3 2 a (3):\sum_{x=1}^{\infty}{\frac{1}{x^3}}=\sum_{x=0}^{\infty}{\frac{1}{7(x+1/2)^3}}=\int_{0}^{\infty}\frac{1}{7\left( x+\sqrt{\frac{12.897}{7\pi^3}}\right)^3}dx=\frac{\pi^3}{25.79\cdot\cdot}=\frac{\pi^3}{2a} (3):x=1∑∞​x31​=x=0∑∞​7(x+1/2)31​=∫0∞​7(x+7π312.897​

​)31​dx=25.79⋅⋅π3​=2aπ3​

2 a = 12.89717508287582 × 2 = 25.79435016575164 ⋯ 2a=12.89717508287582\times2=25.79435016575164\cdots 2a=12.89717508287582×2=25.79435016575164⋯

∣ B 3 ∣ = 3 ! ( 3 − 1 ) 2 3 − 1 a = 3 4 a = 3 ζ ( 3 ) 2 π 3 |B_{3}|=\frac{3!}{(3-1)2^{3-1}a}=\frac{3}{4a}=\frac{3\zeta(3)}{2\pi^3} ∣B3​∣=(3−1)23−1a3!​=4a3​=2π33ζ(3)​

= 0.05815228940633... = 1 17.19623344383443 ⋯ =0.05815228940633...=\frac{1}{17.19623344383443\cdots} =0.05815228940633...=17.19623344383443⋯1​

( 4 ) : ∑ x = 1 ∞ 1 x 4 = ∑ x = 0 ∞ 1 15 ( x + 1 / 2 ) 4 = ∫ 0 ∞ 1 15 ( x + 2 π 4 3 ) 4 d x = π 4 90 (4):\sum_{x=1}^{\infty}{\frac{1}{x^4}}=\sum_{x=0}^{\infty}{\frac{1}{15(x+1/2)^4}}=\int_{0}^{\infty}\frac{1}{15\left( x+\sqrt[3]{\frac{2}{\pi^4}}\right)^4}dx=\frac{\pi^4}{90} (4):x=1∑∞​x41​=x=0∑∞​15(x+1/2)41​=∫0∞​15(x+3π42​

​)41​dx=90π4​

∫ 0 ∞ 1 15 ( x + 30 15 π 4 3 ) 4 d x = π 4 90 = π 4 3 a \int_{0}^{\infty}\frac{1}{15\left( x+\sqrt[3]{\frac{30}{15\pi^4}} \right)^4}dx=\frac{\pi^4}{90}=\frac{\pi^4}{3a} ∫0∞​15(x+315π430​

​)41​dx=90π4​=3aπ4​

30 × 3 = 90 30\times3=90 30×3=90

∣ B 4 ∣ = 4 ! ( 4 − 1 ) 2 4 − 1 30 = 1 30 |B_{4}|=\frac{4!}{(4-1)2^{4-1}30}=\frac{1}{30} ∣B4​∣=(4−1)24−1304!​=301​

( 5 ) : ∑ x = 1 ∞ 1 x 5 = ∑ x = 0 ∞ 1 31 ( x + 1 / 2 ) 5 = ∫ 0 ∞ 1 31 ( x + 73.780377 ⋯ 31 π 5 4 ) 5 d x = π 5 295.12 ⋯ (5):\sum_{x=1}^{\infty}{\frac{1}{x^5}}=\sum_{x=0}^{\infty}{\frac{1}{31(x+1/2)^5}}=\int_{0}^{\infty}\frac{1}{31\left( x+\sqrt[4]{\frac{73.780377\cdots}{31\pi^5}}\right)^5}dx=\frac{\pi^5}{295.12\cdots} (5):x=1∑∞​x51​=x=0∑∞​31(x+1/2)51​=∫0∞​31(x+431π573.780377⋯​

​)51​dx=295.12⋯π5​

4 a = 73.7803774853556 × 4 = 295.1215099414224 ⋯ 4a=73.7803774853556\times4=295.1215099414224\cdots 4a=73.7803774853556×4=295.1215099414224⋯

ζ ( 5 ) = 2968539 π 5 876079712 = 1.0369277551 ⋯ \zeta(5)=\frac{2968539\pi^5}{876079712}=1.0369277551\cdots ζ(5)=8760797122968539π5​=1.0369277551⋯

ζ ( 5 ) = ∫ 0 ∞ 1 31 ( x + 1 124 ζ ( 5 ) 4 ) 5 d x = π 5 4 a \zeta(5)=\int_{0}^{\infty}\frac{1}{31\left( x+\sqrt[4]{\frac{1}{124\zeta(5)} }\right)^{5} }dx=\frac{\pi^{5}}{4a} ζ(5)=∫0∞​31(x+4124ζ(5)1​

​)51​dx=4aπ5​

∣ B 5 ∣ = 5 ! ( 5 − 1 ) 2 5 − 1 a = 15 8 a = 15 ζ ( 5 ) 2 π 5 |B_{5}|=\frac{5!}{(5-1)2^{5-1}a}=\frac{15}{8a}=\frac{15\zeta(5)}{2\pi^5} ∣B5​∣=(5−1)25−1a5!​=8a15​=2π515ζ(5)​

= 0.025413261139416 ⋯ = 1 39.34953465885632 ⋯ =0.025413261139416\cdots=\frac{1}{39.34953465885632\cdots} =0.025413261139416⋯=39.34953465885632⋯1​

( 6 ) : ∑ x = 1 ∞ 1 x 6 = ∑ x = 0 ∞ 1 63 ( x + 1 / 2 ) 6 = ∫ 0 ∞ 1 63 ( x + 3 π 6 5 ) 6 d x = π 6 945 (6):\sum_{x=1}^{\infty}{\frac{1}{x^6}}=\sum_{x=0}^{\infty}{\frac{1}{63(x+1/2)^6}}=\int_{0}^{\infty}\frac{1}{63\left( x+\sqrt[5]{\frac{3}{\pi^6}}\right)^6}dx=\frac{\pi^6}{945} (6):x=1∑∞​x61​=x=0∑∞​63(x+1/2)61​=∫0∞​63(x+5π63​

​)61​dx=945π6​

∫ 0 ∞ 1 63 ( x + 189 63 π 6 5 ) 6 d x = π 6 945 = π 6 5 a \int_{0}^{\infty}\frac{1}{63\left( x+\sqrt[5]{\frac{189}{63\pi^6}}\right)^6}dx=\frac{\pi^6}{945}=\frac{\pi^6}{5a} ∫0∞​63(x+563π6189​

​)61​dx=945π6​=5aπ6​

189 × 5 = 945 189\times5=945 189×5=945

∣ B 6 ∣ = 6 ! ( 6 − 1 ) 2 6 − 1 a = 9 2 a = 1 42 |B_{6}|=\frac{6!}{(6-1)2^{6-1}a}=\frac{9}{2a}=\frac{1}{42} ∣B6​∣=(6−1)26−1a6!​=2a9​=421​

( 7 ) ： ζ ( 7 ) = ∫ 0 ∞ 1 ( 2 7 − 1 ) ( x + a ( 2 7 − 1 ) π 7 6 ) 7 d x = π 7 6 a (7)：\zeta(7)=\int_{0}^{\infty}\frac{1}{(2^7-1)\left( x+\sqrt[6]{\frac{a}{(2^7-1)\pi^{7}} }\right)^{7} }dx=\frac{\pi^{7}}{6a} (7)：ζ(7)=∫0∞​(27−1)(x+6(27−1)π7a​

​)71​dx=6aπ7​

∑ x = 1 ∞ 1 x 7 = ∫ 0 ∞ 1 127 ( x + 499.21 ⋯ 127 π 7 6 ) 7 d x = π 7 2995.28 ⋯ \sum_{x=1}^{\infty}{\frac{1}{x^7}}=\int_{0}^{\infty}\frac{1}{127\left( x+\sqrt[6]{\frac{499.21\cdots}{127\pi^{7}} }\right)^{7} }dx=\frac{\pi^{7}}{2995.28\cdots} x=1∑∞​x71​=∫0∞​127(x+6127π7499.21⋯​

​)71​dx=2995.28⋯π7​

∣ B 7 ∣ = k ! ζ ( k ) 2 k − 1 π k = 7 ! ζ ( 7 ) 2 6 π 7 = 105 8 a = 1 38.0353621 ⋯ ≠ 0 |B_{7}|=\frac{k!\zeta(k)}{2^{k-1}\pi^k}=\frac{7!\zeta(7)}{2^{6}\pi^7}=\frac{105}{8a}=\frac{1}{38.0353621\cdots}\ne0 ∣B7​∣=2k−1πkk!ζ(k)​=26π77!ζ(7)​=8a105​=38.0353621⋯1​​=0

( 8 ) : ∑ x = 1 ∞ 1 x 8 = ∑ x = 0 ∞ 1 255 ( x + 1 / 2 ) 8 = ∫ 0 ∞ 1 255 ( x + 90 17 π 8 7 ) 8 d x = π 8 9450 (8):\sum_{x=1}^{\infty}{\frac{1}{x^8}}=\sum_{x=0}^{\infty}{\frac{1}{255(x+1/2)^8}}=\int_{0}^{\infty}\frac{1}{255\left( x+\sqrt[7]{\frac{90}{17\pi^8}}\right)^8}dx=\frac{\pi^8}{9450} (8):x=1∑∞​x81​=x=0∑∞​255(x+1/2)81​=∫0∞​255(x+717π890​

​)81​dx=9450π8​

∫ 0 ∞ 1 255 ( x + 1350 255 π 8 7 ) 8 d x = π 8 9450 \int_{0}^{\infty}\frac{1}{255\left( x+\sqrt[7]{\frac{1350}{255\pi^8}} \right)^8}dx=\frac{\pi^8}{9450} ∫0∞​255(x+7255π81350​

​)81​dx=9450π8​

7 a = 1350 × 7 = 9450 7a=1350\times7=9450 7a=1350×7=9450

∣ B 8 ∣ = 8 ! ( 8 − 1 ) 2 7 a = 45 a = 1 30 |B_{8}|=\frac{8!}{(8-1)2^{7}a}=\frac{45}{a}=\frac{1}{30} ∣B8​∣=(8−1)27a8!​=a45​=301​

( 9 ) : ζ ( 9 ) = ∫ 0 ∞ 1 ( 2 9 − 1 ) ( x + a ( 2 9 − 1 ) π 9 8 ) 9 d x = π 9 8 a (9):\zeta(9)=\int_{0}^{\infty}\frac{1}{(2^9-1)\left( x+\sqrt[8]{\frac{a}{(2^9-1)\pi^{9}} }\right)^{9} }dx=\frac{\pi^{9}}{8a} (9):ζ(9)=∫0∞​(29−1)(x+8(29−1)π9a​

​)91​dx=8aπ9​

∣ B 9 ∣ = k ! ζ ( k ) 2 k − 1 π k = 9 ! ζ ( 9 ) 2 8 π 9 ≠ 0 |B_{9}|=\frac{k!\zeta(k)}{2^{k-1}\pi^k}=\frac{9!\zeta(9)}{2^{8}\pi^9}\ne0 ∣B9​∣=2k−1πkk!ζ(k)​=28π99!ζ(9)​​=0

( 10 ) : ∑ x = 1 ∞ 1 x 10 = ∫ 0 ∞ 1 1023 ( x + 10395 1023 π 10 9 ) 10 d x = π 10 93555 (10):\sum_{x=1}^{\infty}{\frac{1}{x^{10}}}=\int_{0}^{\infty}\frac{1}{1023\left( x+\sqrt[9]{\frac{10395}{1023\pi^{10}}}\right)^{10}}dx=\frac{\pi^{10}}{93555} (10):x=1∑∞​x101​=∫0∞​1023(x+91023π1010395​

​)101​dx=93555π10​

10395 × 9 = 93555 10395\times9=93555 10395×9=93555

∫ 0 ∞ 1 1023 ( x + 315 31 π 10 9 ) 10 d x = π 10 93555 \int_{0}^{\infty}\frac{1}{1023\left( x+\sqrt[9]{\frac{315}{31\pi^{10}}} \right)^{10}}dx=\frac{\pi^{10}}{93555} ∫0∞​1023(x+931π10315​

​)101​dx=93555π10​

∣ B 10 ∣ = 10 ! ( 10 − 1 ) 2 9 a = 1575 2 a = 5 66 |B_{10}|=\frac{10!}{(10-1)2^{9}a}=\frac{1575}{2a}=\frac{5}{66} ∣B10​∣=(10−1)29a10!​=2a1575​=665​

結論：負幂伯努利數 B 3 = B 5 = B 7 = ⋯ = 0 B_{3}=B_{5}=B_{7}=\cdots=0 B3​=B5​=B7​=⋯=0 是錯誤的!

可知：黎曼猜想的平凡零點不存在！

下面網站有些内容是錯誤的：

https://www.bernoulli.org/

http://mathworld.wolfram.com/BernoulliNumber.html

http://mathworld.wolfram.com/BernoulliPolynomial.html

http://mathworld.wolfram.com/RiemannZetaFunction.html

黎曼ζ函數(中文維基百科)

發散級數 (中文維基百科)

∑ x = 1 ∞ 1 x k = ∑ x = 0 ∞ 1 ( 2 k − 1 ) ( x + 1 / 2 ) k = π k ( k − 1 ) a {\color{blue}{\sum_{x=1}^{\infty}{\frac{1}{x^{k}}}=\sum_{x=0}^{\infty}{\frac{1}{(2^k-1)(x+1/2)^{k}}}=\frac{\pi^{k}}{(k-1)a}}} x=1∑∞​xk1​=x=0∑∞​(2k−1)(x+1/2)k1​=(k−1)aπk​

ζ ( k ) = ∫ 0 ∞ 1 ( 2 k − 1 ) ( x + k ! 2 k − 1 ( 2 k − 1 ) ( k − 1 ) ∣ B k ∣ π k k − 1 ) k d x \zeta(k)=\int_{0}^{\infty}\frac{1}{(2^k-1)\left( x+\sqrt[k-1]{\frac{k!}{2^{k-1}(2^k-1)(k-1) |B_{k}|\pi^k}}\right)^{k} }dx ζ(k)=∫0∞​(2k−1)(x+k−12k−1(2k−1)(k−1)∣Bk​∣πkk!​

​)k1​dx

∑ x = 1 n x k = B k + 1 ( n ) − B k + 1 k + 1 = ∫ 0 n B k ( x ) d x \sum_{x=1}^{n}{x^k}=\frac{B_{k+1}(n)-B_{k+1}}{k+1}=\int_{0}^{n}B_{k}(x)dx x=1∑n​xk=k+1Bk+1​(n)−Bk+1​​=∫0n​Bk​(x)dx