莱昂哈德·欧拉 | 无穷级数若干注记（2）

注：本文为 "无穷级数" 相关合辑。

英文引文，机翻未校。

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Theorem 7

定理 7

If we take to the infinity the continuation of these fractions

考察下述无穷连乘积：

2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋯ 1 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 10 ⋅ 12 ⋅ 16 ⋅ 18 ⋯ , \frac {2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot 12\cdot 16\cdot 18\cdots }, 1⋅2⋅4⋅6⋅10⋅12⋅16⋅18⋯2⋅3⋅5⋅7⋅11⋅13⋅17⋅19⋯,

where the numerators are all the prime numbers and the denominators are the numerators less one unit, the result is the same as the sum of the series

其分子为全体素数，分母为对应分子减 1 1 1。该连乘积的值等于调和级数

1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + ⋯ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots 1+21+31+41+51+61+⋯

which is certainly infinity.

的和，而调和级数为无穷大。

Proof

证明

If we have

设

x = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + ⋯ x=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots x=1+21+31+41+51+61+⋯

then we will have

则

1 2 x = 1 2 + 1 4 + 1 6 + 1 8 + ⋯ , \frac{1}{2} x=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots, 21x=21+41+61+81+⋯,

which subtracted from the first will leave us with

两式相减得：

1 2 x = 1 + 1 3 + 1 5 + 1 7 + ⋯ , \frac{1}{2} x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots, 21x=1+31+51+71+⋯,

series where there are no even denominators. From this one, we subtract again this series

该级数分母均为奇数。再减去

1 2 ⋅ 1 3 x = 1 3 + 1 9 + 1 15 + 1 21 + ⋯ ; \frac{1}{2} \cdot \frac{1}{3} x=\frac{1}{3}+\frac{1}{9}+\frac{1}{15}+\frac{1}{21}+\cdots ; 21⋅31x=31+91+151+211+⋯;

and we will have

得到：

1 2 ⋅ 2 3 x = 1 + 1 5 + 1 7 + 1 11 + 1 13 + ⋯ , \frac{1}{2} \cdot \frac{2}{3} x=1+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\cdots, 21⋅32x=1+51+71+111+131+⋯,

where among the denominators we cannot find any divisible either by 2 2 2 or by 3 3 3. In order to remove the numbers divisible by 5 5 5, we subtract the following series

此时分母中已无 2 2 2 和 3 3 3 的倍数。为消去 5 5 5 的倍数，减去

1 ⋅ 2 2 ⋅ 3 ⋅ 1 5 x = 1 5 + 1 25 + 1 35 + ⋯ \frac{1 \cdot 2}{2 \cdot 3} \cdot \frac{1}{5} x=\frac{1}{5}+\frac{1}{25}+\frac{1}{35}+\cdots 2⋅31⋅2⋅51x=51+251+351+⋯

and we will have

可得：

1 ⋅ 2 ⋅ 4 2 ⋅ 3 ⋅ 5 x = 1 + 1 7 + 1 11 + 1 13 + ⋯ \frac{1 \cdot 2 \cdot 4}{2 \cdot 3 \cdot 5} x=1+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\cdots 2⋅3⋅51⋅2⋅4x=1+71+111+131+⋯

And proceeding in the same way, subtracting all the terms divisible now by 7 7 7, now by 11 11 11, and now by all the prime numbers we finally have

按照该方法，依次消去分母为 7 , 11 7,11 7,11 及所有素数倍数的项，最终得到：

1 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 10 ⋅ 12 ⋅ 16 ⋅ 18 ⋅ 22 ⋯ 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋯ x = 1. \frac{1 \cdot 2 \cdot 4 \cdot 6 \cdot 10 \cdot 12 \cdot 16 \cdot 18 \cdot 22 \cdots}{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdots} x=1 . 2⋅3⋅5⋅7⋅11⋅13⋅17⋅19⋅23⋯1⋅2⋅4⋅6⋅10⋅12⋅16⋅18⋅22⋯x=1.

结合

x = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + ⋯ , x=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots, x=1+21+31+41+51+61+71+⋯,

we have

整理可得：

1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + ⋯ = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋯ 1 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 10 ⋅ 12 ⋅ 16 ⋅ 18 ⋅ 22 ⋯ , 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+\cdots ={\frac {2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdots }{1\cdot 2\cdot 4\cdot 6\cdot 10\cdot 12\cdot 16\cdot 18\cdot 22\cdots }}, 1+21+31+41+51+61+71+⋯=1⋅2⋅4⋅6⋅10⋅12⋅16⋅18⋅22⋯2⋅3⋅5⋅7⋅11⋅13⋅17⋅19⋅23⋯,

expression whose numerators constitute the sequence of prime numbers and the denominators are the same less one unit. Q .E.D.

该式分子为全体素数，分母为对应素数减 1 1 1。证毕。

Corollary 1

推论 1

Thus, the value of the expression

因此连乘积

2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋯ 1 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 10 ⋅ 12 ⋯ \frac{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdots}{1 \cdot 2 \cdot 4 \cdot 6 \cdot 10 \cdot 12 \cdots} 1⋅2⋅4⋅6⋅10⋅12⋯2⋅3⋅5⋅7⋅11⋅13⋯

is infinity, and if we denote the absolute infinity as ∞ \infty ∞, the value of this expression is log ⁡ ∞ \log\infty log∞, which is the minimum among all the powers of the infinity.

的值为无穷大。记绝对无穷大为 ∞ \infty ∞，则该式取值为 log ⁡ ∞ \log\infty log∞，是无穷大各阶量级中最小的一类。

Corollary 2

推论 2

As the expression

连乘积

4 ⋅ 9 ⋅ 16 ⋅ 25 ⋅ 36 ⋅ 49 ⋯ 3 ⋅ 8 ⋅ 15 ⋅ 24 ⋅ 35 ⋅ 48 ⋯ \frac{4 \cdot 9 \cdot 16 \cdot 25 \cdot 36 \cdot 49 \cdots}{3 \cdot 8 \cdot 15 \cdot 24 \cdot 35 \cdot 48 \cdots} 3⋅8⋅15⋅24⋅35⋅48⋯4⋅9⋅16⋅25⋅36⋅49⋯

has a finite value, which is 2 2 2, it follows that the prime numbers are infinitely many times more numerous than the squares in the sequence of all numbers.

取值为有限数 2 2 2。由此可知：在全体自然数中，素数的数量远多于平方数。

Corollary 3

推论 3

Also from here it is also true that the prime numbers are infinitely many less numerous than the whole numbers as the value of the expression

同理可证，素数的数量远少于全体自然数。连乘积

2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋯ 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋯ \frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdots}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots} 1⋅2⋅3⋅4⋅5⋅6⋯2⋅3⋅4⋅5⋅6⋅7⋯

is the absolute infinity, and the similar value from the prime numbers is the logarithm of this value.

为绝对无穷大，而素数对应连乘积的量级为该无穷大的对数量级。

Theorem 8

定理 8

The expression formed from the sequence of prime numbers

由素数构造的无穷连乘积

2 n ⋅ 3 n ⋅ 5 n ⋅ 7 n ⋅ 11 n ⋯ ( 2 n − 1 ) ( 3 n − 1 ) ( 5 n − 1 ) ( 7 n − 1 ) ( 11 n − 1 ) ⋯ \frac{2^{n} \cdot 3^{n} \cdot 5^{n} \cdot 7^{n} \cdot 11^{n} \cdots}{\left(2^{n}-1\right)\left(3^{n}-1\right)\left(5^{n}-1\right)\left(7^{n}-1\right)\left(11^{n}-1\right) \cdots} (2n−1)(3n−1)(5n−1)(7n−1)(11n−1)⋯2n⋅3n⋅5n⋅7n⋅11n⋯

has the same value as the sum of the series

与级数

1 + 1 2 n + 1 3 n + 1 4 n + 1 5 n + 1 6 n + 1 7 n + ⋯ 1+\frac{1}{2^{n}}+\frac{1}{3^{n}}+\frac{1}{4^{n}}+\frac{1}{5^{n}}+\frac{1}{6^{n}}+\frac{1}{7^{n}}+\cdots 1+2n1+3n1+4n1+5n1+6n1+7n1+⋯

的和相等。

Proof

证明

Let

设

x = 1 + 1 2 n + 1 3 n + 1 4 n + 1 5 n + 1 6 n + ⋯ ; x=1+\frac{1}{2^{n}}+\frac{1}{3^{n}}+\frac{1}{4^{n}}+\frac{1}{5^{n}}+\frac{1}{6^{n}}+\cdots ; x=1+2n1+3n1+4n1+5n1+6n1+⋯;

we will have

则

1 2 n x = 1 2 n + 1 4 n + 1 6 n + 1 8 n + ⋯ , \frac{1}{2^{n}} x=\frac{1}{2^{n}}+\frac{1}{4^{n}}+\frac{1}{6^{n}}+\frac{1}{8^{n}}+\cdots, 2n1x=2n1+4n1+6n1+8n1+⋯,

from where

两式相减得：

2 n − 1 2 n x = 1 + 1 3 n + 1 5 n + 1 7 n + 1 9 n + ⋯ . \frac{2^{n}-1}{2^{n}} x=1+\frac{1}{3^{n}}+\frac{1}{5^{n}}+\frac{1}{7^{n}}+\frac{1}{9^{n}}+\cdots . 2n2n−1x=1+3n1+5n1+7n1+9n1+⋯.

We also have

又有

2 n − 1 2 n ⋅ 1 3 n x = 1 3 n + 1 9 n + 1 15 n + 1 21 n + ⋯ , \frac{2^{n}-1}{2^{n}} \cdot \frac{1}{3^{n}} x=\frac{1}{3^{n}}+\frac{1}{9^{n}}+\frac{1}{15^{n}}+\frac{1}{21^{n}}+\cdots, 2n2n−1⋅3n1x=3n1+9n1+15n1+21n1+⋯,

from where

相减后得：

( 2 n − 1 ) ( 3 n − 1 ) 2 n ⋅ 3 n x = 1 + 1 5 n + 1 7 n + ⋯ . \frac{\left(2^{n}-1\right)\left(3^{n}-1\right)}{2^{n} \cdot 3^{n}} x=1+\frac{1}{5^{n}}+\frac{1}{7^{n}}+\cdots . 2n⋅3n(2n−1)(3n−1)x=1+5n1+7n1+⋯.

Consequently, carrying out the same procedure with each prime number one by one, all the terms of the series will disappear except for the first, and we will have

对每一个素数重复上述运算，级数中除首项外其余项全部消去，最终得到：

1 = ( 2 n − 1 ) ( 3 n − 1 ) ( 5 n − 1 ) ( 7 n − 1 ) ( 11 n − 1 ) ⋯ 2 n ⋅ 3 n ⋅ 5 n ⋅ 7 n ⋅ 11 n ⋯ x , 1=\frac {(2^{n}-1)(3^{n}-1)(5^{n}-1)(7^{n}-1)(11^{n}-1)\cdots }{2^{n}\cdot 3^{n}\cdot 5^{n}\cdot 7^{n}\cdot 11^{n}\cdots }x, 1=2n⋅3n⋅5n⋅7n⋅11n⋯(2n−1)(3n−1)(5n−1)(7n−1)(11n−1)⋯x,

and replacing x x x by its series,

将 x x x 代回原级数形式，即证：

2 n ⋅ 3 n ⋅ 5 n ⋅ 7 n ⋅ 11 n ⋯ ( 2 n − 1 ) ( 3 n − 1 ) ( 5 n − 1 ) ( 7 n − 1 ) ( 11 n − 1 ) ⋯ = 1 + 1 2 n + 1 3 n + 1 4 n + 1 5 n + 1 6 n + ⋯ . \frac{2^{n} \cdot 3^{n} \cdot 5^{n} \cdot 7^{n} \cdot 11^{n} \cdots}{\left(2^{n}-1\right)\left(3^{n}-1\right)\left(5^{n}-1\right)\left(7^{n}-1\right)\left(11^{n}-1\right) \cdots}=1+\frac{1}{2^{n}}+\frac{1}{3^{n}}+\frac{1}{4^{n}}+\frac{1}{5^{n}}+\frac{1}{6^{n}}+\cdots . (2n−1)(3n−1)(5n−1)(7n−1)(11n−1)⋯2n⋅3n⋅5n⋅7n⋅11n⋯=1+2n1+3n1+4n1+5n1+6n1+⋯.

Q.E.D.

证毕。

Corollary 1

推论 1

If we make n = 2 n=2 n=2, as

令 n = 2 n=2 n=2，已知直径为 1 1 1 的圆周长为 π \pi π，且

1 + 1 4 + 1 9 + 1 16 + ⋯ = π 2 6 , 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots=\frac{\pi^{2}}{6}, 1+41+91+161+⋯=6π2,

where π \pi π denotes the perimeter of the circle whose diameter is one, we will have

代入定理 8 得：

4 ⋅ 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋯ 3 ⋅ 8 ⋅ 24 ⋅ 48 ⋅ 120 ⋅ 168 ⋯ = π 2 6 \frac{4 \cdot 9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdots}{3 \cdot 8 \cdot 24 \cdot 48 \cdot 120 \cdot 168 \cdots}=\frac{\pi^{2}}{6} 3⋅8⋅24⋅48⋅120⋅168⋯4⋅9⋅25⋅49⋅121⋅169⋯=6π2

亦可写作：

π 2 6 = 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 7 ⋅ 7 ⋅ 11 ⋅ 11 ⋯ 1 ⋅ 3 ⋅ 2 ⋅ 4 ⋅ 4 ⋅ 6 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋯ . \frac{\pi^{2}}{6}=\frac{2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 11 \cdot 11 \cdots}{1 \cdot 3 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdots} . 6π2=1⋅3⋅2⋅4⋅4⋅6⋅6⋅8⋅10⋅12⋯2⋅2⋅3⋅3⋅5⋅5⋅7⋅7⋅11⋅11⋯.

Corollary 2

推论 2

As before if we make n = 4 n=4 n=4 as

同理，令 n = 4 n=4 n=4，已知

1 + 1 2 4 + 1 3 4 + 1 4 4 + 1 5 4 + ⋯ = π 4 90 , 1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\frac{1}{5^{4}}+\cdots=\frac{\pi^{4}}{90}, 1+241+341+441+541+⋯=90π4,

we will have

可得：

π 4 90 = 4 ⋅ 4 ⋅ 9 ⋅ 9 ⋅ 25 ⋅ 25 ⋅ 49 ⋅ 49 ⋅ 121 ⋅ 121 ⋯ 3 ⋅ 5 ⋅ 8 ⋅ 10 ⋅ 24 ⋅ 26 ⋅ 48 ⋅ 50 ⋅ 120 ⋅ 122 ⋯ . \frac{\pi^{4}}{90}=\frac{4 \cdot 4 \cdot 9 \cdot 9 \cdot 25 \cdot 25 \cdot 49 \cdot 49 \cdot 121 \cdot 121 \cdots}{3 \cdot 5 \cdot 8 \cdot 10 \cdot 24 \cdot 26 \cdot 48 \cdot 50 \cdot 120 \cdot 122 \cdots} . 90π4=3⋅5⋅8⋅10⋅24⋅26⋅48⋅50⋅120⋅122⋯4⋅4⋅9⋅9⋅25⋅25⋅49⋅49⋅121⋅121⋯.

This expression divided by the former will give us

将该式与上一条推论中的式子相除，得到：

π 2 15 = 4 ⋅ 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋯ 5 ⋅ 10 ⋅ 26 ⋅ 50 ⋅ 122 ⋅ 170 ⋯ . \frac{\pi^{2}}{15}=\frac{4 \cdot 9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdots}{5 \cdot 10 \cdot 26 \cdot 50 \cdot 122 \cdot 170 \cdots} . 15π2=5⋅10⋅26⋅50⋅122⋅170⋯4⋅9⋅25⋅49⋅121⋅169⋯.

Theorem 9

定理 9

If the squares of all the prime numbers are separated in two different parts with one unit of difference between them, and we take the odd parts as numerators and the even parts as denominators of the series formed by these factors, the value of this expression will be

将全体素数的平方拆分为相差 1 1 1 的两个数，取其中较大奇数作为连乘积分子、较小偶数作为分母，则该无穷连乘积的值为：

5 ⋅ 13 ⋅ 25 ⋅ 61 ⋅ 85 ⋅ 145 ⋯ 4 ⋅ 12 ⋅ 24 ⋅ 60 ⋅ 84 ⋅ 144 ⋯ = 3 2 . \frac{5 \cdot 13 \cdot 25 \cdot 61 \cdot 85 \cdot 145 \cdots}{4 \cdot 12 \cdot 24 \cdot 60 \cdot 84 \cdot 144 \cdots}=\frac{3}{2} . 4⋅12⋅24⋅60⋅84⋅144⋯5⋅13⋅25⋅61⋅85⋅145⋯=23.

Proof

证明

By Corollary 1 of the previous Theorem we have

由上一定理推论 1 可得：

π 2 6 = 4 ⋅ 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋅ 289 ⋯ 3 ⋅ 8 ⋅ 24 ⋅ 48 ⋅ 120 ⋅ 168 ⋅ 288 ⋯ \frac{\pi^{2}}{6}=\frac{4 \cdot 9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdot 289 \cdots}{3 \cdot 8 \cdot 24 \cdot 48 \cdot 120 \cdot 168 \cdot 288 \cdots} 6π2=3⋅8⋅24⋅48⋅120⋅168⋅288⋯4⋅9⋅25⋅49⋅121⋅169⋅289⋯

But in Corollary 2 we deduced the following equation

结合推论 2 推导出的等式：

π 2 15 = 4 ⋅ 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋅ 289 ⋯ 5 ⋅ 10 ⋅ 26 ⋅ 50 ⋅ 122 ⋅ 170 ⋅ 290 ⋯ . \frac{\pi^{2}}{15}=\frac{4 \cdot 9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdot 289 \cdots}{5 \cdot 10 \cdot 26 \cdot 50 \cdot 122 \cdot 170 \cdot 290 \cdots} . 15π2=5⋅10⋅26⋅50⋅122⋅170⋅290⋯4⋅9⋅25⋅49⋅121⋅169⋅289⋯.

If we divide this expression by the other one we will have

两式相除可得：

5 2 = 5 ⋅ 10 ⋅ 26 ⋅ 50 ⋅ 122 ⋅ 170 ⋅ 290 ⋯ 3 ⋅ 8 ⋅ 24 ⋅ 48 ⋅ 120 ⋅ 168 ⋅ 288 ⋯ , \frac{5}{2}=\frac{5\cdot 10\cdot 26\cdot 50\cdot 122\cdot 170\cdot 290\cdots }{3\cdot 8\cdot 24\cdot 48\cdot 120\cdot 168\cdot 288\cdots }, 25=3⋅8⋅24⋅48⋅120⋅168⋅288⋯5⋅10⋅26⋅50⋅122⋅170⋅290⋯,

expression whose numerators are the squares of the prime numbers increased by one unit and the denominators decreased by one unit.Dividing both sides by 5 / 3 5/3 5/3 and simplifying by 2 2 2 each fraction, one by one, we have

该式分子为素数平方加 1 1 1，分母为素数平方减 1 1 1。将等式两侧同除以 5 3 \dfrac{5}{3} 35，并对每一项依次约去因子 2 2 2，化简后得到：

3 2 = 5 ⋅ 13 ⋅ 25 ⋅ 61 ⋅ 85 ⋅ 145 ⋯ 4 ⋅ 12 ⋅ 24 ⋅ 60 ⋅ 84 ⋅ 144 ⋯ \frac{3}{2}=\frac{5 \cdot 13 \cdot 25 \cdot 61 \cdot 85 \cdot 145 \cdots}{4 \cdot 12 \cdot 24 \cdot 60 \cdot 84 \cdot 144 \cdots} 23=4⋅12⋅24⋅60⋅84⋅144⋯5⋅13⋅25⋅61⋅85⋅145⋯

where the numerators are one unit greater than their corresponding denominators and each numerator added with its denominator is the square of an odd prime number, as in the simplification process the square of the even prime number 2 2 2 has disappeared. Q. E. D.

式中每一组分子均比分母大 1 1 1，且每组分子与分母之和为奇素数的平方；化简过程中，偶素数 2 2 2 的平方项已被消去。证毕。

Theorem 10

定理 10

If π \pi π denotes the perimeter of the circle whose diameter is one, we have

设直径为 1 1 1 的圆的周长为 π \pi π，则有：

π 3 32 = 80 ⋅ 224 ⋅ 440 ⋅ 624 ⋅ 728 ⋯ 81 ⋅ 225 ⋅ 441 ⋅ 625 ⋅ 729 ⋯ , \frac{\pi^{3}}{32}=\frac{80 \cdot 224 \cdot 440 \cdot 624 \cdot 728 \cdots}{81 \cdot 225 \cdot 441 \cdot 625 \cdot 729 \cdots}, 32π3=81⋅225⋅441⋅625⋅729⋯80⋅224⋅440⋅624⋅728⋯,

expression whose denominators are the squares of the odd numbers not prime and whose numerators are these same numbers less one.

该连乘积的分母为非素奇数的平方，分子为对应平方数减 1 1 1。

Proof

证明

We owe the following expression for π \pi π to WALLIs

沃利斯给出了关于 π \pi π 的下述表达式：

π 4 = 8 ⋅ 24 ⋅ 48 ⋅ 80 ⋅ 120 ⋅ 168 ⋯ 9 ⋅ 25 ⋅ 49 ⋅ 81 ⋅ 121 ⋅ 169 ⋯ , \frac {\pi }{4}=\frac {8\cdot 24\cdot 48\cdot 80\cdot 120\cdot 168\cdots }{9\cdot 25\cdot 49\cdot 81\cdot 121\cdot 169\cdots }, 4π=9⋅25⋅49⋅81⋅121⋅169⋯8⋅24⋅48⋅80⋅120⋅168⋯,

whose fractions are formed only by all the squares of the odd numbers.

该连乘积由全体奇数的平方构造而成。

By Corollary 1 of Theorem 8 we have

结合定理 8 推论 1：

π 2 6 = 4 ⋅ 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋯ 3 ⋅ 8 ⋅ 24 ⋅ 48 ⋅ 120 ⋅ 168 ⋯ \frac{\pi^{2}}{6}=\frac{4 \cdot 9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdots}{3 \cdot 8 \cdot 24 \cdot 48 \cdot 120 \cdot 168 \cdots} 6π2=3⋅8⋅24⋅48⋅120⋅168⋯4⋅9⋅25⋅49⋅121⋅169⋯

变形可得：

π 2 8 = 9 ⋅ 25 ⋅ 49 ⋅ 121 ⋅ 169 ⋅ 289 ⋯ 8 ⋅ 24 ⋅ 48 ⋅ 120 ⋅ 168 ⋅ 288 ⋯ , \frac{\pi^{2}}{8}=\frac{9 \cdot 25 \cdot 49 \cdot 121 \cdot 169 \cdot 289 \cdots}{8 \cdot 24 \cdot 48 \cdot 120 \cdot 168 \cdot 288 \cdots}, 8π2=8⋅24⋅48⋅120⋅168⋅288⋯9⋅25⋅49⋅121⋅169⋅289⋯,

whose fractions are formed only by the squares of the odd numbers.

此式同样由全体奇数的平方构造。

Now, if we multiply these two expressions we have

将以上两个表达式相乘，可得：

whose fractions happen to be the squares of the odd numbers not prime. Q. E. D.

该式各项由非素奇数的平方构造而成。证毕。

Theorem 11

定理 11

If we take π \pi π as the perimeter of the circle whose diameter is one, we have

设直径为 1 1 1 的圆的周长为 π \pi π，则：

π 4 = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋯ 4 ⋅ 4 ⋅ 8 ⋅ 12 ⋅ 12 ⋅ 16 ⋅ 20 ⋅ 24 ⋯ , \frac {\pi }{4}=\frac {3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdots }{4\cdot 4\cdot 8\cdot 12\cdot 12\cdot 16\cdot 20\cdot 24\cdots }, 4π=4⋅4⋅8⋅12⋅12⋅16⋅20⋅24⋯3⋅5⋅7⋅11⋅13⋅17⋅19⋅23⋯,

expression whose numerators are the sequence of the prime numbers and whose denominators are multiples of four one unit more or less than the corresponding numerators.

该连乘积分子为全体奇素数，分母为 4 4 4 的倍数，且与对应分子相差 1 1 1。

Proof

证明

已知：

π 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + 1 13 − ⋯ \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\cdots 4π=1−31+51−71+91−111+131−⋯

we will have

两侧同乘 1 3 \dfrac{1}{3} 31：

1 3 ⋅ π 4 = 1 3 − 1 9 + 1 15 − 1 21 + ⋯ , \frac{1}{3} \cdot \frac{\pi}{4}=\frac{1}{3}-\frac{1}{9}+\frac{1}{15}-\frac{1}{21}+\cdots, 31⋅4π=31−91+151−211+⋯,

and adding up both series,

将两式相加：

4 3 ⋅ π 4 = 1 + 1 5 − 1 7 − 1 11 + 1 13 + ⋯ \frac{4}{3} \cdot \frac{\pi}{4}=1+\frac{1}{5}-\frac{1}{7}-\frac{1}{11}+\frac{1}{13}+\cdots 34⋅4π=1+51−71−111+131+⋯

We also have

对上式两侧同乘 1 5 \dfrac{1}{5} 51：

1 5 ⋅ 4 3 ⋅ π 4 = 1 5 + 1 25 − 1 35 − 1 55 + ⋯ , \frac{1}{5} \cdot \frac{4}{3} \cdot \frac{\pi}{4}=\frac{1}{5}+\frac{1}{25}-\frac{1}{35}-\frac{1}{55}+\cdots, 51⋅34⋅4π=51+251−351−551+⋯,

that subtracted from the former leads to

两式相减得：

4 5 ⋅ 4 3 ⋅ π 4 = 1 − 1 7 − 1 11 + 1 13 + ⋯ , \frac{4}{5} \cdot \frac{4}{3} \cdot \frac{\pi}{4}=1-\frac{1}{7}-\frac{1}{11}+\frac{1}{13}+\cdots, 54⋅34⋅4π=1−71−111+131+⋯,

series where there are no denominators divisible either by 3 3 3 or by 5 5 5.

该级数分母已不含 3 3 3 和 5 5 5 的倍数。

In a similar way we remove all those divisible by 7 7 7,

同理消去 7 7 7 的倍数，构造下式：

1 7 ⋅ 4 ⋅ 4 5 ⋅ 3 ⋅ π 4 = 1 7 − 1 49 − 1 77 + ⋯ ; \frac{1}{7} \cdot \frac{4 \cdot 4}{5 \cdot 3} \cdot \frac{\pi}{4}=\frac{1}{7}-\frac{1}{49}-\frac{1}{77}+\cdots ; 71⋅5⋅34⋅4⋅4π=71−491−771+⋯;

which once subtracted gives

相减后得到：

8 ⋅ 4 ⋅ 4 7 ⋅ 5 ⋅ 3 ⋅ π 4 = 1 − 1 11 + 1 13 + 1 17 − ⋯ . \frac{8 \cdot 4 \cdot 4}{7 \cdot 5 \cdot 3} \cdot \frac{\pi}{4}=1-\frac{1}{11}+\frac{1}{13}+\frac{1}{17}-\cdots . 7⋅5⋅38⋅4⋅4⋅4π=1−111+131+171−⋯.

We notice that the denominators which are divisible by a prime number of the form 4 n − 1 4 n-1 4n−1 are removed by addition, from which this new factor gets added, 4 n 4 n − 1 \frac{4 n}{4 n-1} 4n−14n , while the denominators divisible by a prime number of the form 4 n + 1 4 n+1 4n+1 are removed by subtraction from which this new factor gets added, 4 n 4 n + 1 \frac{4 n}{4 n+1} 4n+14n .

可以发现：对于形如 4 n − 1 4n-1 4n−1 的素数，采用加法消去其倍数项，引入因子 4 n 4 n − 1 \dfrac{4n}{4n-1} 4n−14n；对于形如 4 n + 1 4n+1 4n+1 的素数，采用减法消去其倍数项，引入因子 4 n 4 n + 1 \dfrac{4n}{4n+1} 4n+14n。

Thus the denominators of these new factors successively added will be prime numbers while the numerators will be multiples of four, one unit more or less than the denominators.

这些因子的分母为素数，分子为 4 4 4 的倍数，且分子与分母相差 1 1 1。

Consequently if all the terms of the series considered at the beginning are subtracted in this way, we will finally have

按此方式消去原级数所有项后，最终得到：

⋯ 24 ⋅ 20 ⋅ 16 ⋅ 12 ⋅ 12 ⋅ 8 ⋅ 4 ⋅ 4 ⋯ 23 ⋅ 19 ⋅ 17 ⋅ 13 ⋅ 11 ⋅ 7 ⋅ 5 ⋅ 3 ⋅ π 4 = 1. \frac{\cdots 24 \cdot 20 \cdot 16 \cdot 12 \cdot 12 \cdot 8 \cdot 4 \cdot 4}{\cdots 23 \cdot 19 \cdot 17 \cdot 13 \cdot 11 \cdot 7 \cdot 5 \cdot 3} \cdot \frac{\pi}{4}=1 . ⋯23⋅19⋅17⋅13⋅11⋅7⋅5⋅3⋯24⋅20⋅16⋅12⋅12⋅8⋅4⋅4⋅4π=1.

From this we will get

整理后即得：

π 4 = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋯ 4 ⋅ 4 ⋅ 8 ⋅ 12 ⋅ 12 ⋅ 16 ⋅ 20 ⋅ 24 ⋯ . \frac{\pi}{4}=\frac{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdots}{4 \cdot 4 \cdot 8 \cdot 12 \cdot 12 \cdot 16 \cdot 20 \cdot 24 \cdots} . 4π=4⋅4⋅8⋅12⋅12⋅16⋅20⋅24⋯3⋅5⋅7⋅11⋅13⋅17⋅19⋅23⋯.

Q.E. D.

证毕。

Theorem 12

定理 12

If all the odd prime numbers are separated in two parts, one of the parts one unit greater than the other and we take the even parts as numerators and the odd parts as denominators, we will get the continued product

将全体奇素数拆分为相差 1 1 1 的两个数，取其中偶数作为连乘积分子、奇数作为分母，则该无穷连乘积的值为：

2 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋯ 1 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 9 ⋅ 11 ⋯ = 2. \frac{2 \cdot 2 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 9 \cdot 11 \cdots}=2 . 1⋅3⋅3⋅5⋅7⋅9⋅9⋅11⋯2⋅2⋅4⋅6⋅6⋅8⋅10⋅12⋯=2.

Proof

证明

As by the previous Theorem

由上一定理：

π 4 = 3 ⋅ 5 ⋅ 7 ⋯ 4 ⋅ 4 ⋅ 8 ⋯ \frac{\pi}{4}=\frac{3 \cdot 5 \cdot 7 \cdots}{4 \cdot 4 \cdot 8 \cdots} 4π=4⋅4⋅8⋯3⋅5⋅7⋯

we will have

两侧取倒数并平方可得：

16 π 2 = 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 8 ⋅ 8 ⋅ 12 ⋅ 12 ⋅ 12 ⋅ 12 ⋅ 12 ⋅ 16 ⋅ 16 ⋯ 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 7 ⋅ 7 ⋅ 11 ⋅ 11 ⋅ 13 ⋅ 13 ⋅ 13 ⋅ 17 ⋅ 17 ⋯ \frac{16}{\pi ^{2}}=\frac{4\cdot 4\cdot 4\cdot 4\cdot 8\cdot 8\cdot 12\cdot 12\cdot 12\cdot 12\cdot 12\cdot 16\cdot 16\cdots }{3\cdot 3\cdot 5\cdot 5\cdot 7\cdot 7\cdot 11\cdot 11\cdot 13\cdot 13\cdot 13\cdot 17\cdot 17\cdots } π216=3⋅3⋅5⋅5⋅7⋅7⋅11⋅11⋅13⋅13⋅13⋅17⋅17⋯4⋅4⋅4⋅4⋅8⋅8⋅12⋅12⋅12⋅12⋅12⋅16⋅16⋯

But, by Corollary 1 of Theorem 8, if multiplied by 3 / 4 3/4 3/4 we have

结合定理 8 推论 1，将其乘以 3 4 \dfrac{3}{4} 43 得：

π 2 8 = 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 7 ⋅ 7 ⋅ 11 ⋅ 11 ⋅ 13 ⋅ 13 ⋯ 2 ⋅ 4 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 12 ⋅ 14 ⋯ , \frac{\pi^{2}}{8}=\frac{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 11 \cdot 11 \cdot 13 \cdot 13 \cdots}{2 \cdot 4 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 12 \cdot 14 \cdots}, 8π2=2⋅4⋅4⋅6⋅8⋅10⋅12⋅12⋅14⋯3⋅3⋅5⋅5⋅7⋅7⋅11⋅11⋅13⋅13⋯,

and both expressions are formed by odd prime numbers.

以上两式均由奇素数相关项构造。

If we multiply them reciprocally, the denominator of the first one will cancel the numerator of the second one and, moreover, the central part of the terms of both the numerator of the former and the denominator of the latter will vanish.

将两式交叉相乘，第一式分母与第二式分子相互抵消，两式中间项也一并消去。

It will remain,

化简后得到：

2 = 4 ⋅ 4 ⋅ 8 ⋅ 12 ⋅ 12 ⋅ 16 ⋅ 20 ⋅ 24 ⋯ 2 ⋅ 6 ⋅ 6 ⋅ 10 ⋅ 14 ⋅ 18 ⋅ 18 ⋅ 22 ⋯ , 2=\frac{4\cdot 4\cdot 8\cdot 12\cdot 12\cdot 16\cdot 20\cdot 24\cdots }{2\cdot 6\cdot 6\cdot 10\cdot 14\cdot 18\cdot 18\cdot 22\cdots }, 2=2⋅6⋅6⋅10⋅14⋅18⋅18⋅22⋯4⋅4⋅8⋅12⋅12⋅16⋅20⋅24⋯,

where the numerators are multiples of four and the denominators even numbers not divisible by four, both one unit greater or less than the odd prime numbers.

式中分子为 4 4 4 的倍数，分母为非 4 4 4 倍数的偶数，分子、分母均与对应奇素数相差 1 1 1。

If we now simplify by two the fractions one by one, the numerators will be even numbers and the denominators odd; matching couples will differ by one and added together will be prime numbers.

对每一项依次约去因子 2 2 2，此时分子变为一般偶数，分母变为奇数；每一组配对项差值为 1 1 1，两项之和为素数。

Consequently we will have

最终可得：

2 = 2 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋯ 1 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 9 ⋅ 11 ⋯ 2=\frac{2 \cdot 2 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 9 \cdot 11 \cdots} 2=1⋅3⋅3⋅5⋅7⋅9⋅9⋅11⋯2⋅2⋅4⋅6⋅6⋅8⋅10⋅12⋯

Q.E.D.

证毕。

Theorem 13

定理 13

If all the odd prime numbers are separated into two parts, one a unit greater than the other and we use the even parts as numerators and the odd parts as denominators we will have

将全体奇素数拆分为相差 1 1 1 的两个数，取偶数部分为分子、奇数部分为分母，可得等式：

π 4 = 4 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 14 ⋅ 16 ⋅ 18 ⋅ 20 ⋅ 22 ⋅ 24 ⋯ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 13 ⋅ 17 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 25 ⋯ \frac {\pi }{4}=\frac {4\cdot 8\cdot 10\cdot 12\cdot 14\cdot 16\cdot 18\cdot 20\cdot 22\cdot 24\cdots }{5\cdot 7\cdot 11\cdot 13\cdot 13\cdot 17\cdot 17\cdot 19\cdot 23\cdot 25\cdots } 4π=5⋅7⋅11⋅13⋅13⋅17⋅17⋅19⋅23⋅25⋯4⋅8⋅10⋅12⋅14⋅16⋅18⋅20⋅22⋅24⋯

Proof

证明

By WALLis' quadrature of the circle we have

根据沃利斯圆求积公式：

π 2 = 2 ⋅ 2 ⋅ 4 ⋅ 4 ⋅ 6 ⋅ 6 ⋅ 8 ⋅ 8 ⋅ 10 ⋅ 10 ⋅ 12 ⋅ 12 ⋯ 1 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 7 ⋅ 7 ⋅ 9 ⋅ 9 ⋅ 11 ⋅ 11 ⋅ 13 ⋯ , \frac {\pi }{2}=\frac {2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdot 8\cdot 10\cdot 10\cdot 12\cdot 12\cdots }{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\cdot 7\cdot 9\cdot 9\cdot 11\cdot 11\cdot 13\cdots }, 2π=1⋅3⋅3⋅5⋅5⋅7⋅7⋅9⋅9⋅11⋅11⋅13⋯2⋅2⋅4⋅4⋅6⋅6⋅8⋅8⋅10⋅10⋅12⋅12⋯,

expression in which if we add all numerators one by one to their corresponding denominators all odd numbers result.

该式中每一组分子与分母相加，结果为全体奇数。

Now, as from the previous Theorem a similar expression formed only by odd prime numbers has the value 2 2 2,

由上一定理可知，仅由奇素数构造的同类连乘积值为 2 2 2：

2 = 2 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 6 ⋅ 8 ⋅ 10 ⋅ 12 ⋯ 1 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 ⋅ 9 ⋅ 11 ⋯ , 2=\frac{2 \cdot 2 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdots}{1 \cdot 3\cdot 3\cdot 5\cdot 7\cdot 9\cdot 9\cdot 11\cdots }, 2=1⋅3⋅3⋅5⋅7⋅9⋅9⋅11⋯2⋅2⋅4⋅6⋅6⋅8⋅10⋅12⋯,

dividing the former expression by this one we have

将沃利斯公式除以该式，得到：

π 4 = 4 ⋅ 8 ⋅ 10 ⋅ 12 ⋅ 14 ⋅ 16 ⋅ 18 ⋅ 20 ⋅ 22 ⋅ 24 ⋯ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 13 ⋅ 17 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 25 ⋯ , \frac {\pi }{4}=\frac {4\cdot 8\cdot 10\cdot 12\cdot 14\cdot 16\cdot 18\cdot 20\cdot 22\cdot 24\cdots }{5\cdot 7\cdot 11\cdot 13\cdot 13\cdot 17\cdot 17\cdot 19\cdot 23\cdot 25\cdots }, 4π=5⋅7⋅11⋅13⋅13⋅17⋅17⋅19⋅23⋅25⋯4⋅8⋅10⋅12⋅14⋅16⋅18⋅20⋅22⋅24⋯,

which, similarly, is formed by odd numbers not prime.

此式由非素奇数构造。

That is to say, the numerators are even numbers and the denominators are odd numbers one unit from the numerators and such that, added together one by one, we get all the odd numbers not prime.

分子为偶数，分母为奇数，二者相差 1 1 1；每组分子与分母相加，得到全体非素奇数。

Q. E. D.

证毕。

Theorem 14

定理 14

If π \pi π denotes the perimeter of the circle whose diameter is one, we have

设直径为 1 1 1 的圆的周长为 π \pi π，则有：

π 2 = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 29 ⋅ 31 ⋯ 2 ⋅ 6 ⋅ 6 ⋅ 10 ⋅ 14 ⋅ 18 ⋅ 18 ⋅ 22 ⋅ 30 ⋅ 30 ⋯ , \frac {\pi }{2}=\frac {3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\cdots }{2\cdot 6\cdot 6\cdot 10\cdot 14\cdot 18\cdot 18\cdot 22\cdot 30\cdot 30\cdots }, 2π=2⋅6⋅6⋅10⋅14⋅18⋅18⋅22⋅30⋅30⋯3⋅5⋅7⋅11⋅13⋅17⋅19⋅23⋅29⋅31⋯,

expression whose numerators constitute the sequence of all odd prime numbers and the denominators are even numbers not divisible by four one unit greater or less than the corresponding numerators.

该连乘积分子为全体奇素数，分母为非 4 4 4 倍数的偶数，且分母与对应分子相差 1 1 1。

Proof

证明

By Corollary 1 of Theorem 8, if multiplied by 3 / 4 3/4 3/4 we have

将定理 8 推论 1 的式子乘以 3 4 \dfrac{3}{4} 43，得：

where the numerators are twice the odd prime numbers and the denominators are either multiples of four or even numbers not divisible by four one unit greater or lesser than the same prime numbers.

该式分子为奇素数的二次项，分母分为两类： 4 4 4 的倍数、与素数相差 1 1 1 的非 4 4 4 倍数偶数。

Besides, by Theorem 11 we had

结合定理 11 结论：

expression whose numerators are the odd prime numbers once and whose denominators are multiples of four one unit apart from the prime numbers.

其分子为一次奇素数，分母为与素数相差 1 1 1 的 4 4 4 的倍数。

Thus this expression is contained in the former and if we divide the former by this one we have

用前式除以此式，化简得：

π 2 = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋯ 2 ⋅ 6 ⋅ 6 ⋅ 10 ⋅ 14 ⋅ 18 ⋅ 18 ⋯ \frac{\pi}{2}=\frac{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdots}{2 \cdot 6 \cdot 6 \cdot 10 \cdot 14 \cdot 18 \cdot 18 \cdots} 2π=2⋅6⋅6⋅10⋅14⋅18⋅18⋯3⋅5⋅7⋅11⋅13⋅17⋅19⋯

where the odd prime numbers are the numerators and the denominators are even numbers not divisible by four one unit greater or less than the numerators.

式中分子为奇素数，分母为与分子相差 1 1 1 的非 4 4 4 倍数偶数。

Q.E. D.

证毕。

Theorem 15

定理 15

Being π \pi π the perimeter of the circle whose diameter is one, we have

设直径为 1 1 1 的圆的周长为 π \pi π，则：

π 2 = 1 + 1 3 − 1 5 + 1 7 + 1 9 + 1 11 − 1 13 − 1 15 − 1 17 + 1 19 + 1 21 + 1 23 + 1 25 + 1 27 − 1 29 + 1 31 + 1 33 − 1 35 − 1 35 − 1 37 − ⋯ , \begin{aligned} \frac{\pi}{2}=1 & +\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{13}-\frac{1}{15}-\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23} \\ & +\frac{1}{25}+\frac{1}{27}-\frac{1}{29}+\frac{1}{31}+\frac{1}{33}-\frac{1}{35}-\frac{1}{35}-\frac{1}{37}-\cdots, \end{aligned} 2π=1+31−51+71+91+111−131−151−171+191+211+231+251+271−291+311+331−351−351−371−⋯,

series whose denominators are all the odd numbers and whose signs obey to the following law: consider those prime numbers of the form 4 n − 1 4n -1 4n−1 with the sign + + + and those prime numbers of the form 4 n + 1 4n +1 4n+1 with the sign − - −. Composite numbers bear the sign that correspond to them according the rules of product applied to their prime number factorization.

该级数分母为全体奇数，符号遵循如下规则：形如 4 n − 1 4n-1 4n−1 的素数对应项取正号，形如 4 n + 1 4n+1 4n+1 的素数对应项取负号；合数的符号由其素因数分解后，按照乘积符号规则确定。

Proof

证明

In the same way that, with the operations we use, this series

借助前文运算方法，可将级数

1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + 1 13 − ⋯ , 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\cdots, 1−31+51−71+91−111+131−⋯,

becomes the expression

转化为连乘积

3 ⋅ 5 ⋅ 7 ⋅ 11 ⋯ 4 ⋅ 4 ⋅ 8 ⋅ 12 ⋯ \frac{3 \cdot 5 \cdot 7 \cdot 11 \cdots}{4 \cdot 4 \cdot 8 \cdot 12 \cdots} 4⋅4⋅8⋅12⋯3⋅5⋅7⋅11⋯

we can again devise another method to transform the expression

反之，也可将该连乘积还原为原级数。

3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋯ 4 ⋅ 4 ⋅ 8 ⋅ 12 ⋅ 12 ⋯ \frac{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdots}{4 \cdot 4 \cdot 8 \cdot 12 \cdot 12 \cdots} 4⋅4⋅8⋅12⋅12⋯3⋅5⋅7⋅11⋅13⋯

into the series

1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + 1 13 − ⋯ . 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\cdots . 1−31+51−71+91−111+131−⋯.

Applying this method to the expression found in the previous Theorem

将此方法应用于上一定理的连乘积表达式：

π 2 = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋯ 2 ⋅ 6 ⋅ 6 ⋅ 10 ⋅ 14 ⋅ 18 ⋯ \frac{\pi}{2}=\frac{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdots}{2 \cdot 6 \cdot 6 \cdot 10 \cdot 14 \cdot 18 \cdots} 2π=2⋅6⋅6⋅10⋅14⋅18⋯3⋅5⋅7⋅11⋅13⋅17⋯

this expression will become the series we contend

便可得到本定理所给出的级数：

1 + 1 3 − 1 5 + 1 7 + 1 9 + 1 11 − ⋯ , 1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots, 1+31−51+71+91+111−⋯,

whose sum is π / 2 \pi / 2 π/2.

其和为 π 2 \dfrac{\pi}{2} 2π。

The same result can be reached a posterior writing

下面通过直接推导验证：

x = 1 + 1 3 − 1 5 + 1 7 + 1 9 + 1 11 − 1 13 − 1 15 − 1 17 + ⋯ , x=1+{\frac{1}{3}}-{\frac{1}{5}}+{\frac{1}{7}}+{\frac{1}{9}}+{\frac{1}{11}}-{\frac{1}{13}}-{\frac{1}{15}}-{\frac{1}{17}}+\cdots , x=1+31−51+71+91+111−131−151−171+⋯,

from where

两侧同乘 1 3 \dfrac{1}{3} 31：

1 3 x = 1 3 + 1 9 − 1 15 + 1 21 + 1 27 + 1 33 − ⋯ . \frac{1}{3} x=\frac{1}{3}+\frac{1}{9}-\frac{1}{15}+\frac{1}{21}+\frac{1}{27}+\frac{1}{33}-\cdots . 31x=31+91−151+211+271+331−⋯.

By subtraction we have

两式相减：

2 3 x = 1 − 1 5 + 1 7 + 1 11 − ⋯ . \frac{2}{3} x=1-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}-\cdots . 32x=1−51+71+111−⋯.

Now, in the same way,

对上式两侧同乘 1 5 \dfrac{1}{5} 51：

1 5 ⋅ 2 3 x = 1 5 − 1 25 + 1 35 + 1 55 − ⋯ , \frac{1}{5} \cdot \frac{2}{3} x=\frac{1}{5}-\frac{1}{25}+\frac{1}{35}+\frac{1}{55}-\cdots, 51⋅32x=51−251+351+551−⋯,

which added to the former,

两式相加：

6 ⋅ 2 5 ⋅ 3 x = 1 + 1 7 + 1 11 − 1 13 − ⋯ . \frac{6 \cdot 2}{5 \cdot 3} x=1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}-\cdots . 5⋅36⋅2x=1+71+111−131−⋯.

Removing in the same way all the terms except for the first 1 1 1, we have

按此方式消去除首项 1 1 1 以外的所有项，最终可得：

x = 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 19 ⋯ 2 ⋅ 6 ⋅ 6 ⋅ 10 ⋅ 14 ⋅ 18 ⋅ 18 ⋯ = π 2 . x=\frac{3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdots}{2 \cdot 6 \cdot 6 \cdot 10 \cdot 14 \cdot 18 \cdot 18 \cdots}=\frac{\pi}{2} . x=2⋅6⋅6⋅10⋅14⋅18⋅18⋯3⋅5⋅7⋅11⋅13⋅17⋅19⋯=2π.

And from here, at the same time we will obtain the law of signs described in the statement of the Theorem.

同时也推导出了定理中所述的符号规则。

Q.E. D.

证毕。

Corollary 1

推论 1

The sum of the proposed series

级数

1 + 1 3 − 1 5 + 1 7 + 1 9 + 1 11 − 1 13 − ⋯ 1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{13}-\cdots 1+31−51+71+91+111−131−⋯

is twice the sum of this series

的和，是级数

1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + ⋯ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots 1−31+51−71+91−111+⋯

sum's twice.

和的两倍。

As the fractions are the same in both series, the only reason for one being double than the other must be found in the signs.

两个级数的项完全一致，和产生倍数关系的原因仅在于各项符号不同。

Theorem 16

定理 16

Let π \pi π be the perimeter of the circle whose diameter is one. We have

设直径为 1 1 1 的圆的周长为 π \pi π，则：

π 2 = 1 + 1 2 − 1 6 + 1 6 + 1 10 − 1 14 − 1 16 − 1 18 + 1 18 + 1 20 + ⋯ . \frac {\pi }{2}=1+\frac {1}{2}-\frac {1}{6}+\frac {1}{6}+\frac {1}{10}-\frac {1}{14}-\frac {1}{16}-\frac {1}{18}+\frac {1}{18}+\frac {1}{20}+\cdots . 2π=1+21−61+61+101−141−161−181+181+201+⋯.

The denominators of the positive fractions are one unit less than the odd numbers which are not powers and the denominators of the negative fractions are one unit greater. The sign of each fraction is in accordance with the sign of the odd number one unit greater or less not a power obtained through the law of the previous Theorem.

正项的分母为非幂次奇数减 1 1 1，负项的分母为非幂次奇数加 1 1 1；每一项的符号，由与之相差 1 1 1 的非幂次奇数、依照上一定理的符号规则确定。

Proof

证明

This very series is obtained from the conversion of the previous following the method of Theorems 1, 2 and 3 according to which geometrical progressions are added or subtracted from an expression until only the first term remains.

沿用定理 1、2、3 中加减几何级数、最终仅保留首项的方法，对上一定理的级数进行变换，即可得到本级数。

Q E.D.

证毕。

Theorem 17

定理 17

If odd primes of the form 4 n − 1 4n -1 4n−1 are given a + + + sign, and the rest of the form 4 n + 1 4 n+1 4n+1 are given the sign − - − and to the composite numbers are given the signs that are obtained from their prime number factorization obeying the rule of product, we will have

规定：形如 4 n − 1 4n-1 4n−1 的奇素数取正号，形如 4 n + 1 4n+1 4n+1 的奇素数取负号；合数的符号由其素因数分解结果，按乘积符号规则判定。由此构造级数：

3 π 8 = 1 + 1 9 − 1 15 + 1 21 + 1 25 + 1 33 − 1 35 − 1 39 + 1 49 − 1 51 − ⋯ , \frac {3\pi }{8}=1+{\frac {1}{9}}-{\frac {1}{15}}+{\frac {1}{21}}+{\frac {1}{25}}+{\frac {1}{33}}-{\frac {1}{35}}-{\frac {1}{39}}+{\frac {1}{49}}-{\frac {1}{51}}-\cdots , 83π=1+91−151+211+251+331−351−391+491−511−⋯,

series whose denominators are constituted by two, four, six, etc. prime factors.

该级数各项分母包含两个、四个、六个乃至更多个素因子。

Proof

证明

By Theorem 15 we have

由定理 15：

π 2 = 1 + 1 3 − 1 5 + 1 7 + 1 9 + 1 11 − 1 13 − 1 15 − 1 17 + 1 19 + ⋯ , \frac {\pi }{2}=1+\frac {1}{3}-\frac {1}{5}+\frac {1}{7}+\frac {1}{9}+\frac {1}{11}-\frac {1}{13}-\frac {1}{15}-\frac {1}{17}+\frac {1}{19}+\cdots , 2π=1+31−51+71+91+111−131−151−171+191+⋯,

where the denominators are all the odd numbers and the signs follow the law we have stated above;

该级数分母为全体奇数，符号遵循既定规则。

moreover we have

同时已知：

π 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + 1 13 − ⋯ ; \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\cdots ; 4π=1−31+51−71+91−111+131−⋯;

the terms of these series have the signs that their denominators command, whether they are made of two or four or six prime numbers, etc.

两个级数的项均按照分母对应的规则判定符号，无论分母含有两个、四个、六个素因子。

Thus adding them up only those terms will remain and then dividing by two, we will have

将两式相加，保留对应项后再整体除以 2 2 2，可得：

3 π 8 = 1 + 1 9 − 1 15 + 1 21 + 1 25 + 1 33 − ⋯ , \frac{3 \pi}{8}=1+\frac{1}{9}-\frac{1}{15}+\frac{1}{21}+\frac{1}{25}+\frac{1}{33}-\cdots, 83π=1+91−151+211+251+331−⋯,

which is the very series proposed.

即为待证级数。

And by the rule of signs it follows that those terms of the form 4 n + 1 4 n+1 4n+1,have a + a + a+ sign,and the rest, have sign − - −.

根据符号规则，形如 4 n + 1 4n+1 4n+1 的数对应项取正号，其余项取负号。

Q .E. D.

证毕。

Corollary 1

推论 1

If from the series of the Theorem we subtract this one,

用本定理的级数减去

π 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + 1 13 − ⋯ , \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\cdots, 4π=1−31+51−71+91−111+131−⋯,

we will have the series

可得新级数：

π 8 = 1 3 − 1 5 + 1 7 + 1 11 − 1 13 − 1 17 + 1 19 + ⋯ , \frac{\pi}{8}=\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}-\frac{1}{17}+\frac{1}{19}+\cdots, 8π=31−51+71+111−131−171+191+⋯,

whose denominators are either the prime numbers themselves or are constituted by three, five, etc. prime factors; and those terms of the form 4 n − 1 4 n-1 4n−1 have a + a + a+ sign and the rest of the form 4 n + 1 4 n+1 4n+1 have sign − - −.

该级数分母为素数，或含有三个、五个等奇数个素因子；形如 4 n − 1 4n-1 4n−1 的数对应项取正号，形如 4 n + 1 4n+1 4n+1 的数对应项取负号。

Theorem 18

定理 18

If we assign a − - − sign to all the prime numbers and composite numbers are assigned the sign that correspond to them according to the rule of signs in the product and with all the numbers we form the series

规定全体素数对应项取负号，合数的符号由其素因子的符号按乘积规则确定。依此构造无穷级数：

1 − 1 2 − 1 3 + 1 4 − 1 5 + 1 6 − 1 7 − 1 8 + 1 9 + 1 10 − 1 11 − 1 12 − ⋯ 1-{\frac {1}{2}}-{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}-{\frac {1}{7}}-{\frac {1}{8}}+{\frac {1}{9}}+{\frac {1}{10}}-{\frac {1}{11}}-{\frac {1}{12}}-\cdots 1−21−31+41−51+61−71−81+91+101−111−121−⋯

will have, once infinitely continued, sum 0 0 0.

该无穷级数的和为 0 0 0。

Proof

证明

Let x x x be the sum of the series, that is

设该级数和为 x x x，即：

x = 1 − 1 2 − 1 3 + 1 4 − 1 5 + 1 6 − 1 7 − 1 8 + ⋯ ; x=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\cdots ; x=1−21−31+41−51+61−71−81+⋯;

we will have, operating as we did in previous Theorems,

沿用前文定理的运算方法，可得：

3 2 x = 1 − 1 3 − 1 5 − 1 7 + 1 9 − 1 11 − ⋯ ; \frac{3}{2} x=1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\cdots ; 23x=1−31−51−71+91−111−⋯;

and, similarly,

同理：

3 2 ⋅ 4 3 x = 1 − 1 5 − 1 7 − 1 11 − 1 13 − ⋯ . \frac{3}{2} \cdot \frac{4}{3} x=1-\frac{1}{5}-\frac{1}{7}-\frac{1}{11}-\frac{1}{13}-\cdots . 23⋅34x=1−51−71−111−131−⋯.

Finally, repeating the same operations infinitely many times,

对全体素数重复该运算至无穷次，最终得到：

3 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 12 ⋅ 14 ⋯ 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋯ x = 1. \frac{3 \cdot 4 \cdot 6 \cdot 8 \cdot 12 \cdot 14 \cdots}{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdots} x=1 . 2⋅3⋅5⋅7⋅11⋅13⋯3⋅4⋅6⋅8⋅12⋅14⋯x=1.

Now, by Theorem 7,

由定理 7 可知：

2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋯ 1 ⋅ 2 ⋅ 4 ⋅ 6 ⋅ 10 ⋯ = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ = ∞ , \frac{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots}{1 \cdot 2 \cdot 4 \cdot 6 \cdot 10 \cdots}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots=\infty, 1⋅2⋅4⋅6⋅10⋯2⋅3⋅5⋅7⋅11⋯=1+21+31+41+51+⋯=∞,

and it is easily seen that also the coefficient of our x x x is infinitely great.

显然上式中 x x x 的系数也为无穷大。

Thus, in order to be able to equal 1 1 1, x x x must be 0 0 0 and so

无穷大数与 x x x 相乘结果为 1 1 1，可推得 x = 0 x=0 x=0，即：

0 = 1 − 1 2 − 1 3 + 1 4 − 1 5 + 1 6 − 1 7 − 1 8 + ⋯ , 0=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\cdots, 0=1−21−31+41−51+61−71−81+⋯,

whose denominators which either prime numbers themselves or which are formed by 3 , 5 , 7 , 3, 5, 7, 3,5,7, etc. prime factors have a − - − sign and the rest a + + + sign.

该级数中，分母为素数、或含有 3 , 5 , 7 3,5,7 3,5,7 等素因子的项取负号，其余项取正号。

Q.E. D.

证毕。

Corollary 1

推论 1

It is thus obvious the way in which in the harmonic progression the signs have to be distributed in order to have 0 0 0 as the sum of the whole series.

由此可以明确：对调和级数各项配置符号，使其整体和为 0 0 0 的具体规则。

Corollary 2

推论 2

If we have found x = 0 x=0 x=0, also 3 2 x = 0 \frac{3}{2} x=0 23x=0 and, consequently, we will have

已知 x = 0 x=0 x=0，则 3 2 x = 0 \dfrac{3}{2}x=0 23x=0，因此有：

0 = 1 − 1 3 − 1 5 − 1 7 + 1 9 − 1 11 − 1 13 + 1 15 − ⋯ , 0=1-\frac {1}{3}-\frac {1}{5}-\frac {1}{7}+\frac {1}{9}-\frac {1}{11}-\frac {1}{13}+\frac {1}{15}-\cdots , 0=1−31−51−71+91−111−131+151−⋯,

where only the odd numbers occur and the law of signs is as described above.

该级数仅含奇数分母，符号规则与前文一致。

Theorem 19

定理 19

The sum of the reciprocals of the prime numbers,

全体素数的倒数和：

1 2 + 1 3 + 1 5 + 1 7 + 1 11 + 1 13 + ⋯ \frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\cdots 21+31+51+71+111+131+⋯

is infinitely great but is infinitely times less than the sum of the harmonic series

为无穷大，但该无穷大的量级远小于调和级数

1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots 1+21+31+41+51+⋯

And the sum of the former is as the logarithm of the sum of the latter.

的无穷大；素数倒数和等价于调和级数和的对数量级。

Proof

证明

If we write

记：

1 2 + 1 3 + 1 5 + 1 7 + 1 11 + ⋯ = A \frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots=A 21+31+51+71+111+⋯=A

and

1 2 2 + 1 3 2 + 1 5 2 + 1 7 2 + 1 11 2 + ⋯ = B \frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\frac{1}{11^{2}}+\cdots=B 221+321+521+721+1121+⋯=B

and

1 2 3 + 1 3 3 + 1 5 3 + 1 7 3 + 1 11 3 + ⋯ = C \frac{1}{2^{3}}+\frac{1}{3^{3}}+\frac{1}{5^{3}}+\frac{1}{7^{3}}+\frac{1}{11^{3}}+\cdots=C 231+331+531+731+1131+⋯=C

and thus successively we denote all the powers by their corresponding letters, we will have, calling e e e that number whose natural logarithm is 1 1 1,

以此类推，用字母依次标记素数各次幂的倒数和。设自然对数的底数为 e e e，则有：

e A + 1 2 B + 1 3 C + 1 4 D + ⋯ = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + ⋯ . e^{A+\frac {1}{2}B+\frac {1}{3}C+\frac {1}{4}D+\cdots }=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\cdots . eA+21B+31C+41D+⋯=1+21+31+41+51+61+71+⋯.

由对数性质可得：

A + 1 2 B + 1 3 C + 1 4 D + ⋯ = ln ⁡ 2 1 + ln ⁡ 3 2 + ln ⁡ 5 4 + ln ⁡ 7 6 + ⋯ , A+\frac{1}{2} B+\frac{1}{3} C+\frac{1}{4} D+\cdots=\ln \frac{2}{1}+\ln \frac{3}{2}+\ln \frac{5}{4}+\ln \frac{7}{6}+\cdots, A+21B+31C+41D+⋯=ln12+ln23+ln45+ln67+⋯,

and consequently

结合定理 7 结论：

e A + 1 2 B + 1 3 C + 1 4 D + ⋯ = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋯ 1 ⋅ 2 ⋅ 4 ⋅ 6 ⋯ = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + ⋯ e^{A+\frac {1}{2}B+\frac {1}{3}C+\frac {1}{4}D+\cdots }={\frac {2\cdot 3\cdot 5\cdot 7\cdots }{1\cdot 2\cdot 4\cdot 6\cdots }}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+\cdots eA+21B+31C+41D+⋯=1⋅2⋅4⋅6⋯2⋅3⋅5⋅7⋯=1+21+31+41+51+61+⋯

But not only B , C , D B,C,D B,C,D, etc will have finite values, but also

B , C , D B,C,D B,C,D 以及 1 2 B + 1 3 C + 1 4 D + ⋯ \dfrac{1}{2} B+\dfrac{1}{3} C+\dfrac{1}{4} D+\cdots 21B+31C+41D+⋯ 均为有限值。

1 2 B + 1 3 C + 1 4 D + ⋯ \frac{1}{2} B+\frac{1}{3} C+\frac{1}{4} D+\cdots 21B+31C+41D+⋯

will have a finite value.

上述组合项取值有限。

Thus in order to have

由于等式右侧调和级数和为无穷大，因此 A A A 必须为无穷大；相较于无穷大的 A A A，其余有限项可以忽略，于是近似得到：

e A + 1 2 B + 1 3 C + 1 4 D + ⋯ = 1 + 1 2 + 1 3 + 1 4 + ⋯ = ∞ e^{A+\frac{1}{2} B+\frac{1}{3} C+\frac{1}{4} D+\cdots}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=\infty eA+21B+31C+41D+⋯=1+21+31+41+⋯=∞

it is necessary that A A A be an infinitely great quantity and thus with respect to it, the following terms

要使等式成立， A A A 必须是无穷大量；相对于 A A A，后面的项

1 2 B + 1 3 C + 1 4 D + ⋯ \frac{1}{2} B+\frac{1}{3} C+\frac{1}{4} D+\cdots 21B+31C+41D+⋯

will vanish and we will have

可以忽略不计，于是有：

e A = e 1 2 + 1 3 + 1 5 + 1 7 + 1 11 + ⋯ = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ . e^{A}=e^{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots }=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots . eA=e21+31+51+71+111+⋯=1+21+31+41+51+⋯.

Hence, we will have

两侧取自然对数：

1 2 + 1 3 + 1 5 + 1 7 + 1 11 + 1 13 + 1 17 + ⋯ = ln ⁡ ( 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 7 + ⋯ ) \begin{aligned} \frac{1}{2}+ & \frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\cdots \\ & =\ln\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\cdots\right) \end{aligned} 21+31+51+71+111+131+171+⋯=ln(1+21+31+41+51+71+⋯)

and the sum of the former series' will be infinitely times less than this one whose sum is ∞ \infty ∞ and finally,

这说明：素数倒数和为无穷大，但其量级远小于调和级数的无穷大，即：

1 2 + 1 3 + 1 5 + 1 7 + 1 11 + ⋯ = ln ⁡ ∞ . \frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots=\ln \infty . 21+31+51+71+111+⋯=ln∞.

Q.E.D.

证毕。

莱昂哈德·欧拉 | 无穷级数若干注记（1）-CSDN博客
https://blog.csdn.net/u013669912/article/details/162045184

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欧拉的真面目：是严谨的宗师，还是直觉的赌徒？_ 大可数学人生工作室
2025-12-04 07:33 发布于：甘肃省
https://www.sohu.com/a/961156890_348129
E72 -- Variae observationes circa series infinitas...
http://eulerarchive.maa.org/backup/E072.html