### Various representations for famous mathematical constants

In this unusual post, much like in the older post, The beauty of Infinity, we're listing the most famous mathematical constants as representations of infinite series, infinite products and limits.

The list of constants considered in this post, are:

`\zeta(3) = \sum_{n=1}^(\infty) ((30 n - 11) (n!)^4)/(4 n^3 (2 n - 1) ((2 n)!)^2)`

`\zeta(3) = π \sum_{n=1}^(\infty) ((30 n - 11) Γ(n)^2)/(n 16^n (2 n - 1)^3 Γ(n - 1/2)^2)`

` \zeta(3) = -π^2 / 3 \sum_{n=0}^(\infty) (2^(-2 n) (2 n + 5) ζ(2 n))/((2 n + 1) (2 n + 2) (2 n + 3))`

`\zeta(3) = -(4 π^2)/7 \sum_{n=0}^(\infty) (2^(-2 n) ζ(2 n))/((2 n + 1) (2 n + 2))`

`\zeta(3) = -(4 π^2)/7 [log(27/16) + \sum_{n=0}^(\infty) (2^(-2 n) (ζ(2 n) - 1))/((2 n + 1) (2 n + 2))]`

`\zeta(3) = (4 π^2)/35 [(2 G)/π + 1/2 + \sum_{n=1}^(\infty) (16^(-n) ζ(2 n))/((n + 1) (2 n + 1))]`

` \zeta(3) = (4 π^2)/35 [3/2 - log(π/2) + \sum_{n=1}^(\infty) (16^(-n) ζ(2 n))/(n (n + 1) (2 n + 1))]`

`\zeta(3) = (2 π^2)/35 [9 + 138 log(2) - 18 log(3) - 50 log(5) - 2 log(π) + \sum_{n=1}^(\infty) (2^(1 - 4 n) (ζ(2 n) - 1))/(n (n + 1) (2 n + 1))]`

`\zeta(3) = (2 π^2)/9 [log(2) + \sum_{n=0}^(\infty) (2^(1 - 2 n) ζ(2 n))/(2 n + 3)]`

`\zeta(3) = \sum_{n=1}^(\infty) ((-1)^(n - 1) (205 n^2 - 160 n + 32) (n!)^10)/(2 n^5 ((2 n)!)^5)`

`π = [4 + \sum_{n=1}^(\infty) 2^(3 - 2 n) (2 n - 1) ζ(2 n)]^(1/2)`

`γ = \lim_{n -> \infty}[ζ((n+1)/n) - n]`

`γ = \sum_{n=2}^(\infty) ((-1)^n ζ(n))/n`

`γ = \lim_{n -> \infty}[H_n - log(n) - 3/(6 n + 1)]`

`γ = \lim_{n -> \infty}[H_n - log(n) -1/(2 n) + 1/(12 n^2) - 1/(120 n^4) + 1/(252 n^6) - 1/(240 n^8) + 1/(132 n^10)]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - \sum_{k=1}^n (-B_(2 k))/(n^(2k) * 2k)]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - \sum_{k=1}^n ((-1)^k (2 k)!)/(k (2 π)^(2 k) n^(2 k))]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - 2 \sum_{k=1}^n ((-1)^k k^(2 k - 1/2))/(e^(2 k) n^(2 k) π^(2 k - 1/2))]`

`γ = 3/4 - log(2)/2 + \sum_{n=1}^(\infty) [(1 - 1/(2 n + 1)) [ζ(2 n + 1) - 1]]`

`γ = log(2) - \sum_{n=1}^(\infty) (4^(-n) ζ(2 n + 1))/(2 n + 1)`

`γ = \sum_{n=1}^(\infty) ((-1)^n floor((log(n))/(log(2))))/n`

`γ = \sum_{n=1}^(\infty) ((-1)^n * 2 π * log(n + 1/2) - log(2) (i log(1 - (2 n + 1)^(-(2 i π)/(log(2)))) - i log(1 - (2 n + 1)^((2 i π)/(log(2)))) + π))/(π n log(4))`

`G = [3/8 \sum_{n=0}^(\infty) ((n!)^2)/((2 n + 1)^2 (2 n)!)] + π/8 log(2 + sqrt(3))`

`log(2) = \sum_{n=1}^(\infty) 2^(-n)/n`

`log(2) = \sum_{n=0}^(\infty) (n!)/((2 n + 2)!!)`

`log(2) = \sum_{n=0}^(\infty) 1/((2 n + 1) (2 n + 2))`

`log(2) = 2 \sum_{n=0}^(\infty) (3^(-2 n - 1))/(2 n + 1)`

`log(2) = 3 \sum_{n=0}^(\infty) ((-1)^n 2^(-n - 2) (n!)^2)/((2 n + 1)!)`

`log(2) = 1/2 + \sum_{n=0}^(\infty) (-1)^n/(2 (n + 1) (n + 2))`

`log(2) = 1/2 + \sum_{n=1}^(\infty) 1/(2 n (2 n - 1) (2 n + 1))`

`log(2) = 1 + \sum_{n=1}^(\infty) (-1)^n/(n (2 n - 1) (2 n + 1))`

`log(2) = 3/4 + \sum_{n=1}^(\infty) (-1)^n/(2 n (3 n - 1) (3 n + 1))`

`log(2) = 1/2 + \sum_{n=2}^(\infty) (ζ(n) - 1)/(2^n)`

`log(2) = 1 - \sum_{n=1}^(\infty) (2^(1 - 2 n) ζ(2 n))/(2 n + 1)`

`log(2) = \sum_{n=1}^(\infty) (3^(-n) + 4^(-n))/n`

`log(2) = 2/3 + \sum_{n=1}^(\infty) [2^(-4 n - 1) (1/(4 n + 1) + 1/(8 n + 4) + 1/(16 n + 12) + 1/(2 n))]`

`log(2) = \sum_{n=0}^(\infty) (10 * 7^(-4 n - 2) + 14 * 31^(-2 n - 1) + 6 * 161^(-2 n - 1))/(2 n + 1)`

The list of constants considered in this post, are:

- Apéry's constant: `\zeta(3)`
**π**mathematical constant**e**mathematical constant- Euler-Mascheroni constant: `γ`
- Catalan's constant: `\G` or `\beta(2)`
- Natural logarithm of 2: `\log(2)`

Most of this representations have a very fast convergence.

### # Representations for Apéry's constant

`\zeta(3) = 5/2 \sum_{n=1}^(\infty) ((-1)^(n - 1) (n!)^2)/(n^3 (2 n)!)``\zeta(3) = \sum_{n=1}^(\infty) ((30 n - 11) (n!)^4)/(4 n^3 (2 n - 1) ((2 n)!)^2)`

`\zeta(3) = π \sum_{n=1}^(\infty) ((30 n - 11) Γ(n)^2)/(n 16^n (2 n - 1)^3 Γ(n - 1/2)^2)`

` \zeta(3) = -π^2 / 3 \sum_{n=0}^(\infty) (2^(-2 n) (2 n + 5) ζ(2 n))/((2 n + 1) (2 n + 2) (2 n + 3))`

`\zeta(3) = -(4 π^2)/7 \sum_{n=0}^(\infty) (2^(-2 n) ζ(2 n))/((2 n + 1) (2 n + 2))`

`\zeta(3) = (4 π^2)/35 [(2 G)/π + 1/2 + \sum_{n=1}^(\infty) (16^(-n) ζ(2 n))/((n + 1) (2 n + 1))]`

`\zeta(3) = (2 π^2)/9 [log(2) + \sum_{n=0}^(\infty) (2^(1 - 2 n) ζ(2 n))/(2 n + 3)]`

### # Representations for `π`

`π = [\sum_{n=0}^(\infty) (2^(3/2 - 8 n) 99^(-4 n - 2) (26390 n + 1103) (4 n)!)/((n!)^4)]^(-1)`

`π = [\sum_{n=0}^(\infty) (16^(-3 n - 1) (42 n + 5) ((2 n)!)^3)/(n!)^6]^(-1)`

`π = [1/(3335 sqrt(10005)) \sum_{n=0}^(\infty) ((-1)^n (545140134 n + 13591409) (6 n)!)/(2^(18 n + 7) 10005^(3 n) (n!)^3 (3 n)!)]^(-1)`

`π = \exp(1 + \sum_{n=1}^(\infty) (4^(-n) ζ(2 n))/(n (2 n + 1)))`

`π = 4 - \sum_{n=1}^(\infty) 2^(3 - 4 n) ζ(2 n)`

`π = [8 + \sum_{n=0}^(\infty) 16^(1 - n) (2 n - 1) ζ(2 n)]^(1/2)`

`π = [32 - \sum_{n=1}^(\infty) 4^(3 - 2 n) (n - 1) (2 n - 1) ζ(2 n)]^(1/3)`

`π = [\sum_{n=1}^(\infty) 4^(2 - n) n ζ(2 n)]^(1/2)`

`π = 2 \sum_{n=0}^(\infty) (n!)/((2 n + 1)!!)`

`π = [1/3 \sum_{n=1}^(\infty) 2^(5 - 2 n) n^2 ζ(2 n)]^(1/2)`

`π = 2sqrt(2)\exp(\sum_{n=1}^(\infty) (16^(-n) ζ(2 n))/n)`

`π = [8 + \sum_{n=1}^(\infty) (2^(n + 4) (n!)^2)/((2 n + 2)!)]^(1/2)`

`π = [\sum_{n=0}^(\infty) 8/(2 n + 1)^2]^(1/2)`

`π = 2 \prod_{n=1}^(\infty) (4 n^2)/(4 n^2 - 1)`

`π = 4 \prod_{n=1}^(\infty) ((2 n + 1)^2 - 1)/(2 n + 1)^2`

`π = \prod_{n=1}^(\infty) (4 n^3)/(4 n^3 - 3 n + 1)`

`π = \sum_{n=0}^(\infty) (3 sqrt(3) (n!)^2)/(2 (2 n + 1)!)`

`π = [12 - \sum_{n=0}^(\infty) 6/(n (n + 1)^2)]^(1/2)`

`π = [\sum_{n=0}^(\infty) 6/((4 n + 1) (n + 1))] - log(64)`

`π = \sum_{n=0}^(\infty) ((-1)^n 2^(1 - 2 n) (20 n^2 + 21 n + 5))/((2 n + 1) (4 n + 1) (4 n + 3))`

`π = [32 \sum_{n=0}^(\infty) ((-1)^n)/(2 n + 1)^3]^(1/3)`

`π = \lim_{n -> \infty} 1/2 e^(2 n) n^(-2 n - 1) (n!)^2`

`π = \lim_{n -> \infty} (4^(2 n + 1) (n!)^4 * n)/((2 n + 1)!)^2`

`π = \lim_{n -> \infty} (2^(4 n + 1) (n!)^4)/((2 n + 1) ((2 n)!)^2)`

`π = \lim_{n -> \infty} (4^(n+1) (n!)^2 * n)/(((2 n + 1)!!)^2)`

`π = \lim_{n -> \infty} (2^(2 n + 1) (n!)^2)/((2 n - 1)!! (2 n + 1)!!)`

`π = \lim_{n -> \infty} (4^(2 n + 1) (n + 1) (n!)^4)/((2 n + 1)^2 ((2 n)!)^2)`

`π = \lim_{n -> \infty} (16^n (n!)^4)/(n * ((2 n)!)^2)`

`π = \lim_{n -> \infty} sqrt((4^(4 n + 2) n (n!)^7 (n + 1)!)/((2 n + 1)!)^4)`

`π = \lim_{n -> \infty} sqrt((2^(8 n + 7) (n + 1)^4 (2 n^2 + 2 n + 1) (n!)^8)/((2 n + 2)!)^4)`

`π = \lim_{n -> \infty} 1/2 e^(2 n) n^(-2 n - 1) (n!)^2`

`π = \lim_{n -> \infty} (4^(2 n + 1) (n!)^4 * n)/((2 n + 1)!)^2`

`π = \lim_{n -> \infty} (2^(4 n + 1) (n!)^4)/((2 n + 1) ((2 n)!)^2)`

`π = \lim_{n -> \infty} (4^(n+1) (n!)^2 * n)/(((2 n + 1)!!)^2)`

`π = \lim_{n -> \infty} (2^(2 n + 1) (n!)^2)/((2 n - 1)!! (2 n + 1)!!)`

`π = \lim_{n -> \infty} (4^(2 n + 1) (n + 1) (n!)^4)/((2 n + 1)^2 ((2 n)!)^2)`

`π = \lim_{n -> \infty} (16^n (n!)^4)/(n * ((2 n)!)^2)`

`π = \lim_{n -> \infty} sqrt((4^(4 n + 2) n (n!)^7 (n + 1)!)/((2 n + 1)!)^4)`

`π = \lim_{n -> \infty} sqrt((2^(8 n + 7) (n + 1)^4 (2 n^2 + 2 n + 1) (n!)^8)/((2 n + 2)!)^4)`

`π = [6 log^2(2) + 3 \sum_{n=1}^(\infty) (2^(2 - n))/n^2]^(1/2)`

`π = [exp(γ/2)/2 * \prod_{n=1}^(\infty)[exp(-1/(2 n)) (1/(2 n) + 1)]]^(-2)`

`π = 2 \sum_{n=0}^(\infty) ((-1)^n 3^(1/2 - n))/(2 n + 1)`

`π = 4 \sum_{n=0}^(\infty) ((-1)^(n + 1) 1195^(-2 n - 1) (5^(2 n + 1) - 4 * 239^(2 n + 1)))/(2 n + 1)`

`π = [\sum_{n=0}^(\infty) (18 (n!)^2)/((2 n + 2)!)]^(1/2)`

`π = 4 \sum_{n=0}^(\infty) ((-1)^n (2^(-2 n - 1) + 3^(-2 n - 1)))/(2 n + 1)`

`π = [\sum_{n=1}^(\infty) (3240 (n!)^2)/(17 n^4 (2 n)!)]^(1/4)`

`π = \sum_{n=0}^(\infty) 16^(-n) (-1/(4 n + 2) + 4/(8 n + 1) - 1/(8 n + 6) + 1/(-8 n - 5))`

`π = 1/64 \sum_{n=0}^(\infty) (-1)^n/2^(10 n) (-32/(4 n + 1) - 1/(4 n + 3) + 256/(10 n + 1) - 64/(10 n + 3) - 4/(10 n + 5) - 4/(10 n + 7) + 1/(10 n + 9))`

`π = \sum_{n=0}^(\infty) (3^(-2 n - 13/2) (127169/(12 n + 1) - 1070/(12 n + 5) - 131/(12 n + 7) + 2/(2 n + 11)) ((4 n)!)^2 (6 n)!)/(32 (2 n)! (12 n)!)`

`π = \sum_{n=0}^(\infty) (3^(-2 n - 1/2) (-413/(8 n + 3) - 45/(8 n + 5) + 5/(8 n + 7) + 5717/(8 n + 1)) ((4 n)!)^2)/(1024 (8 n)!)`

# Representations for `e`

`e = \sum_{n=0}^(\infty) 1/(n!)`

`e = [\sum_{n=0}^(\infty) ((2 n + 1)!!)/((2 n + 1)!)]^2`

`e = lim_{n -> \infty} (1 + 1/n)^n`

`e = \lim_{n -> \infty} [\sum_{k=0}^n (n! n^(k - n))/(k! (n - k)!)]`

`e = \lim_{n -> \infty} ((n + 1)/(n - 1))^(n/2)`

`π = 4 \sum_{n=0}^(\infty) ((-1)^(n + 1) 1195^(-2 n - 1) (5^(2 n + 1) - 4 * 239^(2 n + 1)))/(2 n + 1)`

`π = [\sum_{n=0}^(\infty) (18 (n!)^2)/((2 n + 2)!)]^(1/2)`

`π = 4 \sum_{n=0}^(\infty) ((-1)^n (2^(-2 n - 1) + 3^(-2 n - 1)))/(2 n + 1)`

`π = [\sum_{n=1}^(\infty) (3240 (n!)^2)/(17 n^4 (2 n)!)]^(1/4)`

`π = \sum_{n=0}^(\infty) 16^(-n) (-1/(4 n + 2) + 4/(8 n + 1) - 1/(8 n + 6) + 1/(-8 n - 5))`

`π = 1/64 \sum_{n=0}^(\infty) (-1)^n/2^(10 n) (-32/(4 n + 1) - 1/(4 n + 3) + 256/(10 n + 1) - 64/(10 n + 3) - 4/(10 n + 5) - 4/(10 n + 7) + 1/(10 n + 9))`

`π = \sum_{n=0}^(\infty) (3^(-2 n - 13/2) (127169/(12 n + 1) - 1070/(12 n + 5) - 131/(12 n + 7) + 2/(2 n + 11)) ((4 n)!)^2 (6 n)!)/(32 (2 n)! (12 n)!)`

`π = \sum_{n=0}^(\infty) (3^(-2 n - 1/2) (-413/(8 n + 3) - 45/(8 n + 5) + 5/(8 n + 7) + 5717/(8 n + 1)) ((4 n)!)^2)/(1024 (8 n)!)`

# Representations for `e`

`e = \sum_{n=0}^(\infty) 1/(n!)`

`e = [\sum_{n=0}^(\infty) ((2 n + 1)!!)/((2 n + 1)!)]^2`

`e = lim_{n -> \infty} (1 + 1/n)^n`

`e = \lim_{n -> \infty} [\sum_{k=0}^n (n! n^(k - n))/(k! (n - k)!)]`

`e = \lim_{n -> \infty} ((n + 1)/(n - 1))^(n/2)`

### # Representations for the Euler-Mascheroni constant

`γ = 1 - \sum_{n=2}^(\infty) (ζ(n) - 1)/n``γ = \lim_{n -> \infty}[ζ((n+1)/n) - n]`

`γ = \sum_{n=2}^(\infty) ((-1)^n ζ(n))/n`

`γ = \lim_{n -> \infty}[H_n - log(n) - 3/(6 n + 1)]`

`γ = \lim_{n -> \infty}[H_n - log(n) -1/(2 n) + 1/(12 n^2) - 1/(120 n^4) + 1/(252 n^6) - 1/(240 n^8) + 1/(132 n^10)]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - \sum_{k=1}^n (-B_(2 k))/(n^(2k) * 2k)]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - \sum_{k=1}^n ((-1)^k (2 k)!)/(k (2 π)^(2 k) n^(2 k))]`

`γ = \lim_{n -> \infty}[H_n - log(n) - 1/(2n) - 2 \sum_{k=1}^n ((-1)^k k^(2 k - 1/2))/(e^(2 k) n^(2 k) π^(2 k - 1/2))]`

`γ = 3/4 - log(2)/2 + \sum_{n=1}^(\infty) [(1 - 1/(2 n + 1)) [ζ(2 n + 1) - 1]]`

`γ = log(2) - \sum_{n=1}^(\infty) (4^(-n) ζ(2 n + 1))/(2 n + 1)`

`γ = \sum_{n=1}^(\infty) ((-1)^n floor((log(n))/(log(2))))/n`

`γ = \sum_{n=1}^(\infty) ((-1)^n * 2 π * log(n + 1/2) - log(2) (i log(1 - (2 n + 1)^(-(2 i π)/(log(2)))) - i log(1 - (2 n + 1)^((2 i π)/(log(2)))) + π))/(π n log(4))`

### # Representations for the Catalan's constant

`G = π/2 [[\sum_{n=1}^(\infty) (16^(-n) ζ(2 n))/(n (2 n + 1))] - log(π/2) + 1]``G = [3/8 \sum_{n=0}^(\infty) ((n!)^2)/((2 n + 1)^2 (2 n)!)] + π/8 log(2 + sqrt(3))`

### # Representations for the natural logarithm of 2

`log(2) = \sum_{n=1}^(\infty) (ζ(2 n) - 1)/n``log(2) = \sum_{n=1}^(\infty) 2^(-n)/n`

`log(2) = \sum_{n=0}^(\infty) (n!)/((2 n + 2)!!)`

`log(2) = \sum_{n=0}^(\infty) 1/((2 n + 1) (2 n + 2))`

`log(2) = 2 \sum_{n=0}^(\infty) (3^(-2 n - 1))/(2 n + 1)`

`log(2) = 3 \sum_{n=0}^(\infty) ((-1)^n 2^(-n - 2) (n!)^2)/((2 n + 1)!)`

`log(2) = 1/2 + \sum_{n=0}^(\infty) (-1)^n/(2 (n + 1) (n + 2))`

`log(2) = 1/2 + \sum_{n=1}^(\infty) 1/(2 n (2 n - 1) (2 n + 1))`

`log(2) = 1 + \sum_{n=1}^(\infty) (-1)^n/(n (2 n - 1) (2 n + 1))`

`log(2) = 3/4 + \sum_{n=1}^(\infty) (-1)^n/(2 n (3 n - 1) (3 n + 1))`

`log(2) = 1/2 + \sum_{n=2}^(\infty) (ζ(n) - 1)/(2^n)`

`log(2) = 1 - \sum_{n=1}^(\infty) (2^(1 - 2 n) ζ(2 n))/(2 n + 1)`

`log(2) = \sum_{n=1}^(\infty) (3^(-n) + 4^(-n))/n`

`log(2) = 2/3 + \sum_{n=1}^(\infty) [2^(-4 n - 1) (1/(4 n + 1) + 1/(8 n + 4) + 1/(16 n + 12) + 1/(2 n))]`

`log(2) = \sum_{n=0}^(\infty) (10 * 7^(-4 n - 2) + 14 * 31^(-2 n - 1) + 6 * 161^(-2 n - 1))/(2 n + 1)`

### # Implementations

The following Sidef script implements all the above formulas (+ a few more) and can be used for comparing the rate of convergence of each formula.**GitHub:**https://github.com/trizen/sidef-scripts/blob/master/Math/mathematical_formulas.sf