# List of formulae involving π

The following is a list of significant formulae involving the mathematical constant π. Many of these formulae can be found in the article Pi, or the article Approximations of π.

## Euclidean geometry

$\pi ={\frac {C}{d}}$

where C is the circumference of a circle, d is the diameter.

$A=\pi r^{2}$

where A is the area of a circle and r is the radius.

$V={4 \over 3}\pi r^{3}$

where V is the volume of a sphere and r is the radius.

$SA=4\pi r^{2}$

where SA is the surface area of a sphere and r is the radius.

$H={1 \over 2}\pi ^{2}r^{4}$

where H is the hypervolume of a 3-sphere and r is the radius.

$SV=2\pi ^{2}r^{3}$

where SV is the surface volume of a 3-sphere and r is the radius.

## Physics

$\Lambda ={{8\pi G} \over {3c^{2}}}\rho$
$\Delta x\,\Delta p\geq {\frac {h}{4\pi }}$
$R_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }R+\Lambda g_{\mu \nu }={8\pi G \over c^{4}}T_{\mu \nu }$
$F={\frac {|q_{1}q_{2}|}{4\pi \varepsilon _{0}r^{2}}}$
$\mu _{0}\approx 4\pi \cdot 10^{-7}\,\mathrm {N} /\mathrm {A} ^{2}$
• Period of a simple pendulum with small amplitude:
$T\approx 2\pi {\sqrt {\frac {L}{g}}}$
${\frac {R^{3}}{T^{2}}}={\frac {GM}{4\pi ^{2}}}$
$F={\frac {\pi ^{2}EI}{L^{2}}}$

## Formulae yielding π

### Integrals

$2\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx=\pi$  (integrating two halves $y(x)={\sqrt {r^{2}-x^{2}}}$  to obtain the area of a circle of radius $r=1$ )
$\int _{-\infty }^{\infty }\operatorname {sech} (x)\,dx=\pi$
$\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-1/2t^{2}-x^{2}+xt}\,dx\,dt=\int _{-\infty }^{\infty }\int _{t}^{\infty }e^{-t^{2}-1/2x^{2}+xt}\,dx\,dt=\pi$
$\int _{-1}^{1}{\frac {dx}{\sqrt {1-x^{2}}}}=\pi$
$\int _{-\infty }^{\infty }{\frac {dx}{1+x^{2}}}=\pi$  (integral form of arctan over its entire domain, giving the period of tan).
$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$  (see Gaussian integral).
$\oint {\frac {dz}{z}}=2\pi i$  (when the path of integration winds once counterclockwise around 0. See also Cauchy's integral formula).
$\int _{0}^{\infty }\ln \left(1+{\frac {1}{x^{2}}}\right)\,dx=\pi$ 
$\int _{-\infty }^{\infty }{\frac {\sin x}{x}}\,dx=\pi$
$\int _{0}^{1}{x^{4}(1-x)^{4} \over 1+x^{2}}\,dx={22 \over 7}-\pi$  (see also Proof that 22/7 exceeds π).

Note that with symmetric integrands $f(-x)=f(x)$ , formulas of the form $\int _{-a}^{a}f(x)\,dx$  can also be translated to formulas $2\int _{0}^{a}f(x)\,dx$ .

### Efficient infinite series

$\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}={\frac {\pi }{2}}$  (see also Double factorial)
$12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}={\frac {1}{\pi }}$  (see Chudnovsky algorithm)
${\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}={\frac {1}{\pi }}$  (see Srinivasa Ramanujan, Ramanujan–Sato series)

The following are efficient for calculating arbitrary binary digits of π:

$\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{4^{k}}}\left({\frac {2}{4k+1}}+{\frac {2}{4k+2}}+{\frac {1}{4k+3}}\right)=\pi$ 
$\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right)=\pi$  (see Bailey–Borwein–Plouffe formula)
${\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)=\pi$

Plouffe's series for calculating arbitrary decimal digits of π:

$\sum _{k=1}^{\infty }k{\frac {2^{k}k!^{2}}{(2k)!}}=\pi +3$

### Other infinite series

$\zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}$    (see also Basel problem and Riemann zeta function)
$\zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}$
$\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$  , where B2n is a Bernoulli number.
$\sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi$ 
$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}$    (see Leibniz formula for pi)
$\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}(2n+1)}}=1-{\frac {1}{3\cdot 3}}+{\frac {1}{3^{2}\cdot 5}}-{\frac {1}{3^{3}\cdot 7}}+{\frac {1}{3^{4}\cdot 9}}-\cdots ={\sqrt {3}}\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{\sqrt {12}}}$  (Madhava series)
$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}$
$\sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}$
$\sum _{n=0}^{\infty }\left({\frac {(-1)^{n}}{2n+1}}\right)^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}$
$\sum _{n=0}^{\infty }{\binom {\frac {1}{2}}{n}}{\frac {(-1)^{n}}{2n+1}}=1-{\frac {1}{6}}-{\frac {1}{40}}-\cdots ={\frac {\pi }{4}}$
$\sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}$
$\sum _{n=1}^{\infty }(-1)^{(n^{2}+n)/2+1}\left|G_{\left((-1)^{n+1}+6n-3\right)/4}\right|=|G_{1}|+|G_{2}|-|G_{4}|-|G_{5}|+|G_{7}|+|G_{8}|-|G_{10}|-|G_{11}|+\cdots ={\frac {\sqrt {3}}{\pi }}$  (see Gregory coefficients)
$\sum _{n=0}^{\infty }{\frac {(1/2)_{n}^{2}}{2^{n}n!^{2}}}\sum _{n=0}^{\infty }{\frac {n(1/2)_{n}^{2}}{2^{n}n!^{2}}}={\frac {1}{\pi }}$  (where $(x)_{n}$  is the rising factorial)
$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(2n+1)}}=\pi -3$  (Nilakantha series)
$\sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}$  (where $F_{n}$  is the n-th Fibonacci number)
$\pi =\sum _{n=1}^{\infty }{\frac {(-1)^{\varepsilon (n)}}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots$    (where $\varepsilon (n)$  is the number of prime factors of the form $p\equiv 1\,(\mathrm {mod} \,4)$  of $n$ ; Euler, 1748)

Some formulas relating π and harmonic numbers are given here.

### Machin-like formulae

${\frac {\pi }{4}}=\arctan 1$
${\frac {\pi }{4}}=\arctan {\frac {1}{2}}+\arctan {\frac {1}{3}}$
${\frac {\pi }{4}}=2\arctan {\frac {1}{2}}-\arctan {\frac {1}{7}}$
${\frac {\pi }{4}}=2\arctan {\frac {1}{3}}+\arctan {\frac {1}{7}}$
${\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}$  (the original Machin's formula)
${\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}$
${\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}$
${\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}$
${\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}$
${\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots$

where $F_{n}$  is the n-th Fibonacci number.

### Infinite series

Some infinite series involving π are:

 $\pi ={\frac {1}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {((2n)!)^{3}(42n+5)}{(n!)^{6}{16}^{3n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}{441}^{2n+1}{2}^{10n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(6n+1)\left({\frac {1}{2}}\right)_{n}^{3}}{{4^{n}}(n!)^{3}}}$ $\pi ={\frac {32}{Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {{\sqrt {5}}-1}{2}}\right)^{8n}{\frac {(42n{\sqrt {5}}+30n+5{\sqrt {5}}-1)\left({\frac {1}{2}}\right)_{n}^{3}}{{64^{n}}(n!)^{3}}}$ $\pi ={\frac {27}{4Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {2}{27}}\right)^{n}{\frac {(15n+2)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {15{\sqrt {3}}}{2Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(33n+4)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{3}}\right)_{n}\left({\frac {2}{3}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {85{\sqrt {85}}}{18{\sqrt {3}}Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{85}}\right)^{n}{\frac {(133n+8)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {5{\sqrt {5}}}{2{\sqrt {3}}Z}}$ $Z=\sum _{n=0}^{\infty }\left({\frac {4}{125}}\right)^{n}{\frac {(11n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{6}}\right)_{n}\left({\frac {5}{6}}\right)_{n}}{(n!)^{3}}}$ $\pi ={\frac {2{\sqrt {3}}}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(8n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{n}}}$ $\pi ={\frac {\sqrt {3}}{9Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(40n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{49}^{2n+1}}}$ $\pi ={\frac {2{\sqrt {11}}}{11Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(280n+19)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{99}^{2n+1}}}$ $\pi ={\frac {\sqrt {2}}{4Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(10n+1)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{9}^{2n+1}}}$ $\pi ={\frac {4{\sqrt {5}}}{5Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(644n+41)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}5^{n}{72}^{2n+1}}}$ $\pi ={\frac {4{\sqrt {3}}}{3Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(28n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{3^{n}}{4}^{n+1}}}$ $\pi ={\frac {4}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(20n+3)\left({\frac {1}{2}}\right)_{n}\left({\frac {1}{4}}\right)_{n}\left({\frac {3}{4}}\right)_{n}}{(n!)^{3}{2}^{2n+1}}}$ $\pi ={\frac {72}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(260n+23)}{(n!)^{4}4^{4n}18^{2n}}}$ $\pi ={\frac {3528}{Z}}$ $Z=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(4n)!(21460n+1123)}{(n!)^{4}4^{4n}882^{2n}}}$ where $(x)_{n}$  is the Pochhammer symbol for the rising factorial. See also Ramanujan–Sato series.

### Infinite products

${\frac {\pi }{4}}=\left(\prod _{p\equiv 1{\pmod {4}}}{\frac {p}{p-1}}\right)\cdot \left(\prod _{p\equiv 3{\pmod {4}}}{\frac {p}{p+1}}\right)={\frac {3}{4}}\cdot {\frac {5}{4}}\cdot {\frac {7}{8}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdots ,$  (Euler)
where the numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
${\frac {{\sqrt {3}}\pi }{6}}=\left(\displaystyle \prod _{p\equiv 1{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p-1}}\right)\cdot \left(\displaystyle \prod _{p\equiv 5{\pmod {6}} \atop p\in \mathbb {P} }{\frac {p}{p+1}}\right)={\frac {5}{6}}\cdot {\frac {7}{6}}\cdot {\frac {11}{12}}\cdot {\frac {13}{12}}\cdot {\frac {17}{18}}\cdots ,$
$\prod _{n=1}^{\infty }{\frac {4n^{2}}{4n^{2}-1}}={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {4}{3}}\cdot {\frac {16}{15}}\cdot {\frac {36}{35}}\cdot {\frac {64}{63}}\cdots ={\frac {\pi }{2}}$  (see also Wallis product)
${\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdot \cdots ={\frac {2}{\pi }}$

A double infinite product formula involving the Thue-Morse sequence:

$\prod _{m\geq 1}\prod _{n\geq 1}\left({\frac {(4m^{2}+n-2)(4m^{2}+2n-1)^{2}}{4(2m^{2}+n-1)(4m^{2}+n-1)(2m^{2}+n)}}\right)^{\epsilon _{n}}={\frac {\pi }{2}},$
where $\epsilon _{n}=(-1)^{t_{n}}$  and $t_{n}$  is the Thue-Morse sequence (Tóth 2020).

### Arctangent formulas

${\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2$
${\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},$

where $a_{k}={\sqrt {2+a_{k-1}}}$  such that $a_{1}={\sqrt {2}}$ .

$\pi =\arctan a+\arctan b+\arctan c$

whenever $a+b+c=abc$  and $a$ , $b$ , $c$  are positive real numbers (see List of trigonometric identities). A special case is

$\pi =\arctan 1+\arctan 2+\arctan 3.$

### Continued fractions

$\pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+\ddots \,}}}}}}}}}$
$\pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}$
$\pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}$
$2\pi ={6+{\cfrac {2^{2}}{12+{\cfrac {6^{2}}{12+{\cfrac {10^{2}}{12+{\cfrac {14^{2}}{12+{\cfrac {18^{2}}{12+\ddots }}}}}}}}}}}$

For more on the third identity, see Euler's continued fraction formula.

### Miscellaneous

$n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}$  (Stirling's approximation)
$e^{i\pi }+1=0$  (Euler's identity)
$\sum _{k=1}^{n}\varphi (k)\sim {\frac {3n^{2}}{\pi ^{2}}}$  (see Euler's totient function)
$\sum _{k=1}^{n}{\frac {\varphi (k)}{k}}\sim {\frac {6n}{\pi ^{2}}}$  (see Euler's totient function)
$\pi =\mathrm {B} (1/2,1/2)=\Gamma (1/2)^{2}$  (see also Beta function and Gamma function)
$\pi ={\frac {\Gamma (3/4)^{4}}{\operatorname {agm} (1,1/{\sqrt {2}})^{2}}}={\frac {\Gamma \left({1/4}\right)^{4/3}\operatorname {agm} (1,{\sqrt {2}})^{2/3}}{2}}$  (where agm is the arithmetic–geometric mean)
$\pi =\operatorname {agm} \left(\theta _{2}^{2}(1/e),\theta _{3}^{2}(1/e)\right)$  (where $\theta _{2}$  and $\theta _{3}$  are the Jacobi theta functions)
$\pi =-{\frac {\operatorname {K} (k)}{\operatorname {K} \left({\sqrt {1-k^{2}}}\right)}}\ln q,\quad k={\frac {\theta _{2}^{2}(q)}{\theta _{3}^{2}(q)}}$  (where $q\in (0,1)$  and $\operatorname {K} (k)$  is the complete elliptic integral of the first kind with modulus $k$ ; reflecting the nome-modulus inversion problem)
$\pi =-{\frac {\operatorname {agm} \left(1,{\sqrt {1-k'^{2}}}\right)}{\operatorname {agm} (1,k')}}\ln q,\quad k'={\frac {\theta _{4}^{2}(q)}{\theta _{3}^{2}(q)}}$  (where $q\in (0,1)$ )
$\operatorname {agm} (1,{\sqrt {2}})={\frac {\pi }{\varpi }}$  (due to Gauss, $\varpi$  is the lemniscate constant)
$\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}\sum _{k=1}^{n}(n{\bmod {k}})=1-{\frac {\pi ^{2}}{12}}$  (where ${\textstyle n{\bmod {k}}}$  is the remainder upon division of n by k)
$\pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r\end{cases}}$  (summing a circle's area)
$\pi =\lim _{n\rightarrow \infty }{\frac {4}{n^{2}}}\sum _{k=1}^{n}{\sqrt {n^{2}-k^{2}}}$  (Riemann sum to evaluate the area of the unit circle)
$\pi =\lim _{n\rightarrow \infty }{\frac {2^{4n}}{n{2n \choose n}^{2}}}=\lim _{n\rightarrow \infty }{\frac {1}{n}}\left({\frac {(2n)!!}{(2n-1)!!}}\right)^{2}$  (by Stirling's approximation)
$a_{0}=1,\,a_{n+1}=\left(1+{\frac {1}{2n+1}}\right)a_{n},\,\pi =\lim _{n\to \infty }{\frac {a_{n}^{2}}{n}}$  (recurrence form of the above formula)
$a_{1}=0,\,a_{n+1}={\sqrt {2+a_{n}}},\,\pi =\lim _{n\to \infty }2^{n}{\sqrt {2-a_{n}}}$  (closely related to Viète's formula)
$a_{1}=1,\,a_{n+1}=a_{n}+\sin a_{n},\,\pi =\lim _{n\to \infty }a_{n}$  (cubic convergence)
$a_{0}=2{\sqrt {3}},\,b_{0}=3,\,a_{n+1}=\operatorname {hm} (a_{n},b_{n}),\,b_{n+1}=\operatorname {gm} (a_{n+1},b_{n}),\,\pi =\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }b_{n}$  (Archimedes' algorithm, see also harmonic mean and geometric mean)