Leibniz formula for π
alternating series which converges to π/4
In mathematics, the Leibniz formula for π, named after Gottfried Wilhelm Leibniz, states that
π
4
=
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
=
∑
k
=
0
∞
(
−
1
)
k
2
k
+
1
,
{\displaystyle {\frac {\pi }{4}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}},}
an alternating series.
It is sometimes called the Madhava–Leibniz series as it was first discovered by the Indian mathematician Madhava of Sangamagrama or his followers in the 14th–15th century (see Madhava series), and was later independently rediscovered by James Gregory in 1671 and Leibniz in 1673. The Taylor series for the inverse tangent function, often called Gregory's series, is
arctan
x
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
⋯
=
∑
k
=
0
∞
(
−
1
)
k
x
2
k
+
1
2
k
+
1
.
{\displaystyle \arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots =\sum _{k=0}^{\infty }{\frac {(-1)^{k}x^{2k+1}}{2k+1}}.}
The Leibniz formula is the special case arctan 1 = π/4.
It also is the Dirichlet L-series of the non-principal Dirichlet character of modulus 4 evaluated at s = 1, and therefore the value β(1) of the Dirichlet beta function.
Proofs
Proof 1
π
4
=
arctan
1
=
∫
0
1
1
1
+
x
2
d
x
=
∫
0
1
(
∑
k
=
0
n
(
−
1
)
k
x
2
k
+
(
−
1
)
n
+
1
x
2
n
+
2
1
+
x
2
)
d
x
=
(
∑
k
=
0
n
(
−
1
)
k
2
k
+
1
)
+
(
−
1
)
n
+
1
(
∫
0
1
x
2
n
+
2
1
+
x
2
d
x
)
{\displaystyle {\begin{aligned}{\frac {\pi }{4}}&=\arctan 1\\&=\int _{0}^{1}{\frac {1}{1+x^{2}}}\,dx\\[8pt]&=\int _{0}^{1}\left(\sum _{k=0}^{n}(-1)^{k}x^{2k}+{\frac {(-1)^{n+1}\,x^{2n+2}}{1+x^{2}}}\right)\,dx\\[8pt]&=\left(\sum _{k=0}^{n}{\frac {(-1)^{k}}{2k+1}}\right)+(-1)^{n+1}\left(\int _{0}^{1}{\frac {x^{2n+2}}{1+x^{2}}}\,dx\right)\end{aligned}}}
Considering only the integral in the last term, we have:
0
≤
∫
0
1
x
2
n
+
2
1
+
x
2
d
x
≤
∫
0
1
x
2
n
+
2
d
x
=
1
2
n
+
3
→
0
as
n
→
∞
.
{\displaystyle 0\leq \int _{0}^{1}{\frac {x^{2n+2}}{1+x^{2}}}\,dx\leq \int _{0}^{1}x^{2n+2}\,dx={\frac {1}{2n+3}}\;\rightarrow 0{\text{ as }}n\rightarrow \infty .}
Therefore, by the squeeze theorem, as n → ∞, we are left with the Leibniz series:
π
4
=
∑
k
=
0
∞
(
−
1
)
k
2
k
+
1
{\displaystyle {\frac {\pi }{4}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}}
Proof 2
Let
f
(
z
)
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
z
2
n
+
1
.
{\displaystyle f(z)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}z^{2n+1}.}
When |z| < 1, the series
∑
k
=
0
∞
(
−
1
)
k
z
2
k
{\displaystyle \sum _{k=0}^{\infty }(-1)^{k}z^{2k}}
converges uniformly, so then
arctan
z
=
∫
0
z
1
1
+
t
2
d
t
=
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
z
2
n
+
1
=
f
(
z
)
for
|
z
|
<
1.
{\displaystyle \arctan z=\int _{0}^{z}{\frac {1}{1+t^{2}}}dt=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}z^{2n+1}=f(z)\quad {\text{for }}|z|<1.}
Therefore, if f(z) approaches f(1) so that it is continuous and converges uniformly, the proof is complete, where the series
∑
n
=
0
∞
(
−
1
)
n
2
n
+
1
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}}
converges by Leibniz's test, and also f(z) approaches f(1) from within the Stolz angle, so from Abel's theorem this is correct.
Convergence
Leibniz's formula converges extremely slowly: it exhibits sublinear convergence.
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