Technique
edit
Let
A
=
∑
n
=
1
∞
a
n
{\displaystyle A=\sum _{n=1}^{\infty }a_{n}}
be an infinite sum whose value we wish to compute, and let
B
=
∑
n
=
1
∞
b
n
{\displaystyle B=\sum _{n=1}^{\infty }b_{n}}
be an infinite sum with comparable terms whose value is known.
If the limit
γ
:=
lim
n
→
∞
a
n
b
n
{\displaystyle \gamma :=\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}}
exists, then
a
n
−
γ
b
n
{\displaystyle a_{n}-\gamma \,b_{n}}
is always also a sequence going to zero and the series given by the difference,
∑
n
=
1
∞
(
a
n
−
γ
b
n
)
{\displaystyle \sum _{n=1}^{\infty }(a_{n}-\gamma \,b_{n})}
, converges.
If
γ
≠
0
{\displaystyle \gamma \neq 0}
, this new series differs from the original
∑
n
=
1
∞
a
n
{\displaystyle \sum _{n=1}^{\infty }a_{n}}
and, under broad conditions, converges more rapidly.[1]
We may then compute
A
{\displaystyle A}
as
A
=
γ
B
+
∑
n
=
1
∞
(
a
n
−
γ
b
n
)
{\displaystyle A=\gamma \,B+\sum _{n=1}^{\infty }(a_{n}-\gamma \,b_{n})}
,
where
γ
B
{\displaystyle \gamma B}
is a constant. Where
a
n
≠
0
{\displaystyle a_{n}\neq 0}
, the terms can be written as the product
(
1
−
γ
b
n
/
a
n
)
a
n
{\displaystyle (1-\gamma \,b_{n}/a_{n})\,a_{n}}
.
If
a
n
≠
0
{\displaystyle a_{n}\neq 0}
for all
n
{\displaystyle n}
, the sum is over a component-wise product of two sequences going to zero,
A
=
γ
B
+
∑
n
=
1
∞
(
1
−
γ
b
n
/
a
n
)
a
n
{\displaystyle A=\gamma \,B+\sum _{n=1}^{\infty }(1-\gamma \,b_{n}/a_{n})\,a_{n}}
.
Consider the Leibniz formula for π :
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
=
π
4
.
{\displaystyle 1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,{\frac {1}{9}}\,-\,\cdots \,=\,{\frac {\pi }{4}}.}
We group terms in pairs as
1
−
(
1
3
−
1
5
)
−
(
1
7
−
1
9
)
+
⋯
{\displaystyle 1-\left({\frac {1}{3}}-{\frac {1}{5}}\right)-\left({\frac {1}{7}}-{\frac {1}{9}}\right)+\cdots }
=
1
−
2
(
1
15
+
1
63
+
⋯
)
=
1
−
2
A
{\displaystyle \,=1-2\left({\frac {1}{15}}+{\frac {1}{63}}+\cdots \right)=1-2A}
where we identify
A
=
∑
n
=
1
∞
1
16
n
2
−
1
{\displaystyle A=\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}}
.
We apply Kummer's method to accelerate
A
{\displaystyle A}
, which will give an accelerated sum for computing
π
=
4
−
8
A
{\displaystyle \pi =4-8A}
.
Let
B
=
∑
n
=
1
∞
1
4
n
2
−
1
=
1
3
+
1
15
+
⋯
{\displaystyle B=\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}={\frac {1}{3}}+{\frac {1}{15}}+\cdots }
=
1
2
−
1
6
+
1
6
−
1
10
+
⋯
{\displaystyle \,={\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{6}}-{\frac {1}{10}}+\cdots }
This is a telescoping series with sum value 1 ⁄2 .
In this case
γ
:=
lim
n
→
∞
1
16
n
2
−
1
1
4
n
2
−
1
=
lim
n
→
∞
4
n
2
−
1
16
n
2
−
1
=
1
4
{\displaystyle \gamma :=\lim _{n\to \infty }{\frac {\frac {1}{16n^{2}-1}}{\frac {1}{4n^{2}-1}}}=\lim _{n\to \infty }{\frac {4n^{2}-1}{16n^{2}-1}}={\frac {1}{4}}}
and so Kummer's transformation formula above gives
A
=
1
4
⋅
1
2
+
∑
n
=
1
∞
(
1
−
1
4
1
4
n
2
−
1
1
16
n
2
−
1
)
1
16
n
2
−
1
{\displaystyle A={\frac {1}{4}}\cdot {\frac {1}{2}}+\sum _{n=1}^{\infty }\left(1-{\frac {1}{4}}{\frac {\frac {1}{4n^{2}-1}}{\frac {1}{16n^{2}-1}}}\right){\frac {1}{16n^{2}-1}}}
=
1
8
−
3
4
∑
n
=
1
∞
1
16
n
2
−
1
1
4
n
2
−
1
{\displaystyle ={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}{\frac {1}{4n^{2}-1}}}
which converges much faster than the original series.
Coming back to Leibniz formula, we obtain a representation of
π
{\displaystyle \pi }
that separates
3
{\displaystyle 3}
and involves a fastly converging sum over just the squared even numbers
(
2
n
)
2
{\displaystyle (2n)^{2}}
,
π
=
4
−
8
A
{\displaystyle \pi =4-8A}
=
3
+
6
⋅
∑
n
=
1
∞
1
(
4
(
2
n
)
2
−
1
)
(
(
2
n
)
2
−
1
)
{\displaystyle =3+6\cdot \sum _{n=1}^{\infty }{\frac {1}{(4(2n)^{2}-1)((2n)^{2}-1)}}}
=
3
+
2
15
+
2
315
+
6
5005
+
⋯
{\displaystyle =3+{\frac {2}{15}}+{\frac {2}{315}}+{\frac {6}{5005}}+\cdots }
See also
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References
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External links
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