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In mathematics, specifically in the field of numerical analysis , Kummer's transformation of series is a method used to accelerate the convergence of an infinite series. The method was first suggested by Ernst Kummer in 1837.
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
lim
n
→
∞
a
n
b
n
=
γ
≠
0
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\gamma \neq 0}
then A is more easily computed as
A
=
γ
B
+
∑
n
=
1
∞
(
1
−
γ
b
n
a
n
)
a
n
=
γ
B
+
∑
n
=
1
∞
a
n
−
γ
b
n
.
{\displaystyle A=\gamma B+\sum _{n=1}^{\infty }\left(1-\gamma {\frac {b_{n}}{a_{n}}}\right)a_{n}=\gamma B+\sum _{n=1}^{\infty }a_{n}-\gamma b_{n}.}
Example We apply the method to accelerate 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}}.}
First 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
A
=
∑
n
=
1
∞
1
16
n
2
−
1
{\displaystyle A=\sum _{n=1}^{\infty }{\frac {1}{16n^{2}-1}}}
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 }
which is a telescoping series with sum 1 ⁄2
γ
=
lim
n
→
∞
1
16
n
2
−
1
1
4
n
2
−
1
=
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}}}={\frac {4n^{2}-1}{16n^{2}-1}}={\frac {1}{4}}}
and Kummer's transformation 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}}.}
This simplifies to
A
=
1
8
−
3
4
∑
n
=
1
∞
1
(
16
n
2
−
1
)
(
4
n
2
−
1
)
{\displaystyle A={\frac {1}{8}}-{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {1}{(16n^{2}-1)(4n^{2}-1)}}}
which converges much faster than the original series.
See also References External links
