This series converges for
x
=
1
{\displaystyle x=1}
log
2
{\displaystyle \log 2}
x
=
1
{\displaystyle x=1}
(A ) , giving
log
2
=
1
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
⋯
.
{\displaystyle \log 2=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .}
But this series is not well adapted to numerical computation, because it converges so slowly that it would be necessary to take 1000 terms in order to get the value of
log
2
{\displaystyle \log 2}
By the theory of logarithms,
(B )
log
1
+
x
1
−
x
=
log
(
1
+
x
)
−
log
(
1
−
x
)
{\displaystyle \log {\frac {1+x}{1-x}}=\log \left(1+x\right)-\log \left(1-x\right)}
By 8, §1
Substituting in (B ) the equivalent series for
log
(
1
+
x
)
{\displaystyle \log(1+x)}
l
o
g
(
1
−
x
)
{\displaystyle log(1-x)}
Exs. 6 and 7 §145 , we get[ 1]
(C )
log
1
+
x
1
−
x
=
2
[
x
+
x
3
3
+
x
5
5
+
x
7
7
+
⋯
]
,
{\displaystyle \log {\frac {1+x}{1-x}}=2\left[x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots \right],}
which is convergent when
x
{\displaystyle x}
(D )
1
+
x
1
−
x
=
M
N
,
{\displaystyle {\frac {1+x}{1-x}}={\frac {M}{N}},}
x
=
M
−
N
M
+
N
,
{\displaystyle x={\frac {M-N}{M+N}},}
and we see that
x
{\displaystyle x}
M
{\displaystyle M}
N
{\displaystyle N}
(D ) into (C ) , we get
(E )
log
M
N
=
log
M
−
log
N
=
2
[
M
−
N
M
+
N
+
1
3
(
M
−
N
M
+
N
)
3
+
1
5
(
M
−
N
M
+
N
)
5
+
⋯
]
,
{\displaystyle {\begin{aligned}\log {\frac {M}{N}}&=\log M-\log N\\&=2\left[{\frac {M-N}{M+N}}+{\frac {1}{3}}\left({\frac {M-N}{M+N}}\right)^{3}+{\frac {1}{5}}\left({\frac {M-N}{M+N}}\right)^{5}+\cdots \right],\\\end{aligned}}}
a series which is convergent for all positive values of
M
{\displaystyle M}
N
{\displaystyle N}
M
{\displaystyle M}
N
{\displaystyle N}
Placing
M
=
2
{\displaystyle M=2}
N
=
1
{\displaystyle N=1}
(E ) , we get
log
2
=
2
[
1
3
+
1
3
⋅
1
3
3
+
1
5
⋅
1
3
5
+
1
7
⋅
1
3
7
+
⋯
]
=
0.69314718
⋯
.
{\displaystyle \log 2=2\left[{\frac {1}{3}}+{\frac {1}{3}}\cdot {\frac {1}{3^{3}}}+{\frac {1}{5}}\cdot {\frac {1}{3^{5}}}+{\frac {1}{7}}\cdot {\frac {1}{3^{7}}}+\cdots \right]=0.69314718\cdots .}
[ Since
log
N
=
log
1
=
0
{\displaystyle \log N=\log 1=0}
M
−
N
M
+
N
=
1
3
{\displaystyle {\tfrac {M-N}{M+N}}={\tfrac {1}{3}}}
↑ The student should notice that we have treated the series as if they were ordinary sums, but they are not; they are limits of sums. To justify this step is beyond the scope of this book.
![{\displaystyle \log {\frac {1+x}{1-x}}=2\left[x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+{\frac {x^{7}}{7}}+\cdots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f526ffd7461e141072ffbe45af49ed06e67849f)
![{\displaystyle {\begin{aligned}\log {\frac {M}{N}}&=\log M-\log N\\&=2\left[{\frac {M-N}{M+N}}+{\frac {1}{3}}\left({\frac {M-N}{M+N}}\right)^{3}+{\frac {1}{5}}\left({\frac {M-N}{M+N}}\right)^{5}+\cdots \right],\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b2208f6c8b642c8393dea5ec98feabeb0c062c1)
![{\displaystyle \log 2=2\left[{\frac {1}{3}}+{\frac {1}{3}}\cdot {\frac {1}{3^{3}}}+{\frac {1}{5}}\cdot {\frac {1}{3^{5}}}+{\frac {1}{7}}\cdot {\frac {1}{3^{7}}}+\cdots \right]=0.69314718\cdots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/810bc8151425fa39af24f79cf316ec4f64740d14)
