11 Series Expansion of Reciprocal of Gamma Function
11.1 Functions with Integers as Roots
Function fz with negative integers as roots can be described as follows.

f(z) = 1+
z
1

1+
z
2

1+



z
z
 = Π 1+
3
n
n =1

However, this infinite product diverges. That is, such a function cannot exist alone. So, let us consider the
following function which added compensation to each factor.
Π
n =1 

z
1+
n
e
z
n
-
Then, this infinite product uniformly converges for arbitrary
and results in function
z
e  z/1+ z .
This was shown by Weierstrass. If the functions with various integers as roots are shown, it is as follows.
In addition, these left sides can be expanded to series of
z
. So, we call the formulas factorization around 0.
Formula 11.1.1 (factorization around 0)
When

Π
n =1 

z
1+
n
Π
n =1 

z
1n
e
e
z
n
z
1+
2n
Π
n =1 

z
12n
z
2n
1
=
e
(1+ z/2)
z
2n
1
=
e
(1- z/2)
z
2
(1,2-)
z
2
(1,2+)

z


z


z
2

e
z
n
-1-
(1,4-)
z
2
Weierstrass expression on the gamma function was as follows.
-
(1,3+)
z
Proof

1
z
= e  z z Π 1+
(z)
n
n =1
(1,3-)

z

z
1e 2n =
e
2n -1
(1- z)/2
Π
n =1 

(1,1+)

z
1+
e 2n =
e
2n -1
(1+ z)/2
Π
n =1 

e
e z
(1- z)

z
 +log 2z
1e 2n -1 =
e 2
2n -1
(1- z)/2
Π
n =1 

e
(1,1-)
=
-
the following expressions hold.
e - z
=
(1+ z)
+log 2 z

z


2n -1
2
=
1+
e
e
2n -1
(1+ z)/2
Π
n =1 

z
n
-
Π
n =1 

 z is the Gamma function,
is Euler-Mascheroni constant and
(1,4+)
From this, (1,1-) is obtained immediately. Reversing the sign of
Replacing
z
with
z /2
z
in (1,1-) , we obtain (1,1+) .
in (1,1-) and (1,1+) , we obtain (1,2-) and (1,2+) .
Next, dividing (1,1-) by (1,2-) ,
z
(1+ z/2) e - z
(z/2) z
2n -1
=
=
1+
e
e
2n -1
(1+ z) e - z/2
2(z)
Π
n =1 


z
2
Here, using the Legendre duplication formula,
1
(z/2)

 e -zlog2
=
=
2(z)
(1+ z)/2
2z (1+ z)/2
Substituting this for the above,


z
 +log 2z
1+
e 2n -1 =
e 2
2n -1
(1+ z)/2
Π
n =1 

z

And, reversing the sign of
z
(1,3-)
, we obtain (1,3+) .
Next,
z
2n -1

e
Π
n =1
e
-

=e
z
2n
z 1-
 = e z log2
1 1 1
+ - +-
2 3 4
Multiplying each side of (1,3-) by this,

And, reversing the sign of
Giving
z

z
1+
e 2n =
e
2n -1
(1+ z)/2
Π
n =1 

z
z
2
(1.4-)
, we obtain (1,4+) .
z = 1 to the formulas, we obtain the following special values.
Special Values
Π
n =1 


Π
n =1 

1
1+
2n
Π
n =1 

1
12n
Π
n =1 

e
e
-
= e -
1
2n
=
1
2n
=
1

1
1+
e 2n -1 =
2n -1
Π
n =1 

1
1 -n
1+
e
n

2

1


2
1
1
1+
e 2n
2n -1
=
e
= 0.56145948
(1.11-)
= 0.84550128
(1.21-)
= 0.75294950
(1.21+)
= 0.66405515
(1.31-)
= 1.32811030
(1.41-)

-
2
e

e2
-
e
-

2

2
From Formula 11.1.1 , the following corollary is derived
-2-
Corollary 11.1.1
1
sin  z
=
(1+ z)(1- z)
z
1
z
2
z
z
=
=
1+
1sin
2n
2n
z
2
z
z
 1+
 12
2
z
z
z

=
= cos
1+
12n -1
2n -1
1+ z
1-z
2


2
2
1+
1Π
n  n 
n =1 

z
z
Π
n =1 


Π
n =1 

=




1+
1Π
2n -1   2n 
n =1 


z
1Π
n =1 
2
z
n -1
z

1+
z
2n

=
=


(1.1)
(1.2)


(1.3)




1+ z
z

 12
2
(1-2+)




1- z
z

 1+
2
2
(1+2-)

(1.1) is a function with all Integers as roots. (1.2) is a function with all even numbers as roots, and (1.3) is a
function with all odd numbers as roots. These are all expressed with an elementary function.
(1-2+) is a function with negative odd numbers and positive even numbers as froots. (1+2-) is a function with
positive odd numbers and negative even numbers as froots. These cannot be expressed with an elementary
function.
If both sides of (1+2-) are illustrated, it is as follows. The product is calculated 500 terms. Since both sides are
overlapped exactly, the left side (blue) is not visible.
Compensation Term
We discuss (1,1-) as an example.

z
e - z
= Π 1+
(1+ z) n =1
n

e
-
z
n
(1,1-)
-3-
Although the compensation term is
e -z/n , the compensation term is not necessarily limited to this. First of all,

the complete compensation term for the right side is
Π
n =1 
1+
z
n

-1
. However, if such a compensation term
is adopted, the function of the left side becomes a constant function and the utility value disappears. This is the
reason why the compensation term like
e -z/n
is adopted.
In fact, if the body and the compensation term are illustrated, it is as follows. The products are calculated
10000 terms respectively. Both are very close but they do not overlap.
Even from this figure, it is hard to consider compensation term other than an exponential function. On the other
hand, it seems that it may be other exponential functions. So let us consider the following exponential function.

f(z) = Π
n =1

z
1+
n
c
The necessity for the condition
if fz is illustrated for
-
z
n
c>1
c>1
is obvious. Anything is okay as long as this is satisfied. For example,
c = 2.4 , c = 2.71828 , c = 3.2 , it is as follows.
We can see that the maximum value of fz is minimum at
c=e
. Actually, If
the maximum value of fz becomes large surprising. Considering this,
compensation term.
-4-
e -z/n
c
deviates from
e
greatly,
seems to be optimal as a
11.2 Maclaurin Expansion
Since the reciprocal of the gamma function is holomorphic over the whole complex plane, the Maclaurin
expansion should be possible. In this section, treating the divergent product and divergent series as numbers
or functions, we aim at the Maclaurin expansion of the reciprocal of gamma function.
Formula 11.1.2
When  is Euler-Mascheroni constant ,  z is the Gamma function,
a harmonic series , the following expressions hold.
1
=1 +  z +
1+ z
+

3

+
H2
 2!
2
1
3
2
3
-
3
2
z
3
 3!
3

 2!
H2
+
2
3


H -Hn+1

H -Hn+1
n +1
2
 Hn
H
-Σ
2 n =1 n
+
3
-
n +1

2H
+ΣHn
3
n =1
H 2  Hn
+
2!
2 Σ
n =1 n
3

2H
+
+ΣHn
3!
2
3
n =1
z

2
H2
-
 -H
n
z
3
+
(2.1-)
z
3
+-
(2.1+)
z
3
+
(2.2-)
z
3
+-
(2.2+)
2

2H
+
+ΣHn
3!
2
3
n =1
2
H is
2
3
H2
1
1

=1 - z + 2
1- z/2
2
2
1
z
2
 Hn
H
-Σ
2 n =1 n
+
1
1

=1 + z + 2
1+ z/2
2
2
+
2
 Hn
H
-Σ
2 n =1 n

2H
+
+ΣHn
3!
2
3
n =1
1
=1 -  z +
1- z
-
 2!
2
is a harmonic number and
Hn
-
n +1

n
2
H -Hn+1
z
 -H
-
 -H
n
2
H -Hn+1
n +1
-
 -H
n
Proof
From the Weierstrass expression on the gamma function,


1
z
= e  zΠ 1+
(1+ z)
n
n =1
When
Hn
is a harmonic number and

-
e
Π
n =1
z
n

=e
-zΣ
n =1
1
n
e
-
z
n
(2.1-)
H is a harmonic series, the compensation term is
= e -Hz
Then
e
z

e
Π
n =1
-
z
n
= e -Hz = 1 +
 -H
1!
z1 +
 -H
2!
2
z2 +
 -H
Next, assume that the body is expanded as follows.
1+
Π
n
n =1 

z

= 1+
z
1

1+
z
2

1+
z
3

= 1 + a1z 1 + a2z 2 + a3z 3 +
-5-
1+

z

4
3!
3
z 3+
(2.1e)
Then, from Vieta's formulas,
a1 =
1 1 1
+ + + = H
1 2 3
a2 =
1
1

 
1 1 1
1
+ + + +
2 3 4
2
H -H1
=
1
1 1
1 2
1 1
+
2 3
1 1
+
3 4
a3 =
+
H -H2
2
+
H -H3
3
 
1 1 1
1
+ + + +
3 4 5
3

H-Hn
n =1
n
+ = Σ
 +
1 1 1
+ + +
4 5 6

 
 

1 1 1
1 1 1 1 1
1 1 1 1 1
+ + + +
+ + + +
 4 5 6  2 4  5 6 7  2 5  6 + 7 + 8 + +
1 1 1
1 1 1 1 1
1 1 1 1 1
+ + + +
+ + + +
 5 6 7  3 5  6 7 8  3 6  7 + 8 + 9 + +
1 1 1
1 1
+ + + +
3 4 5
1 3
1 1 1
1 1
+ + + +
4 5 6
1 4
1 1 1
+ + + +
5 6 7
+

1 1
1 1
1 1
=
H -H2+
H -H3+


H -H4+
1 2
1 3
1 4
+
1 1
1 1
1 1
H -H3+
H -H4+


H -H5+
2 3
2 4
2 5
+
1 1
1 1
1 1
H -H4+
H -H5+


H -H6+
3 4
3 5
3 6
+

Rearranging this along the diagonal line,
a3 =
 
= H1
1
1
1
H -H2+
2

1 1
+
1 2

1
H -H3+
3

1 1 1
+ +
1 2 3

1
H -H4+
4
 HnH -Hn+1
1
1
1
H -H2+H2 H -H3+H3 H -H4+ = Σ
n +1
2
3
4
n =1
Thus,

1+
Π
n
n =1 
z
= 1 + zH + z

Σ
n =1
2
H-Hn
n
+z

Σ
n =1
3
HnH -Hn+1
n +1
+
(2.1p)
The right side of (2.1-) is a product of (2.1e) and (2.1p) . That is
e

z


e
Π
n =1
-
z
n
1+
Π
n
n =1 
z
=1+
 -H
1!
z1 +
= 1 + zH + z

Σ
n =1
2
 -H
2!
H-Hn
n
2
z2 +
+z
Σ
n =1
3
Taking the Cauchy product,
-6-

 -H
3!
3
z 3+
HnH -Hn+1
+
n +1
(2.1e)
(2.1p)
The coefficient of
H+
 -H
Σ
n =1
is
z 2 is
H -Hn
n
1
=
1!
The coefficient of

z
2
 Hn
H -H  -H2
 2-2 H +H 2
H H
2
+
+
+
+
= H -Σ
1!
2!
1! 1!
2!
n =1 n
2
=
The coefficient of

Σ
n =1
z
3
2!
H 2  Hn
-Σ
2 n =1 n
+
is
HnH -Hn+1  -H
+
n +1
1!

=Σ

Σ
n =1
HnH -Hn+1
n +1
n =1
H2
3
H -Hn
n
+
 -H
1!
H -H2  -H3
+
2!
3!
+

Σ
n =1
3

2H
+
+ΣHn
=
2
3
3!
n =1

2
3
H   -HHn H -H
 -H
-Σ
+
+
n n =1
n
2!
3!
H -Hn+1
n +1
 -H
-
n

Then
1
=1 +  z +
1+ z
+


2
2
 Hn
H
+
-Σ
2! 2 n =1 n
3
H2
z
2
3

2H
+
+ΣHn
3!
2
3
n =1

H -Hn+1
n +1
-
 -H
n
z
3
+
(2.1-)
(2.1+), (2.1-) and (2.2+) are easily derived from this.
Verification
The ordinary Maclaurin expansion of the left side of (2.1-) is as follows. In addition,
nz
is a polygamma
function.
1
=1 +  z +
1+ z
 2!
2
-
12 
2
2
z +
 3!
3
-
2
12
-
2(1)
6
z
2
3
= 1 + 0.5772157 z - 0.6558781 z - 0.04200264 z
+
3
Therefore, the following equations must hold.
2
H 2  Hn
2  2
=
+
= -0.655878
2!
2 Σ
2! 12
n =1 n
3
H2
3

2H
+
+ΣHn
3!
2
3
n =1

H-Hn+1
n +1
-
 -H
n

=
3
3!
-
2
12
-
2(1)
6
= -0.0420026
Since the harmonic series is infinite, we use the following limit value for the verification.
lim
m 

2
Hm
2
m
Hn
n =1
n
-Σ

-7-
+
Calculating these with the formula manipulation software Mathematica, we obtained the following result.
The coefficients of each term are almost consistent with the above values, so, the correctness of this formula
was numerically verified.
Also, the following special values were obtained.
Special Values

H 2 - 2!Σ
Hn
n =1
=-
n


3
4H - 3!ΣHn
n =1
2
= -(2)
6
H-Hn+1
n +1
+
H
n
(2.3)
: Riemann Zeta
 =  (1) = -2.404113
(2.4)
2
If easier (2.3) is written down, it is as follows.

1+

1 1
+ +
2 3
2
- 2!
     
1
1
1
+
1
2
1+
1
2
+
1
3
1+
1 1
+
2 3
 + = -(2)
Note
A double series of
of
z5 .
H and Hn
appears in the coefficient of
z4 ,
and a triple series appears in the coefficient
These are far more complicated than notation by polygamma function. So, these are not calculated in
this chapter..
-8-
11.3 Factorization around 1
Replacing
z -1 in
ed to power series of z -1 .
z
with
Formula 11.1.1 , we obtain the following formula. These left sides can be expandSo, we call the formulas factorization around 1.
Formula 11.3.1
When

 z is the Gamma function,
is Euler-Mascheroni constant and
1
(z)
=e
1
(2- z)
=e
1
=e
(1+ z)/2
Π
n =1 
2
1
(1-z/2)
1
(1-z/2)

z -1
1e
2n
n =1
e

 z-1
2
e

1
-
e

(3.2-)
z-1
2n
(3.2+)




z -1
1+
e
2n -1
Π
n =1 

z-1

z -1
1e
2n -1
(3.3-)
z-1
z -1
1e 2n -1
2n -1
Π
n =1 

 z-1
2
z-1
2n

n =1
1
(3.1+)
z -1
1+
e 2n -1
2n -1

 2 +log 2z-1Π
-

=
Π
n =1 


1
=

z -1 1+
e
2n

 +log 2z-1Π
e 2

=
1
(z/2)
2

Π
n =1 
(3.1-)
z-1
n
z -1
1e
n

 z-1
1
=

Π
n =1 
 z-1
z-1
n
z -1 1+
e
n

- z-1
1
=e
(3- z)/2
1
(z/2)

 z-1
the following expressions hold.
z-1
2n
(3.2+)
(3.4-)
z-1
2n
(3.4+)
From this formula, the following formula is further derived.
Formula 11.3.2
When

is Euler-Mascheroni constant and
1
(1+ z)
1
(1- z)
 z is the Gamma function,
Π

n =1
= -(z -1)e
1
=
(1+z/2)
2
1
=
(1+z/2)
2



Π
n =1 
- z-1


z -1
1e
n

+log 2 -1 z-1 

 Π
2
e
n =1
 z-1
e
z-1
z -1
1+
e n +1
n +1
 -1z-1 
= e
2
the following expressions hold.
z-1
n

Π

n =1

(3.5-)

z-1
z -1
1+
e 2n +1
2n +1

z -1
1+
e
2n +1
-9-
(3.5+)
z-1
2n
(3.6-)
(3.7-)
Proof
Dividing both sides of (3.1-) by
,
z


1
1
z -1 = e  z-1
1+
e
1
z(z)
z
z-1

1
Π
n =1 
z-1

z -1
1+
e n +1
n +1
i.e.

z-1

1

z -1
= e -1z-1Π 1+
e n +1
(1+ z)
n +1
n =1
Multiplying both sides of (3.1+) by
1- z
(3.5-)
,



1- z
z -1
= -(z -1)e - z-1Π 1e
(2- z)
n
n =1
z-1
n
i.e.



1
z -1
= -(z -1)e - z-1Π 1e
(1- z)
n
n =1
Dividing both sides of (3.3-) by
1
=
(z/2)(z/2)
z-1
n
(3.5+)
z /2 ,

 +log2(z -1) 1
e 2
2

z
1+ 21-1  e
z -1
z -1

21-1
Π
n =1
-

z -1
z -1
e 2n+1
1+
2n +1
i.e.
1
=
(1+z/2)

2

e
2
Dividing both sides of (3.4-) by
1
=
(z/2)(z/2)

Π
n =1 
+log 2 -1 z-1 
z-1

z -1
1+
e 2n +1
2n +1
(3.6-)
z /2 ,
 z-1
2
2
e


1
z -1
1+
z
21-1
 Π

n =1

z -1
1+
e
2n +1
z-1
2n
i.e.
1
=
(1+ z/2)
Giving
z =2
2

 z-1
2
e
Π

n =1

z -1
1+
e
2n +1

z-1
2n
(3.7-)
to (3.5-) , (3.6-) , (3.7-) , we obtain the following special values.
Special Values
Π
n =1 

Π
n =1 

e 1-
=
2
1


1
1
1+
e 2n
2n +1
= 0.763102
(3.52-)
= 0.902544
(3.62-)
= 0.664055
(3.72-)

1
 - 2 -1
1+
e 2n +1 =
e
2n +1
4
Π
n =1 


1
1
1+
e n +1
n +1
=

2
-
e

2
- 10 -
11.4 Taylor Expansion (Part 1)
Formula 11.4.1
When  is Euler-Mascheroni constant ,  z is the Gamma function,
a harmonic series , the following expressions hold.
1
= 1 + (z -1) +
z
+
 3!
3
+
 2!
2
H2
2
1
= 1 - (z -1) +
2- z
-

3
+
H 2  Hn
-Σ
2 n =1 n
 (z -1)

H -Hn+1
3
-

2H
+ΣHn
3
n =1
 2!
2
H2
+
3
1
1

= 1 + z -1 + 2
1+ z/2
2
2
1
+
3
2

3
H2
-
3
2

3
 3!
3
+
H2
2

 2!
2

+
3
-
 -H
n

2H
+ΣHn
3
n =1

  (z -1) +
(4.1-)
  (z -1) +-
(4.1+)
3
2
H -Hn+1
n +1
-
 -H
n
H -Hn+1
n +1
H -Hn+1
n +1
3
 z -1
 -H
z -1
n 


n +1
3
 +
(4.2-)
 z -1
 -H
z -1
n 


2

3
 +- (4.2+)
 (z -1)
 -H
(z -1) -+
n 
H 2  Hn
2 Σ
n =1 n
H -Hn+1
2

H 2  Hn
+
2!
2 Σ
n =1 n
3
H is
 (z -1)
2

2H
+
+ΣHn
3!
2
3
n =1
1
2
= -z -1 + (z -1) 1- z
+

-
H 2  Hn
+
2!
2 Σ
n =1 n
3
H2
n +1
2

2H
+
+ΣHn
3!
2
3
n =1
1
1

= 1 - z -1 + 2
3- z/2
2
2
1


is a harmonic number and
2
H 2  Hn
2 Σ
n =1 n

2H
+
+ΣHn
3!
2
3
n =1
Hn
3
4
(4.5+)
Proof
Replacing
of (4.1+) by
z
with
1- z
z -1 in
Formula 11.2.1 (2.1-) ~ (2.2+) , we obtain (4.1-) ~ (4.2+) . Multiplying both sides
, we obtain (4.5+) .
Formula 11.4.2
When  is Euler-Mascheroni constant ,  z is the Gamma function,
a harmonic series , the following expression holds.
1

= 1 - 1z -1 +
1!
1+ z

-

1-

1!
+
2
2!
-

3
3!
+


Hn
2
is a harmonic number and
 H
H2
n
1+
+
-Σ
1! 2! 2! n =1 n
H 21- 2H 3 
+
- Hn
2
3 Σ
n =1

H -Hn+1
n +1
+

z -1
1- +H
n
H is
2
  z -1


3
+
(4.5-)
- 11 -
Proof
Formula 11.3.2 (3.5-) was as follows.

z-1

1

z -1
= e -1z-1Π 1+
e n +1
(1+ z)
n +1
n =1
(3.5-)
At first, the compensation term is

e
Π
n =1
-
z-1
n +1

=e
-z-1Σ
n =1

1
n +1
z-1-z-1Σ
=e
n =1
1
n
= e -H-1z-1
Then,
 -1z-1 
e
Π
n =1
e
-
z-1
n +1
-
z-1
n +1
= e -1z-1e -H-1z-1 = e -Hz-1
Expanding this,
 -1z-1 
e
Π
n =1
e
= 1+
 -H
1!
1
(z -1) +
 -H
2!
2
2
(z -1) +
 -H
3
3!
(z -1)3 +
(4.1e)
Let the body be
1+
Π
n +1 
n =1 

z -1

= 1+
z -1
2

1+
z -1
3

1+
z -1
4

1+

z -1

5
= 1 + a1(z -1)1 + a2(z -1)2 + a3(z -1)3 +
Then, from Vieta's formulas,
1 1 1
+ + + = H -1
2 3 4
1 1 1 1
1 1 1 1
1 1 1 1
+ + + +
+ + + +
+ + +
a2 =
2 3 4 5
3 4 5 6
4 5 6 7
 H-H
H -H1 H -H2 H -H3
H -1
H -1
n
=Σ
=
+
+
+ 1
n
2
3
1
1
n =1
a1 =
1
2
1
+
3
1
+
4
a3 =

 
 
1
1
H-H3+
2
3
1
1
H-H4+
3
4
1
1
H-H5+
4
5
1
1
H -H4+
2
4
1
1
H -H5+
3
5
1
1
H -H6+
4
6
1
H -H5+
5
1
H -H6+
6
1
H -H7+
7
1
1
H-H2+
1
2
1
1
H-H3+
2
3
1
1
H-H4+
3
4
1
1
H -H3+
1
3
1
1
H -H4+
2
4
1
1
H -H5+
3
5
1
H -H4+
4
1
H -H5+
5
1
H -H6+
6
+

1
1
1
+
2
1
+
3
=
- 12 -
 +
+

-
1 1
1 1
1 1
H-H2H -H3

H -H4-
1 2
1 3
1 4
Rearranging this along the diagonal line,
 
1
1
a3 =
-
1
H -H2+
2

1 1
+
1 2

1
H -H3+
3

1 1 1
+ +
1 2 3

1
H -H4+
4
1 1
1 1
1 1
H-H2H-H3

H -H4-
1 2
1 3
1 4

=Σ
n =1
 H -1H -H
HnH -Hn+1
 H -Hn+1
n
n+1
=Σ
-Σ
n +1
n +1
n +1
n =1
n =1
Thus,

1+
Π
n +1 
n =1 
z -1
= 1 + z -1H -1 + z -12
Σ
n =1

n =1
The right side of (3.5-) is a product of (4.1e) and (4.1p) . That is
z-1
2
 -H
 -H
1
n +1
 -1z-1 
e
Π
n =1


= 1+
1+
Π
n +1 
n =1 
z -1
1!
(z -1) +
2!
2
= 1 + (z -1)H -1 + (z -1)
n
-
H-1
1

Hn-1H -Hn+1
+
n +1

+ z -13Σ
e
H-Hn
2
(z -1) +
Σ
n =1


+ z -13Σ
n =1
H-Hn
n
-
 -H
3!
H -1
1
(4.1p)
3
(z -1)3 +

Hn-1H -Hn+1
+
n +1
Taking the Cauchy product,
The coefficient of z -1
 -H
1!
Σ
n =1

n

n =1
= 1-
1!

is
H -1  H -1 HH -1  -H2
+H +
+
n
1
1!
1!
2!
Hn

1!
+
2
2
2!
The coefficient of z -1
Σ
n =1
2

H -1  -HH-1  -H2
+
+
1
1!
2!
H -Hn
= -Σ

is
+ H-1 = - 1-
The coefficient of z -1

1
3
+
H 2  Hn
-Σ
2 n =1 n
is as follows.
Hn-1H-Hn+1  -H
+
n +1
1!
Σ

H-Hn
n =1
n
Here,
- 13 -
2
H -1
 -H H -1  -H
+
+
3!
1
2!

3

H -Hn
Σ
n =1
n

Hn
n =1
n
= H -Σ
2
 HnH -Hn+1

 Hn+1
Hn-1H-Hn+1
H
-Σ
+Σ
=Σ
n +1
n +1
n =1
n =1 n +1
n =1 n +1

Σ
n =1

=Σ
HnH -Hn+1
n +1
n =1

= ΣHn
n =1

H -Hn+1
n +1
-1 +Σ
Σ
 n =1
n =1 n

-H
1
n
+

1
Hn
n
H1
-
1
 - H + H -1
2
Then,
Hn-1H-Hn+1  -H
+
n +1
1!

Σ
n =1

= ΣHn
n =1

H -Hn+1
n +1
1
n
+

Σ

H-Hn
n =1
n
2
- H + H-1+
3
2
H -1
 -H H -1  -H
+
+
3!
1
2!

 -H

1!

Hn
H 2-Σ
-
n
n =1
H -1
1
 -H H -1

2
+

= ΣHn
n =1

H -Hn+1
n +1
1-r +H
n
+

 -H
2
- H + H-1+
1!
+
2!

H -1
1
H 2-

2
 -H H -1
3
+
2!

H 21- 2H 3 
+
= - 1+ΣHn
2
3
1! 2! 3!
n =1


3
3!

2
+
 -H
H -Hn+1
n +1
 -H
3
3!
1- +H
n

  z -1
3
+
Therefore,
1

= 1 - 1z -1 +
1!
1+ z

-

1-

1!
+

2
2!
-
3
3!
+

1-

1!
2
+
2!
+
H 21- 2H 3 
+
- Hn
2
3 Σ
n =1
 H
H2
n
-Σ
2! n =1 n

H -Hn+1
n +1
 z -1
+


1- +H
n
2


+
Verification
The ordinary Taylor expansion of the left side of (4.5-) is as follows. Where,  nz is a polygamma function.
1

= 1 - 1z -1 +
1+ z
1!

-


1-

1!
+
2
2!
-
3
3!

1-
-

1!
+
2!
1-
12
 z -1
 (1)
z -1
+
6 
2
2
-
2
12
2



2
3
 +-
= 1 - 0.4227843z -1 - 0.2330937z -12 + 0.1910911z -13 +
Therefore, the following equations must hold.
1-

1!
+
2
2!
+
H 2  Hn

2
2
-Σ
+
= 1= - 0.233093
2 n =1 n
1! 2!
12
- 14 -

- 1-

1!
+
2
2!




= - 1+
 1! 2! 3!  +
-
3
3!
H 1- 2H
3
2
2
2
3
3


+ΣHn
n =1
1-
12
H -Hn+1
n +1
2
-
+
2(1)
6
1- +H
n

= 0.191091
Calculating these with the formula manipulation software Mathematica, we obtained the following result.
The coefficients of each term are almost consistent with the above values.
- 15 -
11.5 Taylor Expansion (Part 2)
Odd Harmonic Number & Odd Harmonic Series (definition)
We define odd harmonic number
and odd harmonic series
hn
as follows.
h
Hn-1
1 1
1
= H2n-1 + ++
3 5
2n -1
2
H
+ log 2
h = h =
2
hn = 1+
(5.hn)
(5.h)
Formula 11.5.1
When
h

is Euler-Mascheroni constant ,  z is the Gamma function,
hn
is an odd harmonic number and
is an odd harmonic series , the following expressions hold.

= 1 + c1(z -1)1 + c2(z -1)2 + c3(z -1)3 + 
(5.3-)
(z/2)

= 1 - c1(z -1)1 + c2(z -1)2 - c3(z -1)3 +- 
(1-z/2)
(5.3+)
Where,

c1 = 2 +log 2

1
c2 =
2!
1
2




+log 2
c3 = 3!
2
h2
2
+log 2
+
2
h2
2
3
+

hn
n =1
2n -1
-Σ

2

+log 2 -
 2n +1 - 2n -1  2 +log 2-h 
2h 3 
+ hn
3 Σ
n =1
h -hn+1
1


Proof
From Formula 11.3.1 (3.3-)


(z/2)
+log 2 z-1 
 Π
= e 2
n =1

z-1

z -1
1+
e 2n -1
2n -1
(5.3-)
At first, the compensation term is

e
Π
n =1
Then,
-
z-1
2n -1

=e
-z-1Σ

n =1
1
2n -1
z-1
+log 2 z-1  
 Πe 2n -1
2
e
= e -hz-1


+log 2 z-1
 e -hz-1 = e 2 +log 2 -hz-1
= e 2
n =1
Expanding this,

z -1
 +log2(z -1)  e - 2n-1
e 2
Π
n =1
= 1+
1


1! 2

+log 2 -h z -11 +
+
- 16 -
2
1 
+log 2 -h z -12
2! 2

3
1 
+log 2 -h z -13+
3! 2



(5.1e)
Let the body be
1+
Π
2n -1 
n =1 


z -1
z -1
1
= 1+

1+

z -1
3
1+
z -1
5


z -1

7
1+
= 1 + a1(z -1)1 + a 2(z -1)2 + a3(z -1)3 +
Then, from Vieta's formulas,
a1 =
1 1 1
+ + + = h
1 3 5
a2 =
1
1
=


1 1 1
+ + +
3 5 7
h -h 1
1
+
h -h2
3
+
+
h -h3
5
1
3

 
1 1 1
1
+ + + +
5 7 9
5

h -hn
n =1
2n -1
+ = Σ
a3 =
1 1
1 1
1 1
h -h2+
h -h3+


h -h4+
1 3
1 5
1 7
+
1 1
1 1
1 1
h -h3+
h -h4+


h -h5+
3 5
3 7
3 9
+
1 1
1 1
1 1
h -h4+
h -h5+


h -h6+
5 7
5 9
5 11
 +
1 1 1
+ +
+
7 9 11
+

Rearranging this along the diagonal line,
a3 =
 
1
1
1
h -h2+
3

h nh -hn+1
n =1
2n +1
=Σ

1 1
+
1 3

1
h -h3+
5

1 1 1
+ +
1 3 5
 7 h -h +
1
4
Thus,

1+
Π
2n -1 
n =1 
z -1

h -h n
n =1
2n -1
= 1 + z -1h + z -12Σ

hnh -hn+1
n =1
2n +1
+ z -13Σ
+
(5.1p)
The right side of (5.3-) is a product of (5.1e) and (5.1p) . That is
z -1

e
 2 +log2(z -1)  e - 2n-1
Π
n =1
= 1+
1


1! 2

+log 2 -h z -11 +
+


1+
Π
2n -1 
n =1 
z -1

h -h n
n =1
2n -1
= 1 + z -1h + z -12Σ
Taking the Cauchy product,
The coefficient of z -1
1


1! 2
1

is
+log 2 -h + h =

2
+ log 2
- 17 -
2
1 
+log 2 -h z -12
2! 2


3
1 
+log 2 -h z -13+
3! 2



hnh -hn+1
n =1
2n +1
+ z -13Σ
+
The coefficient of z -1

h -h n
2
is


1
+  +log 2 -h  +  +log 2 -h 
Σ
1! 2
2! 2
n =1 2n -1
h

= -Σ
n =1
1
=
2!
hn
2
2n -1

2


2

+log 2
3

2
+
1
2

+log 2 h -h 2+
h2
2
+log 2
The coefficient of z -1
1
3!
+h +
2

hn
n =1
2n -1
-Σ

2
2

2
 2
+log 2
-

+log 2 h +
h2
2
is as follows in a similar way.

h2
2
3
+
2

2h 3
+Σhn
3
n =1

+log 2 -
 2n +1
h -hn+1
1
2n -1
-
 2 +log 2-h  

Then,

z/2
= 1+
+

2

+log 2 z -1 +
 2

1
3!

3
+log 2

+
+Σhn
n =1
 2

1
2!
h2
2

2
 2n +1

+log 2

+log 2 -
h -hn+1
-
1
2n -1
h2
2
+
2

-Σ
n =1
z -1
2n -1 
hn


2
2h 3
3
 2 +log 2-h    z -1 +

2
And, changing this to alternating series, we obtain (5.3+) .
Verification
The ordinary Taylor expansion of the left side of (5.3-) is as follows. Where, nz is a polygamma function.

z/2
=1 -
   
 
3
1
1
1
1
1
1
+
 
- 
- 
 2  8  2  8  2   z -1
3!  16
1
1
z -1 +
2
2!
1
0
2
1 2
0
4
1
2
2
z -1
2
8
3
0
2
0
3
2
+
= 1 + 0.981755z -1 - 0.134928z -12 - 0.097286z -13 +
Therefore, the following equations must hold.

2
+ log 2 = 
1
2!
2
1
3!

2
1
0
2

2
+ log 2
+
  = 0.981755
1
2

hn
1
h2
-Σ
=
2 n =1 2n -1
2!

1 2

4 0
 
1
2
-
2
8
 = - 0.134928
 2n +1
 2 + log 2-h  
3
1
1
1
1
1
1
=
 
- 
- 
 2  8  2  8  2   = -0.0972863
3!  16

+ log 2
3
+
h2
2

2

+ log 2 -
2
0

2h 3
+Σhn
3
n =1
h -hn+1
3
0
-
1
2n -1

2
Calculating these with the formula manipulation software Mathematica, we obtained the following result.
The coefficients of each term are almost consistent with the above values.
- 18 -
Also, the following special values were obtained.
Special Values

hn
n =1
2n -1
h - 2!Σ
2
8
 2n +1

h -hn+1
4h - 3!Σhn
3
=-
2
n =1
+
= -(2) : Dirichlet Lambda
h
2n -1

=
1

3 2
2
(5.4)
  = -2.10359958
1
2
(5.5)
If easier (5.4) is written down, it is as follows.

1+

1 1
+ +
3 5
2
- 2!
     
1
1
1
1
3
+
1+
1
3
+
1
5
1 1
+
3 5
1+
 + = -(2)
Formula 11.5.2
When
h

is Euler-Mascheroni constant ,  z is the Gamma function,
is an odd harmonic number and
hn
is an odd harmonic series , the following expression holds.

2(1+z/2)
= 1 - c1(z -1)1 + c2(z -1)2 - c3(z -1)3 + 
(5.6-)
Where,
c1 = 1-

2
-log 2
1
c2 = 11!
c3 = 1-
1
1!


1
+log 2 +
2
2!

2


+log 2 +
1
2!

2

2

2

2
+log 2
+
+log 2
+
-
h2
2
1
3!
2h 3 
- hn
3 Σ
n =1

hn
n =1
2n -1
-Σ

2

+log 2
 2n +1
h -hn+1
+
3
+

1

1- -log 2+h
2
2n -1

Proof
In a similar way as the proof of Formula 11.4.2 , we obtain the desired expression.
- 19 -

h2
1- -log 2
2
2



Verification
If the left side of (5.6-) is expanded to Taylor series with the formula manipulation software Mathematica,
it is as follows.

21+z/2
= 1 - 0.01824z -11 - 0.11668z -12 + 0.01939z -13 + 
On the other hand, calculating
c1 , c2 , c3
according to the formula, we obtained the following results.
Both values are approximately equal.
- 20 -
11.6 Euler-Mascheroni Constant Function
Until the previous section, treating the divergent product and divergent series as numbers or functions,
we performed a series expansion of the reciprocal of the gamma function. Although we succeeded in deriving
the power series from the infinite product, we could not get the general formula.
However, there are not few merits of treating the divergent product and divergent series as numbers or functions
In this section, I will present one example.
Formula 11.6.1
When

is Euler-Mascheroni constant and




 z is the Gamma function,
the following expressions hold.
log (1+ z)
 g(z)
z

 
  = 0

1 1
z
- log 1+
z
n
n
n =1

1
1
 =Σ
- log 1+
n
n
n =1

1 1
z
- log 1+
limΣ
z0 n =1
z
n
n
log (1+ z)
= -
lim
z0
z
 =Σ
-
Re(z) > -1
(6.1)
(6.1')
(6.2)
(6.3)
Proof
Formula 11.1.1 (1.1-) was as follows.
z

z
e - z
n
1+
e
Π
n
n =1 
=
(1.1- )
(1+ z)
The left side can be transformed as follows.
Π
n =1 

e
z
1+
n
-
z
n


log Π 1+
n =1
=e
z
n
 e

-Σ
n =1
z
n


log Π 1+
=e
n =1
z
n


-Σ
n =1
z
n
The right side can also be transformed as follows.
e - z
= e -log (1+ z) e - z = e -log (1+ z) -  z
(1+ z)
Then,


log Π 1+
e
n =1
z
n


-Σ
n =1
z
n
= e -log (1+ z) -  z
Taking the logarithm of both sides and inverting the signs,

 z + log (1+ z) = Σ
n =1


z
z
- logΠ 1+
n
n
n =1

From this,

 =Σ
n =1

Especially, when

 =Σ
n =1

1 1
z
- log 1+
z
n
n

-
log (1+ z)
z
Re(z) > -1
(6.1)
z =1 ,


1
1
- log 1+
n
n

(6.1')
- 21 -
Next, replacing
z
with
z
in the left side of (6.3) ,
log1+  z
d
=
lim
log(z)
 z0
z
dz

z=1
 '(1)
=
=  (1) = -
(1)
Then, (6.3) holds. And (6.2) is derived from this and (6.1).
3D figure of (6.1) is as follows. The left figure is the real part and the right figure is the imaginary part.
The domain is
Rez > -1 . Anywhere on this half plane, the real part is 0.57721566
part is 0 . That is, this function
gz
and the imaginary
is a constant function which gives the Euler-Mascheroni constant
.
In order to obtain  , any value on this half plane may be chosen, and (6.1') is one of them. However, considering the convergence speed, (6.1') is not a good choice. The following is a study of the value of z and the number
of terms necessary to obtain 5 significant digits.
According to this, in order to obtain effective 5 digits,
are sufficient when
35,000 terms are required when z =1 , and 3 terms
z =0.0001 . The reasons are (6.2) and (6.3). When this is illustrated clearly, it's as follows.
- 22 -
The first term of the function
around
function
g =  /2 .
gz
Although
is blue and the second term is yellow green. These are the line symmetry
Therefore, the function value of
gz
is

regardless of the value of
z
. So, we call the
Euler-Mascheroni Constant Function .
gz = 
second term is
gz
for any
z s.t. Rez > -1 ,
especially when
z = 0 , the first term is 0
and the
 . That is, in the vicinity of 0 , the role of the first term is small. That is why convergence of
the first term is fast in the vicinity of
By reference, giving
0.
z = 0.000018
to (6.3) , we obtain
-0.577201
with the same precision as the above.
Note
(6.1') can also be obtained directly from Euler's definition.




1
- log m = lim
m 
n =1 n
m 1
m -1
1
= lim Σ - Σ log 1+
m 
n
n =1 n
n =1
m
 = lim Σ
m 
Σ
n =1 n

m
1

=Σ
n =1


2 3 4
m

1 2 3 m -1
1
1
- log 1+
n
n
- log



2016.12.20
Kano. Kono
Alien's Mathematics
- 23 -