The sum of reduced harmonic series generated
by any number of positive integer factors
POTŮČEK, R.
Department of Mathematics and Physics, University of Defence,
Kounicova 65, 662 10 Brno, CZECH REPUBLIC
ORCID iD: 0000-0003-4385-691X
Abstract: This paper is a free continuation of the author’s previous papers dealing with the sums of the reduced
harmonic series generated of reciprocals of all products generated by all prime divisors of the numbers 2002
and 2022, that were inspired by one task on the sum of a special infinite series on the Berkeley Math Circle. We
determined the sums of these series, i.e. the series of all the unit fractions that have denominators with only factors
consisting of all prime divisors of the numbers 2022 and 2002, analytically and also by calculation in computer
algebra system Maple. In this paper, we generalize our considerations and derive two formulas that obviously
hold not only to series generated by nprime numbers, but also to series generated by npositive integers.
Key-Words: CAS Maple 2022, infinite series, geometric series, harmonic series, reduced harmonic series
Received: August 27, 2021. Revised: April 19, 2022. Accepted: May 20, 2022. Published: July 3, 2022.
1 Introduction
In this paper, we generalize the solution of problems
from the papers [1], [2] and [3], which concerned the
sums of reciprocals of all products generated by prime
factors of the numbers 2002 and 2022 based on the
decomposition of these numbers into prime numbers.
The number 2022 (as well as the year 2022) has three
prime divisors 2,3,337, because
2022 = 2 ·3·337,
and the number 2002 (as well as the year 2002) has
four prime divisors 2,7,11,13, because
2002 = 2 ·7·11 ·13.
These prime numbers generate corresponding re-
duced harmonic series
1
2+1
3+1
337 +1
22+1
32+1
3372+
+1
2·3+1
2·337 +1
3·337 +1
23+1
33+1
3373+· · ·
and
1
2+1
7+1
11 +1
13 +1
22+1
72+1
112+1
132+
+1
2·7+1
2·11 +1
2·13 +1
7·11 +1
7·13 +1
11·13 +
+1
23+1
73+1
113+1
133+1
22·7+1
22·11 +· · ·
The generalization will concern the decomposition
into nprime numbers and the derived formula for the
sum of the series generated by nprime numbers will
be proved by induction. The derived formula will ap-
ply not only to the decomposition of numbers into
prime numbers, but also to the decomposition into
compound numbers.
Because our topic concerns the harmonic series
and the so called reduced harmonic series or modi-
fied harmonic series, let us now recall the necessary
notions from infinite series theory.
2 Basic notions
For any sequence {ak}of numbers the associated in-
finite series or more briefly series is defined as the
sum ∞
k=1
ak=a1+a2+a3+· · · .
The sequence of partial sums {sn}associated to a se-
ries
∞
k=1
akis defined for each nas the sum
sn=
n
k=1
ak=a1+a2+· · · +an.
The series
∞
k=1
akconverges to a limit sif and only
if the sequence of partial sums {sn}converges to s,
i.e. lim
n→∞ sn=s. We say that the series
∞
k=1
akhas a
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
112
Volume 2, 2022
sum sand write
∞
k=1
ak=s.
The geometric series is the sum of an infinite num-
ber of terms that have a constant ratio between succes-
sive terms. Any geometric series can be written as
a+aq +aq2+aq3+· · · ,
where ais the coefficient of each term and qis the
common ratio between adjacent terms. Geometric se-
ries are among the simplest examples of infinite series
and for |q|<1have the sum
s=a
1−q.(1)
More information about geometric series can be
found, for example, on [4].
The sum of the reciprocals of some positive inte-
gers is generally the sum of unit fractions. For exam-
ple the sum of the reciprocals of the cube numbers is
the Apéry’s constant ζ(3) which is given by the for-
mula
∞
k=1
1
k3=1
13+1
23+1
33+1
43+· · · .
= 1.202057.
The harmonic series is the infinite series formed
by summing all positive unit fractions:
∞
n=1
1
n= 1 + 1
2+1
3+1
4+1
5+· · · .
The harmonic series is divergent. Its divergence was
first proven in 1350 by Nicole Oresme. More infor-
mation about harmonic series can be found, for exam-
ple, on [5].
The reduced harmonic series is defined as the sub-
series of the harmonic series, which arises by omit-
ting some its terms. As an example of the reduced
harmonic series we can take the series formed by re-
ciprocals of primes and number one
1 + 1
2+1
3+1
5+1
7+1
11 +1
13 +· · · .
This reduced harmonic series is divergent. The first
proof of its divergence was made by Leonhard Euler
(see e.g. book [6]).
3 Six lemmas
Before giving the general formula for the sum of the
reduced harmonic series generated by any number of
positive integer factors, we present six following use-
ful lemmas.
Lemma 1. Let abe a positive integer. The geometric
series in the form of special reduced harmonic series
Ta=1
a+1
a2+1
a3+1
a4+· · · (2)
has the sum
Sa=1
a−1=a
a−1−1.(3)
Proof. Equation (3) follows from the formula (1) for
the sum of a convergent geometric series. Because the
first term is 1/aand common ratio qis also 1/a > 0,
we have
Sa=1/a
1−1/a=1/a
(a−1)/a=1
a−1=
=a−(a−1)
a−1=a
a−1−1.
Lemma 2. Let a < b be positive integers. The series
in the form of special reduced harmonic series
Ta·b=1
ab +1
a2b+1
b2a+1
a3b+1
b3a+
+1
a2b2+1
a4b+1
b4a+1
a3b2+1
b3a2+· · · =
=1
ab1 + 1
a+1
b+1
a2+1
b2+1
ab +· · ·
(4)
has the sum
Sa·b=1
(a−1)(b−1) .(5)
Proof. It is clear that we have
Sa·b=1
ab1 + Sa+Sb+Sa·b,
whence, according to formula (3), we get
(ab −1)Sa·b= 1 + 1
a−1+1
b−1,
i.e.
(ab −1)Sa·b=(a−1)(b−1) + b−1 + a−1
(a−1)(b−1) ,
i.e.
Sa·b=ab −1
(ab −1)(a−1)(b−1) ,
whence
Sa·b=1
(a−1)(b−1) .
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
113
Volume 2, 2022
Lemma 3. Let a < b be positive integeres. The series
in the form of special reduced harmonic series
T(a, b) = 1
a+1
b+1
a2+1
ab +1
b2+
+1
a3+1
a2b+1
b2a+1
b3+
+1
a4+1
a3b+1
b3a+1
a2b2+1
b4+
+1
a5+1
a4b+1
b4a+1
a3b2+1
b3a2+1
b5+· · ·
(6)
has the sum
S(a, b) = a
a−1·b
b−1−1(7)
and also the sum
S(a, b) = a+b−1
(a−1)(b−1) (8)
and as well the sum
S(a, b) = 1
a·a
a−1+1
b·a
a−1·b
b−1.(9)
Proof. We gradually determine the sum S(a, b)of the
series T(a, b)by rearranging it, appropriately divid-
ing it into suitable subseries and using the well-known
formula (1) for the sum of an infinite geometric series.
Assume that its sum S(a, b)is finite and that the
series (6) converges. Because all its terms are posi-
tive, then the series (6) converges absolutely and so
we can rearrange it. For easier determination the sum
S(a, b)it is necessary the series (6) rearrange and di-
vide it into three subseries: Ta,Tband Ta·b. It is clear
that
S(a, b) = Sa+Sb+Sa·b,
whence, according to formulas (3) and (5), we get
S(a, b) = 1
a−1+1
b−1+1
(a−1)(b−1) =
=b−1 + a−1 + 1
(a−1)(b−1) =a+b−1
(a−1)(b−1) ,
S(a, b) = a+b−1
(a−1)(b−1) =
=ab −ab +a+b−1
(a−1)(b−1) =
=ab −(a−1)(b−1)
(a−1)(b−1) =
=ab
(a−1)(b−1) −1,
S(a, b) = a+b−1
(a−1)(b−1) =ab(b−1) + a2b
ab(a−1)(b−1) =
=ab(b−1)
ab(a−1)(b−1) +a2b
ab(a−1)(b−1) =
=1
a·a
a−1+1
b·a
a−1·b
b−1.
Lemma 4. Let a < b < c be positive integers. The
series in the form of special reduced harmonic series
Ta·b|c=1
ab +1
a2b+1
b2a+1
abc +1
a3b+1
b3a+
+1
a2b2+1
a2bc +1
b2ac +1
c2ab +· · · =
=1
ab1 + 1
a+1
b+1
c+1
a2+1
b2+
+1
ab +1
ac +1
bc +1
c2+· · · .
(10)
has the sum
Sa·b|c=c
(a−1)(b−1)(c−1) .(11)
Proof. Clearly, we have
Sa·b|c=1
ab1 + Sa+Sb+Sc+Sa·b|c+Sa·c+Sb·c,
whence, by formulas (3) and (5), we get
(ab −1)Sa·b|c= 1 + 1
a−1+1
b−1+1
c−1+
+1
(a−1)(c−1) +1
(b−1)(c−1) ,
i.e.
(ab −1)Sa·b|c=1
(a−1)(b−1)(c−1) ·
·(a−1)(b−1)(c−1) + (b−1)(c−1) +
+ (a−1)(c−1) + (a−1)(b−1) +
+ (b−1) + (a−1).
So, we have
Sa·b|c=1
(a−1)(b−1)(c−1)(ab −1) ·
·abc −ab −ac −bc +a+b+c−1 + bc −b−
−c+ 1 + ac −a−c+1+ab −a−b+ 1 +
+b−1 + a−1,
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
114
Volume 2, 2022
i.e.
Sa·b|c=abc −c
(a−1)(b−1)(c−1) ,
whence we get
Sa·b|c=c
(a−1)(b−1)(c−1) .
Lemma 5. Let a < b < c be positive integers. The
series in the form of special reduced harmonic series
T(a, b, c) = 1
a+1
b+1
c+1
a2+1
b2+
+1
c2+1
a·b+1
a·c+1
b·c+1
a3+
+1
b3+1
c3+1
a2·b+1
a2·c+1
b2·a+
+1
b2·c+1
c2·a+1
c2·b+1
a·b·c+
+1
a4+1
b4+1
c4+1
a3·b+1
a3·c+
+1
b3·a+1
b3·c+1
c3·a+1
c3·b+
+1
a2·b·c+1
b2·a·c+1
c2·a·b+
+1
a2·b2+1
a2·c2+1
b2·c2+· · ·
(12)
has the sum
S(a, b, c) = a
a−1·b
b−1·c
c−1−1(13)
and also the sum
S(a, b, c) = (a+b−1)(c−1) + ab
(a−1)(b−1)(c−1) (14)
and as well the sum
S(a, b, c) = 1
a·a
a−1+1
b·a
a−1·b
b−1+
+1
c·a
a−1·b
b−1·c
c−1.
(15)
Proof. We determine the sum S(a, b, c)of the series
T(a, b, c)by rearranging it and using the formula (1).
Assume that its sum S(a, b, c)is finite and that the
series (12) converges. Because all its terms are pos-
itive, then the series (12) converges absolutely and
so we can rearrange it. Now, it is necessary the se-
ries (12) rearrange and divide it into six subseries Ta,
Tb,Tc,Ta·b|c,Ta·cand Tb·c. It is clear that
S(a, b, c) = Sa+Sb+Sc+Sa·b|c+Sa·c+Sb·c,
whence, according to formulas (3), (5) and (11), we
get
S(a, b, c) = 1
a−1+1
b−1+1
c−1+
+c
(a−1)(b−1)(c−1) +
+1
(a−1)(c−1) +1
(b−1)(c−1) =
=1
(a−1)(b−1)(c−1) ·
·(b−1)(c−1) + (a−1)(c−1) +
+ (a−1)(b−1) + c+b−1 + a−1=
=1
(a−1)(b−1)(c−1) ·
·bc −b−c+1+ac −a−c+ 1 +
+ab −a−b+1+c+b−1 + a−1=
=ab +ac +bc −a−b−c+ 1
(a−1)(b−1)(c−1) =
=ac +bc −c−a−b+1+ab
(a−1)(b−1)(c−1) =
=(a+b−1)(c−1) + ab
(a−1)(b−1)(c−1) ,
S(a, b, c) = (a+b−1)(c−1) + ab
(a−1)(b−1)(c−1) =
=1
(a−1)(b−1)(c−1) ·
·abc −abc +ab +ac +bc −
−a−b−c+ 1=
=1
(a−1)(b−1)(c−1) ·
·abc −ab(c−1) + a(c−1) +
+b(c−1) −(c−1)=
=abc −(ab −a−b−1)(c−1)
(a−1)(b−1)(c−1) =
=abc −(a−1)(b−1)(c−1)
(a−1)(b−1)(c−1) =
=abc
(a−1)(b−1)(c−1) −1,
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
115
Volume 2, 2022
S(a, b, c) = (a+b−1)(c−1) + ab
(a−1)(b−1)(c−1) =
=abc(a+b−1)(c−1) + a2b2c
abc(a−1)(b−1)(c−1) =
=abc(b−1)(c−1) + a2bc(c−1) + a2b2c
abc(a−1)(b−1)(c−1) =
=1
a·a
a−1+1
b·a
a−1·b
b−1+
+1
c·a
a−1·b
b−1·c
c−1.
Remark 1. Let us note that a number s(n)of partial
summands making the sum S(a1, a2, . . . , an),n≥3,
is
s(n) = n
1+n
2+· · · +n
n−1= 2n−2,
while s(1) = 1 and s(2) = 3, so
s(3) = 23−2 = 3
1+3
2= 3 + 3 = 6,
according to equations
S(a, b) = Sa+Sb+Sa·b
and
S(a, b, c) = Sa+Sb+Sc+Sa·b|c+Sa·c+Sb·c
given in Lemmas 3 and 5. For example, the sum
S(2,3,5,7) has these
s(4) = 24−2 = 4
1+4
2+4
3=
= 4 + 6 + 4 = 14
partial summands:
S2, S3,S5, S7,
S2·3, S2·5, S2·7,S3·5, S3·7, S5·7,
S2·3·5|7, S2·3·7,S2·5·7, S3·5·7
and for example, the sum S(2,3,5,7,11) has these
s(5) = 25−2 = 5
1+5
2+5
3+5
4=
= 5 + 10 + 10 + 5 = 30
partial summands:
S2, S3,S5, S7, S11 ,
S2·3, S2·5,S2·7, S2·11 , S3·5,
S3·7, S3·11 ,S5·7, S5·11 , S7·11 ,
S2·3·5, S2·3·7,S2·3·11 , S2·5·7, S2·5·11 ,
S2·7·11 , S3·5·7,S3·5·11 , S3·7·11 , S5·7·11 ,
S2·3·5·7|11 , S2·3·5·11 ,S2·3·7·11 , S2·5·7·11 , S3·5·7·11 .
Lemma 6. For all positive integer nand for any pos-
itive integers a1, a2, . . . , anit holds the equality
1 + 1
a1−1+a1
(a1−1)(a2−1) +
+a1a2
(a1−1)(a2−1)(a3−1) +· · ·
· · · +a1a2· · · an−1
(a1−1)(a2−1) · · · (an−1) =
=a1
a1−1·a2
a2−1·a3
a3−1· · · · · an
an−1.
(16)
Proof. Let us denote the sum on the left-hand side of
(16) by Ln, the product on the right-hand side by Rn
and use mathematical induction.
1. Base case: The equality (16) holds for n= 1:
L1= 1 + 1
a1−1=a1−1+1
a1−1=a1
a1−1=R1.
2. Inductive step: Suppose that it holds Ln=Rnfor
n≥1and show that it holds Ln+1 =Rn+1. Since
Ln+1 =
=Ln+a1a2· · · an−1an
(a1−1)(a2−1) · · · (an−1)(an+1 −1) ,
then, by the inductive hypothesis, we get
Ln+1 =
=Rn+a1a2· · · an−1an
(a1−1)(a2−1) · · · (an−1)(an+1 −1) =
=a1
a1−1·a2
a2−1·a3
a3−1· · · · · an
an−1+
+a1a2· · · an−1an
(a1−1)(a2−1) · · · (an−1)(an+1 −1) =
=a1a2· · · an−1an(an+1 −1) + a1a2· · · an−1an
(a1−1)(a2−1) · · · (an−1)(an+1 −1) =
=a1a2· · · anan+1
(a1−1)(a2−1) · · · (an+1 −1) =Rn+1 .
Therefore the equality (16) holds for all positive inte-
ger nand for any positive integers a1, a2, . . . , an.
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
116
Volume 2, 2022
4 Analytical Solution
Now, we determine two general formulas for the sum
S(a1, a2, . . . , an), further denoted briefly by S[n], of
the reduced harmonic series
T(a1, a2, . . . , an) = 1
a1
+1
a2
+· · · +1
an
+
+1
a2
1
+1
a2
2
+· · · +1
a2
n
+1
a1a2
+1
a1a3
+· · ·
· · · +1
a1an
+1
a2a3
+· · · +1
a2an
+· · ·
· · · +1
an−1an+1
a3
1
+1
a3
2
+· · · +1
a3
n
+
+1
a2
1a2
+1
a2
1a3
+· · · +1
a2
1an
+1
a2
2a1
+
+1
a2
2a3
+· · · +1
a2
2an
+· · · +1
a2
na1
+1
a2
na2
+· · ·
· · · +1
a2
nan−1
+1
a1a2a3
+1
a1a2a4
+· · ·
· · · +1
an−2an−1an+1
a4
1
+1
a4
2
+· · · +1
a4
n
+
+1
a3
1a2
+1
a3
1a3
+· · · +1
a3
nan−1
+· · ·
· · · +1
an−3an−2an−1an+· · · ,
generated obviously not only by nprime num-
bers a1, a2, . . . , anbut also by npositive integers
a1, a2, . . . , anand prove it by mathematical induc-
tion.
By (3) for S[1] = S(a1)we have
S[1] = Sa1=1
a1−1,
i.e.
S[1] = 1
a1
·a1
a1−1=
1
i=1 1
ai
i
j=1
aj
aj−1.(17)
By (9) for S[2] = S(a1, a2)we get
S[2] = 1
a1
·a1
a1−1+1
a2
·a1
a1−1·a2
a2−1,
so we have
S[2] =
2
i=1 1
ai
i
j=1
aj
aj−1.(18)
By (15) for S[3] = S(a1, a2, a3)we have
S[3] = 1
a1
·a1
a1−1+1
a2
·a1
a1−1·a2
a2−1+
+1
a3
·a1
a1−1·a2
a2−1·a3
a3−1,
i.e.
S[3] =
3
i=1 1
ai
i
j=1
aj
aj−1.(19)
So, we can assume that it holds the following theorem:
Theorem 1. For all positive integer nthe sum S[n]
of the reduced harmonic series generated by positive
integers a1, a2, . . . , anholds the equality
S[n] =
n
i=1 1
ai
i
j=1
aj
aj−1.(20)
Proof. Let us prove the formula (20) by using math-
ematical induction.
1. Base case: This formula holds for n= 1, as was
stated in (17):
S[1] =
1
i=1 1
ai
i
j=1
aj
aj−1.
2. Inductive step: Suppose that the formula (20) holds
for nand show that it also holds for n+ 1. Since
S[n+ 1] = S[n] + 1
an+1
n+1
j=1
aj
aj−1,
then by (16) and by the inductive hypothesis we can
gradually write
S[n+ 1] =
n
i=1 1
ai
i
j=1
aj
aj−1+
+1
an+1
n+1
j=1
aj
aj−1=
=
n
i=1 1
ai1+
i
j=1
a1a2· · · aj−1
(a1−1)(a2−1) · · · (aj−1)+
+1
an+1 1+
n+1
j=1
a1a2· · · aj
(a1−1)(a2−1) · · · (aj+1 −1)=
=
n+1
i=1 1
ai1+
i
j=1
a1a2· · · aj−1
(a1−1)(a2−1) · · · (aj−1)=
=
n+1
i=1 1
ai
i
j=1
aj
aj−1.
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
117
Volume 2, 2022
Therefore the equality (20) holds for all positive inte-
ger nand for any positive integers a1, a2, . . . , an.
Now, let us prove, also by mathematical induction,
another formula for the sum S[n]:
Theorem 2. For all positive integer nthe sum S[n]
of the reduced harmonic series generated by positive
integers a1, a2, . . . , anholds the equality
S[n] =
n
i=1
ai
ai−1−1.(21)
Proof. 1. Base case: It is easy to see that the formula
(21) holds for n= 1, because by (3) we get
S[1] = a1
a1−1−1 =
1
i=1
ai
ai−1−1
2. Inductive step: Suppose that the formula (21) holds
for nand show that it holds for n+ 1. Since by (20)
S[n+ 1] = S[n] + 1
an+1
n+1
i=1
ai
ai−1,
then by the inductive hypothesis we gradually have
S[n+ 1] =
n
i=1
ai
ai−1−1 + 1
an+1
n+1
i=1
ai
ai−1=
=an+1 −1
an+1
n+1
i=1
ai
ai−1−1 +
+1
an+1
n+1
i=1
ai
ai−1=
=an+1 −1
an+1
+1
an+1 n+1
i=1
ai
ai−1−1 =
=
n+1
i=1
ai
ai−1−1.
Therefore for all positive integer nthe sum S[n]of
the reduced harmonic series generated by positive in-
tegers a1, a2, . . . , anhas also the form (21).
5 Three simple examples
Now, we calculate the sums of three specific reduced
harmonic series generated by two, three and four
smallest prime numbers by means of the results of the
Theorem 1 and 2.
Example 1. Determine the sum S[2] = S(2,3) of the
reduced harmonic series in the form
T(2,3) = 1
2+1
3+1
22+1
32+1
2·3+
+1
23+1
33+1
22·3+1
32·2+
+1
24+1
34+1
23·3+1
33·2+1
22·32+· · · .
formed of all the unit fractions that have denominators
with only prime factors from the set {2,3}.
Solution: For n= 2 and a1= 2,a2= 3 by the
formula (20) from Theorem 1 we have
S[2] = S(2,3) =
2
i=1 1
ai
i
j=1
aj
aj−1=
=1
a1
·a1
a1−1+1
a2
·a1
a1−1·a2
a2−1=
=1
2·2
2−1+1
3·2
2−1·3
3−1=
=1
2·2
1+1
3·2
1·3
2= 1 + 1 = 2.
By the formula (21) from Theorem 2 we get the same
result:
S[2] = S(2,3) =
2
i=1
ai
ai−1−1 =
=a1
a1−1·a2
a2−1−1 =
=2
2−1·3
3−1−1 = 2
1·3
2−1 = 2.
Example 2. Determine the sum S[3] = S(2,3,5) of
the reduced harmonic series in the form
T(2,3,5) = 1
2+1
3+1
5+1
22+1
32+
+1
52+1
2·3+1
2·5+1
3·5+1
23+
+1
33+1
53+1
22·3+1
22·5+1
32·2+
+1
32·5+1
52·2+1
52·3+1
2·3·5+· · ·
formed of all the unit fractions that have denominators
with only prime factors from the set {2,3,5}.
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
118
Volume 2, 2022
Solution: For n= 3 and a1= 2,a2= 3,a3= 5
by the formula (20) from Theorem 1 we have
S[3] = S(2,3,5) =
3
i=1 1
ai
i
j=1
aj
aj−1=
=1
a1
·a1
a1−1+1
a2
·a1
a1−1·a2
a2−1+
+1
a3
·a1
a1−1·a2
a2−1·a3
a3−1=
=1
2·2
2−1+1
3·2
2−1·3
3−1+
+1
5·2
2−1·3
3−1·5
5−1=
=1
2·2
1+1
3·2
1·3
2+1
5·2
1·3
2·5
4=
= 1 + 1 + 3
4=11
4= 2.75.
By the formula (21) from Theorem 2 we get the same
result:
S[3] = S(2,3,5) =
3
i=1
ai
ai−1−1 =
=a1
a1−1·a2
a2−1·a3
a3−1−1 =
=2
2−1·3
3−1·5
5−1−1 =
=2
1·3
2·5
4−1 = 11
4= 2.75.
Example 3. Determine the sum S[4] = S(2,3,5,7)
of the reduced harmonic series
T(2,3,5,7) = 1
2+1
3+1
5+1
7+
+1
22+1
32+1
52+1
72+1
2·3+
+1
2·5+1
2·7+1
3·5+1
3·7+1
5·7+
+1
23+1
33+1
53+1
73+1
22·3+
+1
22·5+1
22·7+1
32·2+1
32·5+
+1
32·7+1
52·2+1
52·3+1
52·7+
+1
72·2+1
72·3+1
72·5+1
2·3·5+
+1
2·3·7+1
2·5·7+1
3·5·7+· · ·
Solution: For n= 4 and a1= 2,a2= 3,a3= 5,
a4= 7 by the formula (20) from Theorem 1 we have
S[4] = S(2,3,5,7) =
4
i=1 1
ai
i
j=1
aj
aj−1=
=1
a1
·a1
a1−1+1
a2
·a1
a1−1·a2
a2−1+
+1
a3
·a1
a1−1·a2
a2−1·a3
a3−1+
+1
a4
·a1
a1−1·a2
a2−1·a3
a3−1·a4
a4−1=
=1
2·2
2−1+1
3·2
2−1·3
3−1+
+1
5·2
2−1·3
3−1·5
5−1+
+1
7·2
2−1·3
3−1·5
5−1·7
7−1=
=1
2·2
1+1
3·2
1·3
2+1
5·2
1·3
2·5
4+
+1
7·2
1·3
2·5
4·7
6=
= 1 + 1 + 3
4+5
8=27
8= 3.375.
By the formula (21) from Theorem 2 we get the same
result:
S[4] = S(2,3,5,7) =
4
i=1
ai
ai−1−1 =
=a1
a1−1·a2
a2−1·a3
a3−1·a4
a4−1−1 =
=2
2−1·3
3−1·5
5−1·7
7−1−1 =
=2
1·3
2·5
4·7
6−1 = 27
8= 3.375.
6 Numerical Solution
We will solve the task to determine the sum Sof recip-
rocals of all products generated by npositive integers,
where n= 2,3,4numerically by using the basic pro-
gramming language in the computer algebra system
Maple 2022. For n≥5we would use the general-
ization of the following simple procedures partab,
partabc and partabcd:
partab:= proc(p,a,b)
local de,i,j,n,s;
s:= 0;
for n from 1 to p do
for i from 0 to n do
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
119
Volume 2, 2022
for j from 0 to n-i do
if i+j > n-1 then
de:=aˆi*bˆj;
s:= s+1/de;
end if;
end do;
end do;
end do;
print("sum for integers",a,b,"
and for",p,"summands is",
evalf[4](s));
end proc:
partabc:= proc(p,a,b,c)
local de,i,j,k,n,s;
.
.
.
for k from 0 to n-i-j do
if i+j+k > n-1 then
de:=aˆi*bˆj*cˆk;
s:= s+1/de;
end if;
end do;
.
.
.
print("sum for integers",a,b,c"
and for",p,"summands is",
evalf[4](s));
end proc:
partabcd:= proc(p,a,b,c,d)
local de,i,j,k,l,n,s;
.
.
.
for k from 0 to n-i-j do
for l from 0 to n-i-j-k do
if i+j+k+l > n-1 then
de:=aˆi*bˆj*cˆk*dˆl;
s:= s+1/de;
end if;
end do;
.
.
.
print("sum for integers",a,b,c,d"
and for",p,"summands is",
evalf[4](s));
end proc:
For example, for parametr p= 10, the first pro-
cedure partab generates and sums first 10 numbers
of 1-combinations, 2-combinations, 3-combinations,
... with repetitions or also multisubsets of size n=
1,2,3, . . . from a set {a, b}of size 2. The number of
multisubsets of size nis then the number of nonneg-
ative integer solutions of the Diophantine equation
i+j=n.
So we consider the first 10 summands that have de-
nominators with only factors aibjfrom the set {a, b}
until i+j≤10, i.e. these 10 summands:
1
a1b0+1
a0b1+1
a2b0+1
a0b2+1
a1b1+
+1
a3b0+1
a0b3+1
a2b1+1
a1b2+1
a4b0
Analogously, for p= 10, the second procedure
partabc generates and sums the following 10 sum-
mands, which correspond to the solution of the Dio-
phantine equation
i+j+k=n,
where n= 1,2,3, . . . :
1
a1b0c0+1
a0b1c0+1
a0b0c1+
+1
a2b0c0+1
a0b2c0+1
a0b0c2+
+1
a1b1c0+1
a1b0c1+1
a0b1c1+1
a3b0c0
and the third procedure partabcd generates and sums
the following 10 summands, which correspond to the
solution of the Diophantine equation
i+j+k+l=n,
where n= 1,2,3, . . . :
1
a1b0c0d0+1
a0b1c0d0+1
a0b0c1d0+
+1
a0b0c0d1+1
a2b0c0d0+1
a0b2c0d0+
+1
a2b0c0d0+1
a0b2c0d0+1
a1b1c0d0+1
a1b0c1d0
The approximate values of 36 sums of the type
S[2] = S(a, b), where a∈ {2,3,4,5,6,7,8,9},
b∈ {3,4,5,6,7,8,9,10},a < b, i.e. of the sums
S(2,3), S(2,4), S(2,5), . . .
. . . , S(8,9), S(8,10), S(9,10),
for parameter p= 100, rounded to 6 decimals and
obtained by two for statements
A:={2,3,4,5,6,7,8,9};
B:={3,4,5,6,7,8,9,10};
for a in A do
for b in B do
if a < b then
partab(100,a,b)
end if;
end do;
end do;
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
120
Volume 2, 2022
Table 1: 36 approximate sums S[2] = S(a, b)for a∈
{2,3, . . . , 9},b∈ {3,4, . . . , 10},a < b.
S(a, b)b= 3 b= 4 b= 5 b= 6
a= 2 2.000000 1.666 667 1.500 000 1.400 000
a= 3 ×1.000000 0.875000 0.800000
a= 4 × × 0.666667 0.600000
a= 5 × × × 0.500000
S(a, b)b= 7 b= 8 b= 9 b= 10
a= 2 1.333333 1.285 714 1.250 000 1.222 222
a= 3 0.750000 0.714 286 0.687 500 0.666 667
a= 4 0.555556 0.523 810 0.500 000 0.481 481
a= 5 0.458333 0.428 571 0.406 250 0.388 889
S(a, b)b= 7 b= 8 b= 9 b= 10
a= 6 0.400000 0.371 429 0.350 000 0.333 333
a= 7 ×0.333333 0.312500 0.296296
a= 8 × × 0.285714 0.269841
a= 9 × × × 0.250000
are written into Table 1.
The approximate values of 36 sums of the type
S[3] = S(2, b, c), where b∈ {3,4,5,6,7,8,9,10}
and c∈ {4,5,6,7,8,9,10,11},b<c, i.e. of the
sums
S(2,3,4), S(2,3,5), S(2,3,6), . . .
. . . , S(2,9,10), S(2,9,11), S(2,10,11),
for parameter p= 100, rounded to 6 decimals and
obtained by two for statements
B:={3,4,5,6,7,8,9,10};
C:={4,5,6,7,8,9,10,11};
for b in B do
for c in C do
if b < c then
partabc(100,2,b,c);
end if;
end do;
end do;
are written into Table 2.
The approximate values of 36 sums of the type
S[4] = S(2,3, c, d), where c∈ {4,5,6,7,8,9,
10,11}and d∈ {5,6,7,8,9,10,11,12},c < d, i.e.
of the sums
S(2,3,4,5), S(2,3,4,6), S(2,3,4,7), . . .
. . . , S(2,3,10,11), S(2,3,10,12), S(2,3,11,12),
for parameter p= 100, rounded to 6 decimals and
obtained by two for statements
Table 2: 36 approximate sums S[3] = S(2, b, c)for
b∈ {3,4, . . . , 10},c∈ {4,5, . . . , 11},b < c.
S(2, b, c)c= 4 c= 5 c= 6 c= 7
b= 3 3.000000 2.750000 2.600000 2.500000
b= 4 ×2.333333 2.200000 2.111111
b= 5 × × 2.000000 1.916667
b= 6 ×××1.800000
S(2, b, c)c= 8 c= 9 c= 10 c= 11
b= 3 2.428571 2.375000 2.333333 2.300000
b= 4 2.047619 2.000000 1.962963 1.933333
b= 5 1.857143 1.812500 1.777778 1.750000
b= 6 1.742857 1.700000 1.666667 1.640000
S(2, b, c)c= 8 c= 9 c= 10 c= 11
b= 7 1.666667 1.625000 1.592593 1.566667
b= 8 ×1.571429 1.539683 1.514286
b= 9 × × 1.500000 1.475000
b= 10 ×××1.444444
C:={4,5,6,7,8,9,10,11};
D1:={5,6,7,8,9,10,11,12};
for c in C do
for d in D1 do
if c < d then
partabcd(100,2,3,c,d);
end if;
end do;
end do;
are written into Table 3.
Table 3: 36 approximate sums S[4] = S(2,3, c, d)for
c∈ {4,5, . . . , 11},d∈ {5,6, . . . , 12},c < d.
S(2,3, c, d)d= 5 d= 6 d= 7 d= 8
c= 4 4.000000 3.800000 3.666667 3.571429
c= 5 ×3.500000 3.375000 3.285714
c= 6 × × 3.200000 3.114286
c= 7 ×××3.000000
S(2,3, c, d)d= 9 d= 10 d= 11 d= 12
c= 4 3.500000 3.444444 3.400000 3.363636
c= 5 3.218750 3.166667 3.125000 3.090909
c= 6 3.050000 3.000000 2.960000 2.927273
c= 7 2.937500 2.888889 2.850000 2.818182
S(2,3, c, d)d= 9 d= 10 d= 11 d= 12
c= 8 2.857143 2.809524 2.771429 2.740260
c= 9 ×2.750000 2.712500 2.681818
c= 10 × × 2.666667 2.636364
c= 11 ×××2.600000
Note that the calculation of these 3·36 = 108
approximate values of sums S[2],S[3] and S[4] took
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
121
Volume 2, 2022
about 2,58 and 1443 seconds, respectively.
Note also that the numerical results S(2,3) = 2,
S(2,3,5) = 2.75 and S(2,3,5,7) = 3.375, obtained
by the three above-mentioned procedures, are com-
pletely in agreement with the results of Examples 1,
2 and 3.
7 Conclusion
In this paper two general formulas for the sum
S(a1, a2, . . . , an), briefly denoted by S[n], of the
convergent reduced harmonic series
T(a1, a2, . . . , an) = 1
a1
+1
a2
+· · · +1
an
+
+1
a2
1
+1
a2
2
+· · · +1
a2
n
+1
a1a2
+1
a1a3
+· · ·
· · · +1
a1an
+1
a2a3
+· · · +1
a2an
+· · ·
· · · +1
an−1an+1
a3
1
+1
a3
2
+· · · +1
a3
n
+· · ·
generated by npositive integers a1, a2, . . . , anwere
derived.
The first general formula has the form
S[n] =
n
i=1 1
ai
i
j=1
aj
aj−1
and the briefer second one has the form
S[n] =
n
i=1
ai
ai−1−1.
So, it can be said that the convergent reduced har-
monic series generated by npositive integers belong
to special types of convergent infinite series, such as
geometric and telescoping series, which sum can be
found analytically by means of a finite formula.
Area of Further Development
It would be interesting to try to derive a reverse for-
mula to determine a series with a given sum that
would not be unambiguous. From Tables 1 to 3 we
get, among other, that
S(2,5) = S(2,9,10) = 1.5,
S(2,3) = S(2,5,6) = 2,
S(2,3,4) = S(2,3,7,8) = S(2,3,6,10) = 3.
Furthermore, it would be possible to determine
whether the formula
S(2,3, . . . , n −1, n) = n−1
holds for each integer n≥3, as indicated by the data
in Tables 1 to 3.
Acknowledgement: This research work was sup-
ported by the Project for the Development of the Or-
ganization „DZRO Military autonomous and robotic
systems“.
Conflict of Interest: The author declare no con-
flict of interest.
References:
[1] Rike, T. Infinite Series (Berkeley seminary).
The study material of the Berkeley Math Circle,
March 24, 2002, 6 pp. https://www.docin.com/p-
107850630.html
[2] Potůček, R. Solving one infinite series problem
using CAS Maple and analytically. In Proceed-
ings of International Conference Presentation of
Mathematics ’11. Faculty of Science, Humani-
ties and Education of the Technical University
of Liberec, 2011, pages 107–112. ISBN 978-80-
7372-773-4.
[3] Potůček, R. Formulas for the Sums of the Series
of Reciprocals of the Cubic Polynomials with In-
teger Roots, at least One Zero. Chapter in book
Polynomial Paradigms: Trends and Applications
in Science and Engineering. IOP Publishing Ltd.,
2022, 33 pages, to appear.
[4] Wikipedia contributors. Geometric series —
Wikipedia, The Free Encyclopedia, 2022.
https://en.wikipedia.org/wiki/Geometric series
[5] Wikipedia contributors. Harmonic series (math-
ematics) — Wikipedia, The Free Encyclopedia,
2022. https://en.wikipedia.org/wiki/Harmonic se-
ries (mathematics)
[6] Hardy, G. H., Wright, E. M. An Introduction to the
Theory of Numbers. 4th Edition, Oxford Univer-
sity Press, London, 1975. IBSN 0-19-853310-7.
https://1library.net/document/zx5r2woq-hardy-
wright-introduction-theory-numbers-pdf.html
Creative Commons Attribution
License 4.0 (Attribution 4.0
International, CC BY 4.0)
This article is published under the terms of the Cre-
ative Commons Attribution License 4.0
https://creativecommons.org/licenses/by/4.0/deed.en
EQUATIONS
DOI: 10.37394/232021.2022.2.18
Potůček R.
E-ISSN: 2732-9976
122
Volume 2, 2022
