J ournal of
Mathematical
I nequalities
Volume 11, Number 1 (2017), 209–215
doi:10.7153/jmi-11-20
A RECIPROCAL SUM RELATED TO THE RIEMANN ζ –FUNCTION
L IN X IN AND L I X IAOXUE
(Communicated by J. Pečarić)
Abstract. This paper, we using the elementary method and several new inequalities to study the
computational problem of one kind reciprocal sums related to the Riemann zeta-function at the
point s = 4 , and give an explicit computational formula for it.
1. Introduction
Let complex number s = σ + it , if σ > 1 , then the famous Riemann zeta-function
ζ (s) is defined as the Dirichlet series
ζ (s) =
∞
1
∑ ns ,
n=1
and by analytic continuation to the whole complex plane except for a simple pole at
s = 1 with residue 1 .
About the various properties of ζ (s), many mathematicians had studied them, and
obtained abundant research results, for example, see [1], [2] and [3]. Very recently, the
first author [4] studied the computational problem of the reciprocal sum related to the
Riemann ζ -function, and used the elementary method and some new inequalities to
prove that for any positive integer n , one has the identities
⎡
−1 ⎤
∞
1
⎦ = n−1
⎣ ∑
2
k=n k
and
⎡
−1 ⎤
∞
⎣ ∑ 1
⎦ = 2n(n − 1),
3
k=n k
where function [x] denotes the greatest integer x .
Mathematics subject classification (2010): 11B83, 11M06.
Keywords and phrases: Riemann zeta-function, inequality, function [x] , computational formula, elementary method.
This work is supported by the N. S. F. (11371291) of P. R. China, G. I. C. F. (YZZ15009) of Northwest University, S.
P. D. E. S. F. (15JK1744) of P. R. China.
c
, Zagreb
Paper JMI-11-20
209
210
L. X IN AND L. X IAOXUE
At the same time, Lin Xin [4] also proposed the following open problem: Whether
there exists an explicit computational formula for
⎡
−1 ⎤
∞
1
⎦,
⎣ ∑
(1)
s
k=n k
where s is an integer with s 4 .
This problem is interesting, because it is actually an effective approximation for
the partial sum of the Riemann zeta-function ζ (s).
More contents related to other second-order linear recurrence sequences and polynomials had also been studied by many authors, see references [5]–[14]. For example,
H. Ohtsuka and S. Nakamura [5] studied the properties of the sum of reciprocal Fibonacci numbers, and proved the identities
⎡
−1 ⎤ ∞
1
if n 2 is an even number;
⎦ = Fn−2 ,
⎣ ∑
F
−
1,
if n 1 is an odd number,
F
n−2
k=n k
⎡
∞
⎣ ∑ 1
2
k=n Fk
−1 ⎤
⎦=
Fn−1 Fn − 1, if n 2 is an even number;
if n 1 an odd number.
Fn Fn−1,
Zhang Wenpeng and Wang Tingting [8] considered the computational problem of Pell
numbers, and proved the identity
⎡
−1 ⎤ ∞
if n 2 is an even number;
⎣ ∑ 1
⎦ = Pn−1 + Pn−2,
P
+
P
−
1,
if
n 1 is an odd number.
P
n−1
n−2
k
k=n
where Pell numbers Pn are defined as P0 = 0 , P1 = 1 , and Pn+1 = 2Pn + Pn−1 for all
integers n 1 .
The main purpose of this paper is to study the computational problem of (1) for
s = 4 , and use the elementary method and some new inequalities to give an interesting computational formula for (1) with s = 4 . That is, we shall prove the following
conclusion:
T HEOREM . For any positive integer n , we have the identity
⎧
⎡
−1 ⎤ ⎪
, if n = 2m;
24m3 − 18m2 + 3(5m−1)
⎨
∞
2
⎦=
⎣ ∑ 1
4
⎪
⎩ 24m3 − 54m2 + 3(58m−17) , if n = 2m − 1.
k=n k
4
For general integers s 5 , whether there exists an explicit computational formula
for (1) is an open problem. Perhaps using our method one can study it for the cases
s 5 , but it is very complicated to construct those new inequalities when s is larger.
With the aid of computer and mathematical software “Mathematica”, I believe that we
can solve this problem completely.
211
A RECIPROCAL SUM RELATED TO THE R IEMANN ζ - FUNCTION
2. Several Lemmas
In this section, we shall give some lemmas which are necessary in the proof of our
theorem. First we have the following:
L EMMA 1. For any integer n > 1 , we have the inequality
1
9
24n3 − 18n2 + 15
2 n− 8
<
−
1
9
24(n + 1)3 − 18(n + 1)2 + 15
2 (n + 1) − 8
1
1
+
.
4
(2n)
(2n + 1)4
Proof. In fact, the inequality in Lemma 1 is equivalent to the inequality
1
1
− 33
5
3
3
3
2
3 8n + 18n2 + 29
3 8n − 6n + 2 n − 8
2 n+ 8
<
32n4 + 32n3 + 24n2 + 8n + 1
256n8 + 512n7 + 384n6 + 128n5 + 16n4
or
8n2 + 4n + 32
33
8n3 − 6n2 + 52 n − 38 8n3 + 18n2 + 29
2 n+ 8
<
Let
and
32n4 + 32n3 + 24n2 + 8n + 1
256n8 + 512n7 + 384n6 + 128n5 + 16n4
.
(2)
3 f (n) = 8n2 + 4n +
256n8 + 512n7 + 384n6 + 128n5 + 16n4 ,
2
3
5
g(n) = 8n −6n + n−
2
8
3
2
33 3
2 29
8n +18n + n+
32n4 +32n3 +24n2+8n+1 .
2
8
Then
f (n) = 2048n10 + 5120n9 + 5504n8 + 3328n7 + 1216n6 + 256n5 + 24n4.
g(n) =
(3)
99
19
39
64n6 + 96n5 + 28n4 − 12n3 + n2 + n −
4
8
64
4
3
2
× 32n + 32n + 24n + 8n + 1
= 2048n10 + 5120n9 + 5504n8 + 3328n7 + 1272n6 + 340n5 +
From (3) and (4) we know that f (n) < g(n). So inequality (2) is correct.
This proves Lemma 1. 305 4
n .
2
(4)
212
L. X IN AND L. X IAOXUE
L EMMA 2. For any integer n 1 , we also have the inequality
1
1
1
+
<
4
4
3
3
(2n)
(2n + 1)
24n − 18n2 + 15
2 n− 2
1
−
.
3
3
24(n + 1) − 18(n + 1)2 + 15
2 (n + 1) − 2
Proof. First we have the identity
1
1
−
3
3 − 18(n + 1)2 + 15 (n + 1) − 3
24n3 − 18n2 + 15
n
−
24(n
+
1)
2
2
2
2
8n2 + 4n + 32
= 8n3 − 6n2 + 52 n − 12 8n3 + 18n2 + 29
2 n+4
8n2 + 4n + 32
=
.
(5)
11
2
64n6 + 96n5 + 28n4 − 14n3 + 13
4 n + 4 n−2
11
2
4
3
2
Let h(n) = 64n6 + 96n5 + 28n4 − 14n3 + 13
4 n + 4 n − 2 (32n + 32n + 24n + 8n +
1), then we have
h(n) = 2048n10 + 5120n9 + 5504n8 + 3264n7 + 1160n6 + 176n5
91
53
+18n4 + 14n3 − n2 − n − 2.
4
4
(6)
From (3) and (6) we know that f (n) > h(n). Thus, from the definition of f (n) we have
the inequality
1
1
1
+
<
4
4
3
3
(2n)
(2n + 1)
24n − 18n2 + 15
2 n− 2
1
−
.
3
3
24(n + 1) − 18(n + 1)2 + 15
2 (n + 1) − 2
This proves Lemma 2.
L EMMA 3. For any integer n 1 , we have the inequality
1
1
− 29
17
17
3
2
3
3 8n − 18n + 2 n − 4
3 8(n + 1) − 18(n + 1)2 + 29
2 (n + 1) − 4
1
1
>
+
.
4
(2n − 1)
(2n)4
Proof. It is clear that
1
32n4 − 32n3 + 24n2 − 8n + 1
1
.
+
=
4
4
(2n − 1)
(2n)
16 (16n8 − 32n7 + 24n6 − 8n5 + n4)
(7)
213
A RECIPROCAL SUM RELATED TO THE R IEMANN ζ - FUNCTION
1
1
− 17
3 − 18(n + 1)2 + 29 (n + 1) − 17
3 8n3 − 18n2 + 29
n
−
3
8(n
+
1)
2
4
2
4
8n2 − 4n + 32
= 3
17
8n − 18n2 + 29
8n3 + 6n2 + 52 n + 14
2 n− 4
=
8n2 − 4n + 32
17
2
64n6 − 96n5 + 28n4 + 10n3 + 25
4 n − 7n − 16
.
(8)
For convenience, we let
25
17
U(n) = 64n6 − 96n5 + 28n4 + 10n3 + n2 − 7n −
4
16
4
3
2
× 32n − 32n + 24n − 8n + 1 ,
3
V (n) = 16 16n8 − 32n7 + 24n6 − 8n5 + n4 8n2 − 4n +
.
2
Then by computations, we have
V (n) − U(n) = 64n7 − 168n6 + 248n5 − 264n4 + 174n3 −
17
147 2 3
n − n− .
4
2
16
It is easy to prove that V (n) − U(n) > 0 for all integers n 1 . So combining (7) and
(8) we may immediately deduce Lemma 3. L EMMA 4. For any integer n 1 , we have the inequality
1
1
1
+
> 3
25
(2n − 1)4 (2n)4
3 8n − 18n2 + 29
2 n− 6
1
.
− 25
3
3 8(n + 1) − 18(n + 1)2 + 29
2 (n + 1) − 6
ity.
Proof. From the method of proving Lemma 3 we can easily deduce this inequal
3. Proof of the theorem
In this section, we shall complete the proof of our theorem. If n = 2m is an even
number, then from Lemma 1 and Lemma 2 we have
1
9
24m3 − 18m2 + 15
2 m− 8
∞
1
1
= ∑
−
15
9
9
3
2
3
24(k + 1) − 18(k + 1)2 + 15
k=m 24k − 18k + 2 k − 8
2 (k + 1) − 8
∞
∞
1
1
1
< ∑
+
.
=
∑
4
4
4
(2k)
(2k
+
1)
k
k=m
k=2m
(9)
214
L. X IN AND L. X IAOXUE
Similarly, from Lemma 2 we can deduce that
∞ 1
1
1
∑ 4 = ∑ (2k)4 + (2k + 1)4
k=2m k
k=m
∞
1
1
< ∑
−
15
3
3
3
2
24(k + 1)3 − 18(k + 1)2 + 15
k=m 24k − 18k + 2 k − 2
2 (k + 1) − 2
∞
1
.
3
24m3 − 18m2 + 15
2 m− 2
=
(10)
Combining (9) and (10) we may immediately deduce that
3
15
24m − 18m + m − <
2
2
3
∞
1
∑ k4
k=2m
2
−1
< 24m3 − 18m2 +
9
15
m− ,
2
8
which implies (according to the parity of m):
⎡
∞
⎣ ∑ 1
4
k=2m k
−1 ⎤
⎦ = 24m3 − 18m2 + 3(5m − 1) .
2
(11)
If n = 2m − 1 be an odd number, then from Lemma 3, Lemma 4 and the method of
proving (9) and (10) we have
51
87
<
24m − 54m + m −
2
4
3
2
∞
1
∑ k4
k=2m−1
−1
< 24m3 − 54m2 +
25
87
m− ,
2
2
which implies (according to the parity of m):
⎡
⎣
∞
1
∑ k4
k=2m−1
−1 ⎤
⎦ = 24m3 − 54m2 + 174m − 51 .
4
(12)
Applying (11) and (12) we know that for any integer n 1 , we have the identity
⎡
−1 ⎤ ⎧
3 − 18m2 + 3(5m−1) , if n = 2m;
⎨
∞
24m
2
⎦=
⎣ ∑ 1
174m−51 4
⎩
k
3
2
k=n
24m − 54m +
, if n = 2m − 1.
4
This completes the proof of the theorem.
Acknowledgement. The authors would like to thank the referee for very helpful and
detailed comments, which have significantly improved the presentation of this paper.
A RECIPROCAL SUM RELATED TO THE R IEMANN ζ - FUNCTION
215
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(Received February 13, 2016)
Lin Xin
School of Mathematics, Northwest University
Xi’an, Shaanxi, P. R. China
Li Xiaoxue
School of Mathematics, Northwest University
Xi’an, Shaanxi, P. R. China
e-mail: [email protected]
Journal of Mathematical Inequalities
www.ele-math.com
[email protected]