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On some summation formulas

  • Taekyun Kim EMAIL logo , Dae San Kim , Hyunseok Lee and Jongkyum Kwon
Published/Copyright: March 23, 2022
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Demonstratio Mathematica
From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

In this note, we derive a finite summation formula and an infinite summation formula involving Harmonic numbers of order up to some order by means of several definite integrals.

Keywords: summation formula; harmonic numbers of order r
MSC 2010: 11B83; 26A42

1 Introduction

For s ∈ C with Re ( s ) > 0 , the gamma function is defined by

(1) Γ ( s ) = ∫ 0 ∞ e − t t s − 1 d t , ( see [ 1 ] ) .

From (1), we note that Γ ( s + 1 ) = s Γ ( s ) and Γ ( n + 1 ) = n ! , ( n ∈ N ) , (see [2,3,4]).

As is well known, the Beta function is defined for Re α > 0 , Re β > 0 by

(2) B ( α , β ) = ∫ 0 1 t α − 1 ( 1 − t ) β − 1 d t , ( see [ 1 , 2 ] ) .

From (2) we note that

(3) B ( α , β ) = Γ ( α ) Γ ( β ) Γ ( α + β ) , ( see [ 1 ] ) .

The Harmonic numbers are defined by

(4) H n = 1 + 1 2 + 1 3 + ⋯ + 1 n , ( n ≥ 1 ) ,

and more generally, for any r ∈ N , the Harmonic numbers of order r are given by

H n ( r ) = 1 + 1 2 r + 1 3 r + ⋯ + 1 n r , ( n ≥ 1 ) .

In this note, we derive a finite summation formula in (5) and an infinite summation formula (6), where f n , r ( 0 ) is determined by the recurrence relation in (7) and given by Harmonic numbers of order j , for j = 1 , 2 , … , r . Our results are illustrated for r = 3 and r = 4 in the Example below. It is amusing that the infinite sum in (6) involving Harmonic numbers of order ≤ r boils down to ( − 1 ) r ( r + 1 ) ! . They are derived from several definite integrals in an elementary way. After considering the two summation formulas in (5) and (6) for r = 1 and r = 2 , we will show the results in Theorem 1.

Theorem 1

Let n , r be positive integers. Then we have

(5) ( − 1 ) r r ! ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) r + 1 = 2 2 n ( 2 n + 1 ) 2 n n f n , r ( 0 ) ,

(6) ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n f n , r ( 0 ) = ( − 1 ) r ( r + 1 ) ! ,

where f n , s ( x ) are determined by the recurrence relation

(7) f n , s + 1 ( x ) = β n ( x ) f n , s ( x ) + d d x f n , s ( x ) , ( s ≥ 1 ) , f n , 1 ( x ) = β n ( x ) = − ∑ k = 0 n ( x + 2 k + 1 ) − 1 ,

so that f n , s ( x ) is a polynomial in x involving β n ( x ) , β n ( 1 ) ( x ) , … , β n ( s − 1 ) ( x ) , with

(8) β n ( j ) ( 0 ) = ( − 1 ) j − 1 j ! H 2 n + 1 ( j + 1 ) − 1 2 j + 1 H n ( j + 1 ) , ( j ≥ 0 ) .

Example:

(a)

− 3 ! ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) 4 = 2 2 n ( 2 n + 1 ) 2 n n − H 2 n + 1 − 1 2 H n 3 − 3 H 2 n + 1 − 1 2 H n H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) − 2 H 2 n + 1 ( 3 ) − 1 8 H n ( 3 ) , ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n − H 2 n + 1 − 1 2 H n 3 − 3 H 2 n + 1 − 1 2 H n H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) − 2 H 2 n + 1 ( 3 ) − 1 8 H n ( 3 ) = − 4 ! .

(b)

4 ! ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) 5 = 2 2 n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n 4 + 6 H 2 n + 1 − 1 2 H n 2 H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 3 H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 2 + 8 H 2 n + 1 − 1 2 H n H 2 n + 1 ( 3 ) − 1 8 H n ( 3 ) + 6 ( H 2 n + 1 ( 4 ) − 1 16 H n ( 4 ) ) , 5 ! = ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n 4 + 6 H 2 n + 1 − 1 2 H n 2 H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 3 H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 2 + 8 H 2 n + 1 − 1 2 H n H 2 n + 1 ( 3 ) − 1 8 H n ( 3 ) + 6 ( H 2 n + 1 ( 4 ) − 1 16 H n ( 4 ) ) .

2 Derivation of summation formulas

On one hand, we first observe that

(9) ∫ 0 1 ( 1 − x 2 ) n d x = ∑ k = 0 n n k ( − 1 ) k ∫ 0 1 x 2 k d x = ∑ k = 0 n n k ( − 1 ) k 1 2 k + 1 .

On othe other hand, we also see that

(10) ∫ 0 1 ( 1 − x 2 ) n d x = 1 2 ∫ 0 1 ( 1 − y ) n y − 1 2 d y = 1 2 ∫ 0 1 ( 1 − y ) n + 1 − 1 ⋅ y 1 2 − 1 d y = 1 2 B n + 1 , 1 2 = 1 2 Γ ( n + 1 ) Γ 1 2 n + 1 2 n − 1 2 n − 3 2 ⋯ 1 2 Γ 1 2 = 2 n n ! ( 2 n + 1 ) ( 2 n − 1 ) ( 2 n − 3 ) ⋯ 1 = 2 2 n n ! n ! ( 2 n + 1 ) ( 2 n ) ! = 2 2 n ( 2 n + 1 ) 2 n n .

Thus, from (9) and (10) we note that

(11) 2 n n ∑ k = 0 n n k ( − 1 ) k 2 k + 1 = 2 2 n 2 n + 1 .

From (10), we have

(12) ∑ n = 1 ∞ 2 2 n n ( 2 n + 1 ) 2 n n = ∑ n = 1 ∞ 1 n ∫ 0 1 ( 1 − x 2 ) n d x = − 2 ∫ 0 1 log x d x = 2 .

Let

(13) F ( x ) = ∫ 0 1 ( 1 − t 2 ) n t x d t = ∑ k = 0 n n k ( − 1 ) k ∫ 0 1 t 2 k + x d t = ∑ k = 0 n n k ( − 1 ) k 1 2 k + x + 1 .

Note that

(14) F ′ ( x ) = d d x ∫ 0 1 ( 1 − t 2 ) n t x d t = ∫ 0 1 ( 1 − t 2 ) n t x log t d t = − ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 + x ) 2 .

On the other hand, we also have

(15) ∫ 0 1 ( 1 − t 2 ) n t x d t = 1 2 ∫ 0 1 ( 1 − y ) n y x + 1 2 − 1 d y = 1 2 B n + 1 , x + 1 2 = 1 2 Γ ( n + 1 ) Γ x + 1 2 Γ n + 1 + x + 1 2 = 1 2 n ! Γ x + 1 2 n + x + 1 2 n + x − 1 2 n + x − 3 2 ⋯ x + 1 2 Γ x + 1 2 = n ! 2 n ( 2 n + x + 1 ) ( 2 n + x − 1 ) ( 2 n + x − 3 ) ⋯ ( x + 1 ) = n ! 2 n ∏ k = 0 n ( x + 2 k + 1 ) .

Thus, we have

(16) F ′ ( x ) = d d x F ( x ) = d d x ∫ 0 1 ( 1 − t 2 ) n t x d x = d d x n ! 2 n ∏ k = 0 n ( x + 2 k + 1 ) = − n ! 2 n ∏ k = 0 n ( x + 2 k + 1 ) ∑ k = 0 n 1 x + 2 k + 1 .

Thus, according to (16), we obtain

(17) F ′ ( 0 ) = − n ! n ! 2 2 n ( 2 n + 1 ) ( 2 n ) ! ∑ k = 0 n 1 2 k + 1 = − 2 2 n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n .

From (14) and (17), we note that

(18) 2 n n ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) 2 = 2 2 n 2 n + 1 H 2 n + 1 − 1 2 H n .

By (18), we have

(19) ∑ n = 1 ∞ 2 2 n n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n = ∑ n = 1 ∞ 1 n ∑ k = 0 n n k ( − 1 ) k ( 2 k + 1 ) 2 = ∑ n = 1 ∞ 1 n ∫ 0 1 ∫ 0 1 ( 1 − x 2 y 2 ) n d x d y = − 2 ∫ 0 1 ∫ 0 1 ( log x + log y ) d x d y = 4 .

Thus, we have

(20) ∑ n = 1 ∞ 2 2 n n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n = 4 .

From (13), we note that

(21) F ″ ( x ) = d 2 d x 2 F ( x ) = d 2 d x 2 ∫ 0 1 ( 1 − t 2 ) n t x d t = ∫ 0 1 ( 1 − t 2 ) n ( log t ) 2 t x d t = 2 ! ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 + x ) 3 .

Hence, by (21), we obtain

(22) F ″ ( x ) = ∫ 0 1 ( 1 − t 2 ) n ( log t ) 2 t x d t = 2 ! ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 + x ) 3 .

From (15) and (16), we have

(23) F ″ ( x ) = d 2 d x 2 ∫ 0 1 ( 1 − t 2 ) n t x d t = n ! 2 n ( 2 n + 1 + x ) ( 2 n + x − 1 ) ⋯ ( x + 1 ) ∑ k = 0 n 1 2 k + x + 1 2 + ∑ k = 0 n 1 ( 2 k + x + 1 ) 2 .

Thus, we note that

(24) F ″ ( 0 ) = n ! 2 n ( 2 n + 1 ) ( 2 n − 1 ) ( 2 n − 3 ) ⋯ 1 ∑ k = 0 n 1 2 k + 1 2 + ∑ k = 0 n 1 ( 2 k + 1 ) 2 = n ! n ! 2 2 n ( 2 n + 1 ) ( 2 n ) ! H 2 n + 1 − 1 2 H n 2 + H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) .

Therefore, by (21) and (24), we obtain

(25) 2 ! 2 n n ∑ k = 0 n n k ( − 1 ) k ( 2 k + 1 ) 3 = 2 2 n ( 2 n + 1 ) H 2 n + 1 − 1 2 H n 2 + H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) .

Note that

(26) ∑ n = 1 ∞ 1 n ∑ k = 0 n n k ( − 1 ) k ( 2 k + 1 ) 3 = ∑ n = 1 ∞ 1 n ∫ 0 1 ∫ 0 1 ∫ 0 1 ( 1 − x 2 y 2 z 2 ) n d x d y d z = − 2 ∫ 0 1 ∫ 0 1 ∫ 0 1 ( log x + log y + log z ) d x d y d z = 6 .

From (25) and (26), we have

(27) ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n 2 + H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 2 = 6 .

Now, we begin to prove Theorem 1. Let F ( x ) = ∫ 0 1 ( 1 − t 2 ) n t x d t be as in (13).

Let r be a positive integer. Then repeated integrating by parts gives us

(28) F ( r ) ( x ) = d d x r ∫ 0 1 ( 1 − t 2 ) n t x d t = ∫ 0 1 ( 1 − t 2 ) n ( log t ) r t x d t = ∑ k = 0 n n k ( − 1 ) k ∫ 0 1 ( log t ) r t x + 2 k d t = ∑ k = 0 n n k ( − 1 ) k − r x + 2 k + 1 ∫ 0 1 ( log t ) r − 1 t x + 2 k d t = ∑ k = 0 n n k ( − 1 ) k − r x + 2 k + 1 ⋯ − 1 x + 2 k + 1 ∫ 0 1 t x + 2 k d t = ( − 1 ) r r ! ∑ k = 0 n n k ( − 1 ) k 1 ( x + 2 k + 1 ) r + 1 .

As shown in (15) and (16), F ( x ) is alternatively expressed by

(29) F ( x ) = ∫ 0 1 ( 1 − t 2 ) n t x d t = n ! 2 n ∏ k = 0 n ( x + 2 k + 1 ) , F ( 0 ) = 2 2 n ( 2 n + 1 ) 2 n n ,

and its derivative is given by

(30) F ( 1 ) ( x ) = F ( x ) β n ( x ) , β n = β n ( x ) = − ∑ k = 0 n ( x + 2 k + 1 ) − 1 .

Repeated differentiations give us

(31) F ( 2 ) ( x ) = F ( x ) ( β n 2 + β n ( 1 ) ) , F ( 3 ) ( x ) = F ( x ) ( β n 3 + 3 β n β n ( 1 ) + β n ( 2 ) ) , F ( 4 ) ( x ) = F ( x ) ( β n 4 + 6 β n 2 β n ( 1 ) + 3 ( β n ( 1 ) ) 2 + 4 β n β n ( 2 ) + β n ( 3 ) ) , F ( 5 ) ( x ) = F ( x ) ( β n 5 + 10 β n 3 β n ( 1 ) + 10 β n 2 β ( 2 ) + 15 β ( β n ( 1 ) ) 2 + 5 β n β n ( 3 ) + 10 β n ( 1 ) β n ( 2 ) + β n ( 4 ) ) , … .

In general, we let F ( s ) ( x ) = F ( x ) f n , s ( x ) , for s ≥ 1 . Then further differentiation gives us

F ( s + 1 ) ( x ) = F ( x ) ( β n ( x ) f n , s ( x ) + d d x f n , s ( x ) ) .

Thus, we obtain the recurrence relation for { f n , s ( x ) } s = 1 ∞ :

(32) f n , s + 1 ( x ) = β n ( x ) f n , s ( x ) + d d x f n , s ( x ) , ( s ≥ 1 ) , f n , 1 ( x ) = β n ( x ) .

Now, from (28) and (29), we obtain

(33) ( − 1 ) r r ! ∑ k = 0 n n k ( − 1 ) k 1 ( x + 2 k + 1 ) r + 1 = n ! 2 n ∏ k = 0 n ( x + 2 k + 1 ) f n , r ( x ) .

Letting x = 0 in (33), we obtain

(34) ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) r + 1 = ( − 1 ) r r ! 2 2 n ( 2 n + 1 ) 2 n n f n , r ( 0 ) .

Multiplying both sides of (34) by 1 n and summing over n , we obtain

(35) ( − 1 ) r r ! ∑ n = 1 ∞ 2 2 n n ( 2 n + 1 ) 2 n n f n , r ( 0 ) = ∑ n = 1 ∞ 1 n ∑ k = 0 n n k ( − 1 ) k 1 ( 2 k + 1 ) r + 1 = ∑ n = 1 ∞ 1 n ∫ 0 1 ⋯ ∫ 0 1 ( 1 − x 1 2 x 2 2 ⋯ x r + 1 2 ) n d x 1 d x 2 ⋯ d x r + 1 = − 2 ∫ 0 1 ⋯ ∫ 0 1 ( log x 1 + ⋯ + log x r + 1 ) d x 1 ⋯ d x r + 1 = 2 ( r + 1 ) .

Finally, we note that f n , s ( x ) is a polynomial in x involving β n ( x ) , β n ( 1 ) ( x ) , … , β n ( s − 1 ) ( x ) , as we can see from the recurrence relation in (32). From (30), we see that

(36) β n ( j ) ( 0 ) = ( − 1 ) j − 1 j ! ∑ k = 0 n ( 2 k + 1 ) − ( j + 1 ) = ( − 1 ) j − 1 j ! H 2 n + 1 ( j + 1 ) − 1 2 j + 1 H n ( j + 1 ) .

Combining (32) and (34)–(36) altogether, we obtain our main results in Theorem 1.

Finally, following the suggestion of one of the reviewers, after slight modifications we state the results in (12), (20), and (27) as a theorem, which are special cases of Theorem 1.

Theorem 2

The following identities hold true:

( a ) ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n = 1 . ( b ) ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n 1 2 H n − H 2 n + 1 = − 2 . ( c ) ∑ n = 1 ∞ 2 2 n − 1 n ( 2 n + 1 ) 2 n n H 2 n + 1 − 1 2 H n 2 + H 2 n + 1 ( 2 ) − 1 4 H n ( 2 ) 2 = 6 .

Acknowledgements

The authors would like to thank the referees for their suggestions that helped improve the original manuscript in its present form. We would also like to thank Jangjeon Institute for Mathematical Sciences for the support of this research.

  1. Conflict of interest: The authors state no conflict of interest.

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Received: 2021-11-17
Revised: 2022-02-16
Accepted: 2022-02-17
Published Online: 2022-03-23

© 2022 Taekyun Kim et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0003
Keywords for this article
summation formula; harmonic numbers of order r
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