aa r X i v : . [ m a t h . G M ] F e b Repeated Sums and Binomial Coefficients
Roudy El Haddad
Universit´e La Sagesse, Facult´e de g´enie, Polytech
Abstract
Binomial coefficients have been used for centuries in a variety of fields and have accumulatednumerous definitions. In this paper, we introduce a new way of defining binomial coefficientsas repeated sums of ones. A multitude of binomial coefficient identities will be shown in orderto prove this definition. Using this new definition, we simplify some particular sums such asthe repeated Harmonic sum and the repeated Binomial-Harmonic sum. We derive formulaefor simplifying general repeated sums as well as a variant containing binomial coefficients. Ad-ditionally, we study the m -th difference of a sequence and show how sequences whose m -thdifference is constant can be related to binomial coefficients. Keywords:
Binomial Coefficients, Binomial Sums, m -th Difference, Repeated Sequences,Repeated Sums, Harmonic Sums, Binomial-Harmonic Sums.
1. Introduction
The author is interested in the study of the various types of repetitive sums, that is, devel-oping formulae and identities to better understand these types of sums as well as to simplify theway we work with them. Previously, the author has presented a set of useful formulae for twotypes of repetitive sums: recurrent sums [1] and multiple sums [2]. In this article, we aim to dothe same for a third type of repetitive sums: the repeated sums. For the first two types, parti-tions were key for developing a reduction theorem to simplify these sums. For this type of sums,binomial coefficients are key. To do so, we first need to develop a set of binomial coefficientidentities as well as we need to introduce a new definition for binomial coefficients. Binomialcoefficients appear in many distinct fields of mathematics (including probability, combinatorics,analysis, etc.) and, therefore, they can be defined in many different ways. In 1654, in his article“Trait´e du triangle arithm´etique” [3, 4], Pascal presented what is now known as Pascal’s trian-gle and which constituted the first definition of binomial coefficients. The binomial coefficient (cid:0) nk (cid:1) represents the element in the k -th column and n -th row of Pascal’s triangle. They can alsobe defined by the recurrent relation used to construct Pascal’s triangle which we will call thePascal’s triangle identity and which states that (cid:0) nk (cid:1) = (cid:0) n − k (cid:1) + (cid:0) n − k − (cid:1) . The notation (cid:0) nk (cid:1) was intro-duced by Euler in the 18th century. Similarly, binomial coefficients can be denoted as C kn . Thissecond notation, introduced in the 19th century, is mostly used in combinatorics where binomialcoefficients are defined as the number of ways of choosing an unordered subset of k elements froma set of n elements. Binomial coefficients are also intensively used in probability calculation forbinomial distributions. Let X be a random variable, let n be the number of experiments, let k be an integer, and let p represent the probability of success in a single experiment, the binomialdistribution formula is given by P ( X = k ) = C kn p k (1 − p ) n − k . Additional definitions of thesecoefficients appear in analysis and calculus. First, they appear in the binomial theorem as wellas in the generalized binomial theorem presented by Newton [5, 6], ( u + v ) n = P nk =0 (cid:0) nk (cid:1) u k v n − k Email address: [email protected] (Roudy El Haddad)
Preprint submitted to Elsevier February 25, 2021 nd ( u + · · · + u m ) n = P k + ··· + k m = n (cid:0) nk m (cid:1)(cid:0) n − k m k m − (cid:1) · · · (cid:0) n − k m −···− k k (cid:1) u k u k · · · u k m m . From the simplecase of the theorem, the binomial coefficient (cid:0) nk (cid:1) can be defined as the coefficient of u k v n − k inthe expansion of ( u + v ) n . Likewise, (cid:0) nk (cid:1) appears in Leibniz’s Formula for the n -th derivativeof a product [7, 8, 9], ( uv ) ( n ) = P nk =0 (cid:0) nk (cid:1) u ( k ) v ( n − k ) . Hence, Leibniz’s rule defines the binomialcoefficient (cid:0) nk (cid:1) as the coefficient of u ( k ) v ( n − k ) in the expansion of ( uv ) ( n ) . These coefficients evenappear in number theory: In [1, 2], binomial coefficients are defined as a sum over partitionsof an integer m in several ways. And, finally, binomial coefficients can be defined in terms offactorials: (cid:0) nk (cid:1) = n ! k !( n − k )! . Remark.
Tables of binomial coefficients can be found in [10, 11, 12].In this study, we propose a new definition: we define binomial coefficients as repeated sums(Section 3). This definition is key for the determination of a reduction formula for repeatedsums which allows the reduction of such sums into simple non-repetitive sums. In Section 3, wealso present formulas for the m -th difference of a sequence as well as we show the link betweensequences with a constant m -th difference and binomial coefficients. In Section 2, we presentsome formulae for binomial sums and repeated binomial sums which are needed to prove thedefinition. In Section 4, this new definition is used to derive results related to the harmonicsum such as an expression for the repeated harmonic sum in terms of the simple harmonic sum.We also derive expressions for a modified version of the repeated harmonic sum which we willrefer to as the repeated binomial-harmonic sum. Then, in Section 5, this definition as well asthe results derived will be utilized to develop a reduction formula for the repeated sum of anysequence a N . A similar expression is also derived for a modified version of repeated sums whichwe will refer to as the repeated “binomial-sequence” sum.
2. Sums of binomial coefficients
In this section, we prove some formulas related to the sums of binomial coefficients. Theseformulas are needed in order to produce the proposed definition for binomial coefficients.
We begin by determining a formula for the sum of binomial coefficients.
Theorem 2.1.
For any k, q, n ∈ N such that n ≥ q , we have n X i = q (cid:18) ik (cid:19) = (cid:18) n + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) . Remark.
The coefficients of the form (cid:0) nk (cid:1) where n < k , being undefined according to thefactorial definition, are supposed zero. Hence, if q < k , we can start the sum at k . Proof.
1. Base case: verify true for n = q .From Pascal’s Triangle identity, (cid:18) qk + 1 (cid:19) + (cid:18) qk (cid:19) = (cid:18) q + 1 k + 1 (cid:19) . Hence, q X i = q (cid:18) ik (cid:19) = (cid:18) qk (cid:19) = (cid:18) q + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) .
2. Induction hypothesis: assume the statement is true until n . n X i = q (cid:18) ik (cid:19) = (cid:18) n + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) .
3. Induction step: we will show that this statement is true for ( n + 1).We have to show the following statement to be true: n +1 X i = q (cid:18) ik (cid:19) = (cid:18) n + 2 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) . n +1 X i = q (cid:18) ik (cid:19) = (cid:18) n + 1 k (cid:19) + n X i = q (cid:18) ik (cid:19) = (cid:18) n + 1 k (cid:19) + (cid:18) n + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) . From Pascal’s Triangle identity, we get n +1 X i = q (cid:18) ik (cid:19) = (cid:18) n + 2 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) . Hence, the theorem is proven by induction. (cid:4)
Corollary 2.1.
For q = 0 or q = k , Theorem 2.1 becomes n X i =0 (cid:18) ik (cid:19) = n X i = k (cid:18) ik (cid:19) = (cid:18) n + 1 k + 1 (cid:19) . Corollary 2.2.
The shifted binomial sum can be calculate as follows n + m X i = m (cid:18) ik (cid:19) = n X i =0 (cid:18) i + mk (cid:19) = (cid:18) n + m + 1 k + 1 (cid:19) − (cid:18) mk + 1 (cid:19) . Proof.
Applying Theorem 2.1 for q = m and replacing n by n + m , we obtain the corollary. (cid:4) Using Theorem 2.1, we develop a formula for the repeated sum of binomial coefficients.
Theorem 2.2.
For any m ∈ N ∗ and for any k, q, n ∈ N such that n ≥ q , we have n X N m = q · · · N X N = q (cid:18) N k (cid:19) = (cid:18) n + mk + m (cid:19) − m X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m − j ) m − j (cid:19) = (cid:18) n + mk + m (cid:19) − m X j =1 (cid:18) ( q −
1) + j ( q − − k (cid:19)(cid:18) ( n − q ) + ( m − j ) n − q (cid:19) . Proof.
1. Base case: verify true for m = 1. (cid:18) n + 1 k + 1 (cid:19) − X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + (1 − j )1 − j (cid:19) = (cid:18) n + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) . n X N = q (cid:18) N k (cid:19) = (cid:18) n + 1 k + 1 (cid:19) − (cid:18) qk + 1 (cid:19) .
2. Induction hypothesis: assume the statement is true until m . n X N m = q · · · N X N = q (cid:18) N k (cid:19) = (cid:18) n + mk + m (cid:19) − m X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m − j ) m − j (cid:19) .
3. Induction step: we will show that this statement is true for ( m + 1).We have to show the following statement to be true: n X N m +1 = q · · · N X N = q (cid:18) N k (cid:19) = (cid:18) n + m + 1 k + m + 1 (cid:19) − m +1 X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m + 1 − j ) m + 1 − j (cid:19) . n X N m +1 = q · · · N X N = q (cid:18) N k (cid:19) = n X N m +1 = q N m +1 X N m = q · · · N X N = q (cid:18) N k (cid:19)! = n X N m +1 = q (cid:18) N m +1 + mk + m (cid:19) − n X N m +1 = q m X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( N m +1 − q ) + ( m − j ) m − j (cid:19) = n X N m +1 = q (cid:18) N m +1 + mk + m (cid:19) − m X j =1 n X N m +1 = q (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( N m +1 − q ) + ( m − j ) m − j (cid:19) = n + m X N m +1 = q + m (cid:18) N m +1 k + m (cid:19) − m X j =1 ( n − q )+( m − j ) X N m +1 = m − j (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) N m +1 m − j (cid:19) = n + m X N m +1 = q + m (cid:18) N m +1 k + m (cid:19) − m X j =1 (cid:18) ( q −
1) + jk + j (cid:19) ( n − q )+( m − j ) X N m +1 = m − j (cid:18) N m +1 m − j (cid:19) . By applying Theorem 2.1, n X N m +1 = q · · · N X N = q (cid:18) N k (cid:19) = (cid:18) n + m + 1 k + m + 1 (cid:19) − (cid:18) q + mk + m + 1 (cid:19) − m X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m + 1 − j ) m + 1 − j (cid:19) . Noticing that m +1 X j = m +1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m + 1 − j ) m + 1 − j (cid:19) = (cid:18) q + mk + m + 1 (cid:19) , hence, n X N m +1 = q · · · N X N = q (cid:18) N k (cid:19) = (cid:18) n + m + 1 k + m + 1 (cid:19) − m +1 X j =1 (cid:18) ( q −
1) + jk + j (cid:19)(cid:18) ( n − q ) + ( m + 1 − j ) m + 1 − j (cid:19) . (cid:4)
4n important particular case of Theorem 2.2 is illustrated by the following corollary.
Corollary 2.3.
For any m ∈ N ∗ and for any k, n ∈ N , we have that n X N m =0 · · · N X N =0 (cid:18) N k (cid:19) = (cid:18) n + mk + m (cid:19) . Proof.
From Theorem 2.2 for q = 0, n X N m =0 · · · N X N =0 (cid:18) N k (cid:19) = (cid:18) n + mk + m (cid:19) − m X j =1 (cid:18) j − k + j (cid:19)(cid:18) n + ( m − j ) m − j (cid:19) . We can notice that because k ≥
0, then k + j ≥ j > j −
1, thus, ∀ j ∈ N , (cid:0) j − k + j (cid:1) = 0. Hence, n X N m =0 · · · N X N =0 (cid:18) N k (cid:19) = (cid:18) n + mk + m (cid:19) . (cid:4) The formula for repeated sums of ones, which is needed to prove the definition, is as follows.
Corollary 2.4.
For any m ∈ N ∗ and for any q, n ∈ N such that n ≥ q , we have that n X N m = q · · · N X N = q (cid:18) n − q + mm (cid:19) . Proof. n X N m = q · · · N X N = q n X N m = q · · · N X N = q N − q X N =0 n X N m = q · · · N − q X N =0 N X N =0 · · · = n − q X N m =0 · · · N X N =0 N X N =0 . From Corollary 2.3 for k = 0, n X N m = q · · · N X N = q n − q X N m =0 · · · N X N =0 N X N =0 (cid:18) n − q + mm (cid:19) . (cid:4) The sum of a special product of binomial coefficients can also be derived from Theorem 2.2.
Corollary 2.5.
For any m ∈ N ∗ and for any q, n ∈ N such that n ≥ q , we have that m X j =1 (cid:18) ( q −
1) + jj (cid:19)(cid:18) ( n − q ) + ( m − j ) m − j (cid:19) = m X j =1 (cid:18) ( q −
1) + jq − (cid:19)(cid:18) ( n − q ) + ( m − j ) n − q (cid:19) = (cid:18) n + mm (cid:19) − (cid:18) n − q + mm (cid:19) . Proof.
By applying Theorem 2.2 for k = 0, n X N m = q · · · N X N = q (cid:18) n + mm (cid:19) − m X j =1 (cid:18) ( q −
1) + jj (cid:19)(cid:18) ( n − q ) + ( m − j ) m − j (cid:19) = (cid:18) n + mm (cid:19) − m X j =1 (cid:18) ( q −
1) + jq − (cid:19)(cid:18) ( n − q ) + ( m − j ) n − q (cid:19) . By comparing with the expression given by Corollary 2.4, we get this corollary. (cid:4) . Binomial coefficients as repeated sums and repeated sequences In this section, we use the formulas presented in the previous section in order to introducea new definition of binomial coefficients. First, we prove that a binomial coefficient can bedefined as a repeated sum of ‘1’s. Second, we prove that a binomial coefficient can be definedas a repeated sequence.
In this section, we prove that binomial coefficients can be defined as repeated sums of ones.A repeated sum of ‘1’s can be expressed as a binomial coefficient as is shown by Theorem 3.1.
Theorem 3.1.
A repeated sum of 1s corresponds to the following binomial coefficient, n X N m =1 · · · N X N =1 (cid:18) n + m − m (cid:19) . n X N m =0 · · · N X N =0 (cid:18) n + mm (cid:19) . Proof.
The first equality of this theorem is obtained by applying Corollary 2.4 for q = 1 andthe second equality is obtained by applying Corollary 2.4 for q = 0. (cid:4) Remark.
We can highlight the combinatorical nature of repeated sums of 1s as (cid:0) n − mm (cid:1) repre-sents the number of ways of choosing m elements from n with replacement but without regardto order ([13] pp. 24-25).A binomial coefficient can be defined as a repeated sum of 1s as shown by Theorem 3.2. Theorem 3.2.
A binomial coefficient corresponds to the following repeated sums of 1s, (cid:18) nk (cid:19) = n − k +1 X N k =1 · · · N X N =1 n − k X N k =0 · · · N X N =0 . Proof.
The first part of this theorem is obtained by replacing n by n − k + 1 in the first equationof Theorem 3.1. The second part is obtained by replacing n by n − k in the second equation ofTheorem 3.1. (cid:4) In this section, we will go through a series of lemmas in order to prove that any binomialcoefficient can be defined as a repeated sequence (defined below). In order to do so, we will firstprove that repeated sequences can be explicitly expressed as a repeated sum of ones. Then,using the fact that any repeated sum of ones can be defined as a binomial coefficient (as provenin the previous section), we define these repeated sequences as binomial coefficients.
Definition.
We define the first difference of a sequence ( x n ) , denoted by ∆ x n , as the differencebetween two consecutive terms of the sequence ( x n ) : ∆ x n = x n − x n − . efinition. Let m ∈ N . We define the m -th difference of a sequence ( x n ) , denoted by ∆ m x n ,by the following recurrent relation: ( ∆ m x n = ∆ m − x n − ∆ m − x n − ∀ i ∈ Z , ∆ x i = x i . Definition.
We define a repeated sequence of degree m , denoted by { x n,m } ∞ n =1 , as a sequencewhose m -th difference is a constant ( ∀ i, j ∈ N ∗ , ∆ m x i,m = ∆ m x j,m ) . x n,m is the n -th term ofthis repeated sequence of order m . Remark.
In order to keep the m -th difference always valid we define an extension of thesequence such that ∀ i ≤ , x i,m = 0. However, the properties of the sequence do not extendto this extension. Note that the theorems develop in this section will only apply to repeatedsequences which validate the condition imposed by this extension. Remark.
For a given order m and a given m -th difference, the repeated sequence is unique.The repeated sequence of order m having ∆ m x n,m = δ ∈ C is denoted by { x ( δ ) n,m } ∞ n =1 . Definition.
We define the unity repeated sequence, denoted by { ˆ x n,m } ∞ n =1 , as the sequencewhose m -th difference is 1 ( ∀ i ∈ N ∗ , ∆ m ˆ x i,m = 1) . ˆ x n,m is the n -th term of the unity repeatedsequence of order m . Example 3.1.
The unity repeated sequence of order 0, (ˆ x n, ), is the following:ˆ x , = 1 , ˆ x , = 1 , ˆ x , = 1 , · · · Example 3.2.
The unity repeated sequence of order 1, (ˆ x n, ), is the following:ˆ x , = 1 , ˆ x , = 2(= 1 + 1) , ˆ x , = 3(= 1 + 1 + 1) , · · · Example 3.3.
The unity repeated sequence of order 2, (ˆ x n, ), is the following:ˆ x , = 1 , ˆ x , = 3(= 1 + (1 + 1)) , ˆ x , = 6(= 1 + (1 + 1) + (1 + 1 + 1)) , · · · In this section, using the recurrent definition of the m -th difference, we develop an additionaldefinition for the m -th difference of a sequence in terms of a lower order difference of thissequence. Theorem 3.3.
For any m, n, p ∈ N such that p ≤ m and for any sequence x n , we have that ∆ m x n = p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p x n − j . Proof.
1. Base case: verify true for m = 0.Knowing that 0 ≤ p ≤ m and m = 0, hence, p = 0. X j =0 ( − j (cid:18) j (cid:19) ∆ x n − j = ∆ x n .
2. Induction hypothesis: assume the statement is true until m .∆ m x n = p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p x n − j .
7. Induction step: we will show that this statement is true for ( m + 1).We have to show the following statement to be true:∆ m +1 x n = p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p +1 x n − j . Knowing that ∆ m +1 x n = ∆ m x n − ∆ m x n − and applying the induction hypothesis, we get∆ m +1 x n = p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p x n − j − p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p x n − j − = p X j =0 ( − j (cid:18) pj (cid:19) (∆ m − p x n − j − ∆ m − p x n − j − )= p X j =0 ( − j (cid:18) pj (cid:19) ∆ m − p +1 x n − j . (cid:4) Remark.
As we can see, the binomial coefficient (cid:0) pj (cid:1) can be defined as the coefficient of( − j ∆ m − p x n − j in the expression of the m -th difference of the sequence ( x n ).In particular, the m -th difference of a sequence can be expressed in terms of its elements. Corollary 3.1.
For p = m , Theorem 3.3 becomes ∆ m x n = m X j =0 ( − j (cid:18) mj (cid:19) x n − j . We are interested in the calculation of the sum of such sequences as well as in the definitionof such sequences in terms of binomial coefficients. To do so, in this section, we convert thedefinition of a repeated sequence of degree m which is in terms of the m -th difference into adefinition in terms of a repeated sum of ones. Then, in the later sections, the theorems forrepeated sums of ones are used to prove the proposed definition as well as to develop a formulafor the repeated sum of a repeated sequence of degree m .We start by proving the following set of lemmas. Lemma 3.1.
For repeated sequences of any order k , the difference of any order m is zero forall terms with an index i ≤ , ∀ k ∈ N , ∀ m ∈ N , ∀ i ≤ , ∆ m x ( δ ) i,k = 0 . Proof.
1. Base case: verify true for m = 0.By definition, x ( δ ) i,k = 0 for i ≤
0. Hence, ∀ k ∈ N , ∀ i ≤ , ∆ x ( δ ) i,k = x ( δ ) i,k = 0 .
2. Induction hypothesis: assume the statement is true until m . ∀ k ∈ N , ∀ i ≤ , ∆ m x ( δ ) i,k = 0 .
8. Induction step: we will show that this statement is true for ( m + 1).We have to show the following statement to be true: ∀ k ∈ N , ∀ i ≤ , ∆ m +1 x ( δ ) i,k = 0 . By definition, ∆ m +1 x ( δ ) i,k = ∆ m x ( δ ) i,k − ∆ m x ( δ ) i − ,k .From the induction hypothesis, we get that ∆ m x ( δ ) i,k = 0 and ∆ m x ( δ ) i − ,k = 0 because, respectively, i ≤ i − ≤ − ≤ m +1 x ( δ ) i,k = 0. (cid:4) Lemma 3.2.
For repeated sequences of any order k , the difference of any order m of the firstterm of the sequence is equal to this initial term, ∀ k ∈ N , ∀ m ∈ N , ∆ m x ( δ )1 ,k = x ( δ )1 ,k . Proof.
1. Base case: verify true for m = 0. ∀ k ∈ N , ∆ x ( δ )1 ,k = x ( δ )1 ,k .
2. Induction hypothesis: assume the statement is true until m . ∀ k ∈ N , ∆ m x ( δ )1 ,k = x ( δ )1 ,k .
3. Induction step: we will show that this statement is true for ( m + 1).We have to show the following statement to be true: ∀ k ∈ N , ∆ m +1 x ( δ )1 ,k = x ( δ )1 ,k . By definition, ∆ m +1 x ( δ )1 ,k = ∆ m x ( δ )1 ,k − ∆ m x ( δ )0 ,k .From Lemma 3.1, ∆ m x ( δ )0 ,k = 0.Hence, ∆ m +1 x ( δ )1 ,k = ∆ m x ( δ )1 ,k .By applying the induction hypothesis, we get ∆ m +1 x ( δ )1 ,k = x ( δ )1 ,k . (cid:4) Remark.
From Lemma 3.2 for k = m , we see that the first term of a repeated sequence oforder m is equal to its m -th difference ( x ( δ )1 ,m = ∆ m x ( δ )1 ,m = δ ). Lemma 3.3.
For any m, k, n ∈ N where n ≥ , we have that n X N =1 ∆ m x ( δ ) N,k = ∆ m − x ( δ ) n,k . Proof.
1. Base case: verify true for n = 1.From Lemma 3.2, ∆ m x ( δ )1 ,k = ∆ m − x ( δ )1 ,k = x ( δ )1 ,k , X N =1 ∆ m x ( δ ) N,k = ∆ m x ( δ )1 ,k = ∆ m − x ( δ )1 ,k .
2. Induction hypothesis: assume the statement is true until n . n X N =1 ∆ m x ( δ ) N,k = ∆ m − x ( δ ) n,k .
9. Induction step: we will show that this statement is true for ( n + 1).We have to show the following statement to be true: n +1 X N =1 ∆ m x ( δ ) N,k = ∆ m − x ( δ ) n +1 ,k . n +1 X N =1 ∆ m x ( δ ) N,k = n X N =1 ∆ m x ( δ ) N,k + ∆ m x ( δ ) n +1 ,k . Applying the induction hypothesis to the first term and using the definition of the second term, n +1 X N =1 ∆ m x ( δ ) N,k = ∆ m − x ( δ ) n,k + h ∆ m − x ( δ ) n +1 ,k − ∆ m − x ( δ ) n,k i = ∆ m − x ( δ ) n +1 ,k . (cid:4) Lemma 3.4.
For any m, k, j, n ∈ N such that n ≥ and ≤ j ≤ m , we have that n X N j =1 · · · N X N =1 N X N =1 ∆ m x ( δ ) N ,k = ∆ m − j x ( δ ) n,k . Proof.
1. Base case: verify true for j = 1.Proven true in Lemma 3.3.2. Induction hypothesis: assume the statement is true until j . n X N j =1 · · · N X N =1 ∆ m x ( δ ) N ,k = ∆ m − j x ( δ ) n,k .
3. Induction step: we will show that this statement is true for ( j + 1).We have to show the following statement to be true: n X N j +1 =1 · · · N X N =1 ∆ m x ( δ ) N ,k = ∆ m − j − x ( δ ) n,k . Using the induction hypothesis, n X N j +1 =1 · · · N X N =1 ∆ m x ( δ ) N ,k = n X N j +1 =1 N j +1 X N j =1 · · · N X N =1 ∆ m x ( δ ) N ,k = n X N j +1 =1 ∆ m − j x ( δ ) N j +1 ,k . Applying Lemma 3.3, we get n X N j +1 =1 · · · N X N =1 ∆ m x ( δ ) N ,k = ∆ m − j − x ( δ ) n,k . (cid:4) A particular case of Lemma 3.4 for j = k = m is as follows. This theorem relates therepeated sequence to the repeated sum of ones.10 heorem 3.4. Any element of the m -th degree repeated sequence can be defined as an m -thorder repeated sum of ones, x ( δ ) n,m = ∆ m x ( δ ) n,m n X N m =1 · · · N X N =1 N X N =1 δ n X N m =1 · · · N X N =1 N X N =1 . Proof.
A particular case of Lemma 3.4 for j = k = m is as follows, n X N m =1 · · · N X N =1 N X N =1 ∆ m x ( δ ) N ,m = x ( δ ) n,m . By definition of a repeated sequence of degree m , ∆ m x ( δ ) n,m is a constant. Hence, we can take itout of the summation. Doing so give the desired theorem. (cid:4) For repeated unity sequences, Theorem 3.4 shows that these sequences are equal to repeatedsums of ones.
Corollary 3.2.
Any element of the m -th degree unity repeated sequence can be defined as an m -th order repeated sum of ones, ˆ x n,m = n X N m =1 · · · N X N =1 N X N =1 . Proof.
The unity repeated sequence is a repeated sequence where ∆ m ˆ x n,m = 1. Hence, substi-tuting into Theorem 3.4, we obtain this theorem. (cid:4) We now prove that repeated sequences can be defined as binomial coefficients.
Theorem 3.5.
For any m, n ∈ N where n ≥ , we have that x ( δ ) n,m = ∆ m x ( δ ) n,m (cid:18) n + m − m (cid:19) = δ (cid:18) n + m − m (cid:19) . Proof.
By applying Corollary 2.3 to Theorem 3.4, we obtain the theorem. (cid:4)
Corollary 3.3.
For any m, n ∈ N where n ≥ , we have that ˆ x n,m = (cid:18) n + m − m (cid:19) . Now that the needed expressions for repeated sequences were developed, one can proceedto develop formulae for the sum of the terms of such sequences.
Theorem 3.6.
For any k ∈ N ∗ and for any m, n ∈ N where n ≥ , we have that n X N k =1 · · · N X N =1 x ( δ ) N ,m = ∆ m x ( δ ) n,m n X N k + m =1 · · · N X N =1 m x ( δ ) n,m (cid:18) n + m + k − m + k (cid:19) = x ( δ ) n,m + k . roof. Applying Theorem 3.4, we get n X N k =1 · · · N X N =1 x ( δ ) N ,m = n X N k =1 · · · N X N =1 N X j m =1 · · · j X j =1 j X j =1 ∆ m x ( δ ) N ,m . Knowing that ∀ i, j ∈ N ∗ , ∆ m x ( δ ) i,m = ∆ m x ( δ ) j,m , hence, ∆ m x ( δ ) N ,m = ∆ m x ( δ ) n,m . n X N k =1 · · · N X N =1 x ( δ ) N ,m = ∆ m x ( δ ) n,m n X N k =1 · · · N X N =1 N X j m =1 · · · j X j =1 j X j =1 m x ( δ ) n,m n X N k + m =1 · · · N X N =1 . The first equality is proven. The second equality is obtained by applying Theorem 2.1.Knowing that ∆ m + k x ( δ ) n,m + k = ∆ m x ( δ ) n,m = δ , we apply Theorem 3.4 to get x ( δ ) n,m + k = ∆ m x ( δ ) n,m + k n X N m + k =1 · · · N X N =1 N X N =1 m x ( δ ) n,m n X N m + k =1 · · · N X N =1 N X N =1 . This completes the proof of the third equality. (cid:4)
Corollary 3.4.
For any k ∈ N ∗ and for any m, n ∈ N where n ≥ , we have that n X N k =1 · · · N X N =1 ˆ x N ,m = n X N k + m =1 · · · N X N =1 (cid:18) n + m + k − m + k (cid:19) = ˆ x n,m + k . Proof.
Applying Theorem 3.6 and noting that ∆ m ˆ x n,m = 1, we obtain this corollary. (cid:4) Example 3.4.
The sequence (ˆ x n, ) is a sequence of terms where the difference of two consec-utive terms is 1: 1 , , , , , . . . . Using Corollary 3.4, we calculate the repeated sum of order 3of its first 10 terms as follows: X N =1 N X N =1 N X N =1 ˆ x N , = (cid:18)
10 + 1 + 3 −
11 + 3 (cid:19) = (cid:18) (cid:19) = 715 . Example 3.5.
The sequence (ˆ x n, ) is a sequence of terms where the difference of two consecu-tive terms is increasing by 1: 1 , , , , , . . . . Using Corollary 3.4, we calculate the repeatedsum of order 3 of its first 10 terms as follows: X N =1 N X N =1 N X N =1 ˆ x N , = (cid:18)
10 + 2 + 3 −
12 + 3 (cid:19) = (cid:18) (cid:19) = 2002 .
4. Reduction of Binomial-Harmonic sums
The harmonic sum has been studied independently by Oresme [14], Mengoli [15], JohannBernoulli [16], Jacob Bernoulli [17, 18], and most importantly by Euler. A particularly inter-esting variant of this sum is that combining binomial coefficients with harmonic sums. We willrefer to such sums as binomial-harmonic sums . In this section, we develop various formulas forthe reduction and calculation of such sums. We also present a formula relating the repeatedharmonic sum to the shifted harmonic sum. 12 .1. Binomial-Harmonic sum
We begin by proving the following reduction formula to simplify Binomial-Harmonic sums.
Theorem 4.1.
For any m, p, n ∈ N such that n ≥ p , we have that n X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 1 m + 1 (cid:19) n + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i . Proof.
1. Base case: verify true for n = p . (cid:18) p + m + 1 m + 1 (cid:19) p + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i = (cid:18) p + m + 1 m + 1 (cid:19) p + m + 1 + (cid:20)(cid:18) p + m + 1 m + 1 (cid:19) − (cid:18) p + mm + 1 (cid:19)(cid:21) p + m X i =2+ m i . From the Pascal’s Triangle identity, (cid:0) p + m +1 m +1 (cid:1) − (cid:0) p + mm +1 (cid:1) = (cid:0) p + mm (cid:1) . In addiction, (cid:18) p + m + 1 m + 1 (cid:19) p + m + 1 = ( p + m + 1)!( m + 1)! p ! 1 p + m + 1 = ( p + m )! m ! p ! 1 m + 1 = (cid:18) p + mm (cid:19) m + 1 . Hence, substituting back, we get (cid:18) p + m + 1 m + 1 (cid:19) p + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i = (cid:18) p + mm (cid:19) m + 1 + (cid:18) p + mm (cid:19) p + m X i =2+ m i = (cid:18) p + mm (cid:19) p + m X i =1+ m i . Likewise, p X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) p + mm (cid:19) p + m X i =1+ m i .
2. Induction hypothesis: assume the statement is true until n . n X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 1 m + 1 (cid:19) n + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i .
3. Induction step: we will show that this statement is true for ( n + 1).We have to show the following statement to be true: n +1 X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 2 m + 1 (cid:19) n + m +2 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i . n +1 X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = n X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i + (cid:18) n + m + 1 m (cid:19) n + m +1 X i =1+ m i . n +1 X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 1 m + 1 (cid:19) n + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i + (cid:18) n + m + 1 m (cid:19) n + m +1 X i =1+ m i = (cid:20)(cid:18) n + m + 1 m + 1 (cid:19) + (cid:18) n + m + 1 m (cid:19)(cid:21) n + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i + (cid:18) n + m + 1 m (cid:19) m + 1 . From the Pascal’s Triangle identity, (cid:0) n + m +1 m +1 (cid:1) + (cid:0) n + m +1 m (cid:1) = (cid:0) n + m +2 m +1 (cid:1) . In addition, (cid:18) n + m + 1 m (cid:19) m + 1 = ( n + m + 1)! m ! ( n + 1)! 1 m + 1 = ( n + m + 2)!( m + 1)! ( n + 1)! 1 n + m + 2 = (cid:18) n + m + 2 m + 1 (cid:19) n + m + 2 . Hence, n +1 X N = p (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 2 m + 1 (cid:19) n + m +1 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i + (cid:18) n + m + 2 m + 1 (cid:19) n + m + 2= (cid:18) n + m + 2 m + 1 (cid:19) n + m +2 X i =2+ m i − (cid:18) p + mm + 1 (cid:19) p + m X i =2+ m i . (cid:4) Corollary 4.1.
For p = 1 , Theorem 4.1 becomes n X N =1 (cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + 1 m + 1 (cid:19) n + m +1 X i =2+ m i . Now we develop a formula for simplifying repeated Binomial-Harmonic sums.
Theorem 4.2.
For any k ∈ N ∗ and for any m, p, n ∈ N such that n ≥ p , we have that n X N k = p · · · N X N = p "(cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + km + k (cid:19) n + m + k X i =1+ m + k i − k − X j =0 (cid:18) n − p + jj (cid:19)(cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i . Proof.
1. Base case: verify true for k = 1.For k = 1 this theorem will reduce to Theorem 4.1. Hence, the case for k = 1 is proven.2. Induction hypothesis: assume the statement is true until k . n X N k = p · · · N X N = p "(cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + km + k (cid:19) n + m + k X i =1+ m + k i − k − X j =0 (cid:18) n − p + jj (cid:19)(cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i .
14. Induction step: we will show that this statement is true for ( k + 1).We have to show the following statement to be true: n X N k +1 = p · · · N X N = p "(cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + k + 1 m + k + 1 (cid:19) n + m + k +1 X i =2+ m + k i − k X j =0 (cid:18) n − p + jj (cid:19)(cid:18) p + m + k − jm + k − j + 1 (cid:19) p + m + k − j X i =2+ m + k − j i . Let us denote the first term of the equality as B. B = n X N k +1 = p ( N k +1 X N k = p · · · N X N = p "(cid:18) N + mm (cid:19) N + m X i =1+ m i . Applying the induction hypothesis, we get B = n X N k +1 = p (cid:18) N k +1 + m + km + k (cid:19) N k +1 + m + k X i =1+ m + k i − n X N k +1 = p k − X j =0 (cid:18) N k +1 − p + jj (cid:19)(cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i = n X N k +1 = p (cid:18) N k +1 + m + km + k (cid:19) N k +1 + m + k X i =1+ m + k i − k − X j =0 n X N k +1 = p (cid:18) N k +1 − p + jj (cid:19) (cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i . By applying Theorem 4.1 to the first part, we have that n X N k +1 = p (cid:18) N k +1 + m + km + k (cid:19) N k +1 + m + k X i =1+ m + k i = (cid:18) n + m + k + 1 m + k + 1 (cid:19) n + m + k +1 X i =2+ m + k i − (cid:18) p + m + km + k + 1 (cid:19) p + m + k X i =2+ m + k i . By applying Theorem 2.1 to the second part, we have that n X N k +1 = p (cid:18) N k +1 − p + jj (cid:19) = n − p + j X N k +1 = j (cid:18) N k +1 j (cid:19) = (cid:18) n − p + j + 1 j + 1 (cid:19) − (cid:18) jj + 1 (cid:19) = (cid:18) n − p + j + 1 j + 1 (cid:19) . B = (cid:18) n + m + k + 1 m + k + 1 (cid:19) n + m + k +1 X i =2+ m + k i − (cid:18) p + m + km + k + 1 (cid:19) p + m + k X i =2+ m + k i − k − X j =0 (cid:18) n − p + j + 1 j + 1 (cid:19)(cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i = (cid:18) n + m + k + 1 m + k + 1 (cid:19) n + m + k +1 X i =2+ m + k i − (cid:18) p + m + km + k + 1 (cid:19) p + m + k X i =2+ m + k i − k X j =1 (cid:18) n − p + jj (cid:19)(cid:18) p + m + k − jm + k − j + 1 (cid:19) p + m + k − j X i =2+ m + k − j i = (cid:18) n + m + k + 1 m + k + 1 (cid:19) n + m + k +1 X i =2+ m + k i − k X j =0 (cid:18) n − p + j + 1 j + 1 (cid:19)(cid:18) p − m + k − jm + k − j (cid:19) p − m + k − j X i =1+ m + k − j i . Hence, the theorem holds true for ( k + 1). Thus, the theorem is proven by induction. (cid:4) Corollary 4.2.
For p = 1 , Theorem 4.2 becomes n X N k =1 · · · N X N =1 "(cid:18) N + mm (cid:19) N + m X i =1+ m i = (cid:18) n + m + km + k (cid:19) n + m + k X i =1+ m + k i . A particular case of Corollary 4.2 which is of major interest is the repeated harmonic sum.
Corollary 4.3.
The repeated harmonic sum of order ( m + 1) is related to the shifted harmonicsum of order by the following relation, n X N m +1 =1 · · · N X N =1 N X N =1 N = (cid:18) n + mm (cid:19) n + m X N =1+ m N .
Proof.
Applying Corollary 4.2 with m = 0, we get this corollary. (cid:4)
5. Application to the computation of repeated sums
Let us denote a repeated sum of order m with lower and upper bounds respectively q and n as S m,q,n . In this section, we will present formulas to express the variation of a repeated sum oforder m ( S m,q,n +1 − S m,q,n ) in terms of lower order repeated sums. Equivalently, these formulascan be used to express S m,q,n +1 in terms of S m,q,n and lower order repeated sums.We begin by presenting the variation formula that allows S m,q,n +1 to be expressed in termsof S m,q,n and of repeated sums of order going from 1 to ( m − Theorem 5.1.
For any n, q ∈ N such that n ≥ q , and for any m ∈ N ∗ , we have n +1 X N m = q N m X N m − = q · · · N X N = q a N = m X k =1 n X N k = q N k X N k − = q · · · N X N = q a N + a n +1 . Or using the notation, S m,q,n +1 = m X k =1 ( S k,q,n ) + a n +1 . roof. Proven in [1]. (cid:4)
The variation of a repeated sum can also be expressed in terms of only a certain range oflower order repeated sums. In other words, S m,q,n +1 can be expressed in terms of S m,q,n and ofrepeated sums of order going only from p to ( m − Theorem 5.2.
For any n, q ∈ N such that n ≥ q , and for any m, p ∈ N ∗ , we have n +1 X N m = q N m X N m − = q · · · N X N = q a N = m X k = p +1 n X N k = q N k X N k − = q · · · N X N = q a N + n +1 X N p = q N p X N p − = q · · · N X N = q a N . Or using the notation, S m,q,n +1 = m X k = p +1 ( S k,q,n ) + S p,q,n +1 . Proof.
Proven in [1]. (cid:4)
Now that we have proven the needed binomial coefficient definitions and identities, we derivea formula to simplify the repeated sum of any sequence a N . Theorem 5.3.
For any m ∈ N ∗ , for any q, n ∈ N such that n ≥ q and for any sequence a N defined in the interval [ q, n ] , we have that n X N m = q · · · N X N = q N X N = q a N = n X i = q (cid:18) n + ( m − − im − (cid:19) a i . Proof.
Applying Theorem 3.3 from [1], for a ( m ); N m = · · · = a (2); N = 1 and a (1); N = a N , we get n X N m = q · · · N X N = q N X N = q a N = n X N = q a N n X N m = N · · · N X N = N N X N = N . By applying Corollary 2.4 to the ( m −
1) inner sums, we have n X N m = N · · · N X N = N N X N = N (cid:18) n − N + m − m − (cid:19) . Hence, substituting back, we get the desired theorem. (cid:4)
Theorem 5.3 offers a more computationally efficient way of calculating repeated sums ascomputing the repeated sum directly requires adding up (cid:0) n − q + mm (cid:1) terms while using the simpli-fied form requires adding up only n − q + 1 terms. For example, let us consider a repeated sumof order m = 10 and with lower and upper bounds respectively q = 1 and n = 10: computingthis repeated sum directly requires adding 92378 terms while using the theorem requires addingonly 10 terms. Remark.
From Theorem 5.3, we can deduce that a term a i appears (cid:0) n +( m − − im − (cid:1) times in therepeated sum of the sequence a N . Corollary 5.1.
If the sums start at , Theorem 5.3 becomes n X N m =1 · · · N X N =1 N X N =1 a N = n X i =1 (cid:18) n + ( m − − im − (cid:19) a i . .1.3. Application to harmonic sums and repeated harmonic sums In this section, we apply Theorem 5.3 to develop some formulae for harmonic sums andrepeated harmonic sums.First, we introduce a formula for simplifying repeated harmonic sums.
Theorem 5.4.
For any m, n ∈ N ∗ , we have that n X N m =1 · · · N X N =1 N X N =1 N = n X i =1 (cid:18) n + ( m − − im − (cid:19) i = (cid:18) n + m − m − (cid:19) n + m − X i = m i . Proof.
By equating the expression of the repeated harmonic sum obtained from Corollary 4.3with that obtained from Theorem 5.3 for a N = N , we obtain this theorem. (cid:4) Now we present identities related to the harmonic sum. we start with the following identity.
Theorem 5.5.
For any m, n ∈ N such that n ≥ , we have that n X i =1 (cid:0) ni (cid:1)(cid:0) n + mi (cid:1) i = n X i =1 " i − Y k =0 ( n − k )( n + m − k ) i = n + m X i =1+ m i . Proof.
From Theorem 5.4 with m substituted by ( m + 1), n X i =1 (cid:18) n + m − im (cid:19) i = (cid:18) n + mm (cid:19) n + m X i =1+ m i . Knowing that (cid:0) n + m − im (cid:1)(cid:0) n + mm (cid:1) = (cid:0) ni (cid:1)(cid:0) n + mi (cid:1) = i − Y k =0 ( n − k )( n + m − k ) , hence, substituting back, we obtain the theorem. (cid:4) In order to prove the second identity, we need to first prove the following two lemmas.
Lemma 5.1.
For any k ∈ N ∗ , we have that k X i =1 i = (2 k + 1) k X i =1 k + 1 − i ) i = (2 k + 1)2 k X i =1 k + 1 − i ) i . Proof.
We split the sum, modify its parts, then recombine them to obtain, k X i =1 i = k X i =1 i + k X i = k +1 i = k X i =1 i + k X i =1 k + 1 − i = (2 k + 1) k X i =1 k + 1 − i ) i . Similarly,(2 k + 1)2 k X i =1 k + 1 − i ) i = (2 k + 1)2 " k X i =1 k + 1 − i ) i + k X i = k +1 k + 1 − i ) i = (2 k + 1)2 " k X i =1 k + 1 − i ) i + k X i =1 k + 1 − i ) i = (2 k + 1) k X i =1 k + 1 − i ) i . Equating, we obtain the lemma. (cid:4) emma 5.2. For any k ∈ N ∗ , we have that k − X i =1 i = (2 k ) k X i =1 k − i ) i − k = (2 k ) k − X i =1 k − i ) i + 1 k = ( k ) k − X i =1 k − i ) i . Proof.
We split the sum, modify its parts, then recombine them to obtain, k − X i =1 i = k X i =1 i + k − X i = k i − k = k X i =1 i + k X i =1 k − i − k = (2 k ) k X i =1 k − i ) i − k . Similarly, k − X i =1 i = k − X i =1 i + k − X i = k +1 i + 1 k = k − X i =1 i + k − X i =1 k − i + 1 k = (2 k ) k − X i =1 k − i ) i + 1 k . Finally, to obtain the last part of this lemma, we perform the following,( k ) k − X i =1 k − i ) i = ( k ) " k − X i =1 k − i ) i + 1 k + k − X i = k +1 k − i ) i = ( k ) " k − X i =1 k − i ) i + k − X i =1 k − i ) i + 1 k = (2 k ) k − X i =1 k − i ) i + 1 k . Equating, we obtain the lemma. (cid:4)
Now that both needed lemmas have been proven, we give the second identity.
Theorem 5.6.
For any n ∈ N such that n ≥ , we have that n + 12 n X i =1 n + 1 − i ) i = n X i =1 i . Proof.
For this proof, a proof by case will be performed. Thus, the proof will be divided intoproving two complementary cases: the case when n is even and the case when n is odd. • For n = 2 k (where k ∈ N ∗ ):This case corresponds to Lemma 5.1 that we have already proven. • For n = 2 k − k ∈ N ∗ ):This case corresponds to Lemma 5.2 that we have already proven.The formula holds true for both even and odd values of n , hence, it holds for any n ≥ (cid:4) In this section, we prove a generalization of Theorem 5.3. This generalization is illustratedin the following theorem. The following theorem simplifies the repeated “Binomial-Sequence”sum. 19 heorem 5.7.
For any k ∈ N ∗ , for any m, q, n ∈ N where n ≥ q and for any sequence a N defined in the interval [ q, n ] , we have that n X N k = q · · · N X N = q N X N = q (cid:18) ( N − N ) + mm (cid:19) a N = n X N = q (cid:18) ( n − N ) + m + k − m + k − (cid:19) a N . Proof.
By applying Theorem 5.3 to the inner sum, we have n X N k = q · · · N X N = q N X N = q (cid:18) ( N − N ) + mm (cid:19) a N = n X N k = q · · · N X N = q N X j m +1 = q · · · j X j = q a j = n X i m + k = q · · · i X i = q i X i = q a i = n X N = q (cid:18) ( n − N ) + m + k − m + k − (cid:19) a N (cid:4) Corollary 5.2.
If the summations start at , Theorem 5.7 becomes, n X N k =1 · · · N X N =1 (cid:18) ( N − N ) + mm (cid:19) a N = n X N =1 (cid:18) ( n − N ) + m + k − m + k − (cid:19) a N . A particular case of Theorem 5.7 is as follows.
Theorem 5.8.
For any m ∈ N ∗ , for any q, n ∈ N where n ≥ q and for any sequence a N definedin the interval [ q, n ] , we have that n X N = q N X N = q (cid:18) ( N − N ) + ( m − m − (cid:19) a N = n X N = q (cid:18) ( n − N ) + mm (cid:19) a N . Corollary 5.3.
If the summations start at , Theorem 5.8 becomes, n X N =1 N X N =1 (cid:18) ( N − N ) + ( m − m − (cid:19) a N = n X N =1 (cid:18) ( n − N ) + mm (cid:19) a N . References [1] R. E. Haddad, Recurrent sums and partition identities (2021). arXiv:2101.09089 .[2] R. E. Haddad, Multiple sums and partition identities (2021). arXiv:2102.00821arXiv:2102.00821