Harmonic Sums via Euler’s Transform: Complementing the Approach of Boyadzhiev

Robert Frontczak^{1} Landesbank Baden-Württemberg (LBBW) Am Hauptbahnhof 2 70173 Stuttgart Germany

               robert.frontczak@lbbw.de

Abstract

We prove a new expression for binomial sums with harmonic numbers. Our derivation is based on an alternative argument for the Euler transform of these sums. The findings complement a result of Boyadzhiev. To demonstrate the usefulness of our alternative approach, several examples are discussed. We rediscover some known identities for harmonic numbers and present some new ones. In particular, we derive some new identities involving harmonic numbers, and Fibonacci and Lucas numbers.

1 Motivation

Harmonic numbers (H_{n})_{n≥0} are defined by

H_{0} = 0 and for n ≥ 1 : H_{n} = ∑_{k=1}^{n} \frac{1}{k} = H_{n−1} + \frac{1}{n}.

They have the following integral form

H_{n} = ∫_{0}^{1} \frac{1 − x^{n}}{1 − x} dx.

1Disclaimer: Statements and conclusions made in this article are entirely those of the author. They do not necessarily reflect the views of LBBW.

Harmonic numbers and generalized harmonic numbers have been studied recently by many mathematicians and a considerable amount of research results has been produced (see [2] - [13], to name a few articles). In 2009, Boyadzhiev [2] studied binomial sums with harmonic numbers using the Euler transform. His main result is the following identity valid for n ≥ 1

∑_{k=1}^{n} \binom{n}{k}a^{k}b^{n−k}H_{k} = (a + b)^{n}H_{n} − (b(a + b)^{n−1} + \frac{b^{2}}{2} (a + b)^{n−2} + ⋯ + \frac{b^{n}}{n} ),  (1)

where a and b are arbitrary complex numbers. His proof is based on the Euler transform for power series and the existence of a power series near zero of the form

\frac{ln(1 − cz)}{1 − dz} = − ∑_{n=1}^{∞} (cd^{n−1} + \frac{1}{2}c^{2}d^{n−2} + ⋯ + \frac{1}{n}c^{n})z^{n}.

In this article, we show how Boyadzhiev’s arguments may be modified to derive an alternative expression for the binomial sums on the left-hand side of (1). To demonstrate the usefulness of our alternative approach, several examples will be discussed. We will rediscover some known identities and present some new. In particular, we will derive some new identities involving harmonic numbers, and Fibonacci and Lucas numbers.

2 Results

Let A(z) be the ordinary generating function for the harmonic numbers, i.e.,

A(z) = ∑_{n=0}^{∞} H_{n}z^{n} = −\frac{ln(1 − z)}{1 − z}.

For a, b ∈ ℂ, let further S_{n}(a, b) be defined as

S_{n}(a, b) = ∑_{k=0}^{n} \binom{n}{k}a^{k}b^{n−k}H_{k}.  (2)

Then we have the following theorem.

Theorem 1. For all n ≥ 1, we have the identity

S_{n}(a, b) = ((a + b)^{n} − b^{n})H_{n} − a ∑_{k=0}^{n−1}(a + b)^{k}b^{n−1−k}H_{n−1−k}.  (3)

Proof. Let S(z) be the ordinary generating function for the sum S_{n}(a, b). Then, by Euler’s transform

S(z) = ∑_{n=0}^{∞} S_{n}(a, b)z^{n} = \frac{1}{1 − bz} A(\frac{az}{1 − bz} )
= −\frac{ln(1 − (a + b)z)}{1 − (a + b)z} + \frac{ln(1 − bz)}{1 − (a + b)z} .

Now, instead of searching for a power series for the second summand, we observe that

\frac{ln(1 − bz)}{1 − (a + b)z} = \frac{az}{1 − (a + b)z} \frac{ln(1 − bz)}{1 − bz} + \frac{ln(1 − bz)}{1 − bz}.

Hence,

S(z) = −\frac{ln(1 − (a + b)z)}{1 − (a + b)z} − ( − \frac{ln(1 − bz)}{1 − bz}) − \frac{az}{1 − (a + b)z} ( − \frac{ln(1 − bz)}{1 − bz})
= ∑_{n=0}^{∞} ((a + b)^{n} − b^{n})H_{n}z^{n} − a( ∑_{n=0}^{∞}(a + b)^{n}z^{n+1})( ∑_{n=0}^{∞} b^{n}H_{n}z^{n}).

Using Cauchy’s product rule for power series and comparing the coefficients of z^{n} gives the result.

For (a; b) = (−1; 1) we get as a special case

S_{n}(−1, 1) = ∑_{k=0}^{n} \binom{n}{k}(−1)^{k}H_{k} = −H_{n} + H_{n−1} = −\frac{1}{n},  (4)

which is an old result and has reappeared as Identity 20 in Spivey’s paper [12]. In addition, using the integral form for H_{n} it is not difficult to show that

∑_{k=1}^{n} \binom{n}{k}(−1)^{k−1} \frac{1}{k} = H_{n}.  (5)

This shows that the sequences (H_{n})_{n≥1} and (1/n)_{n≥1} are connected by the binomial transform. More information about the binomial transform can be found in the book [1].

Corollary 2. For n ≥ 1, the harmonic numbers allow the representation

H_{n} = ∑_{k=1}^{n} (\frac{1}{k}2^{n−k} − 2^{k−1}H_{n−k}).  (6)

Proof. Setting (a; b) = (a; a) in (3) yields

∑_{k=0}^{n} \binom{n}{k}H_{k} = (2^{n} − 1)H_{n} − ∑_{k=0}^{n−1} 2^{k}H_{n−1−k}.

Now, compare with the well-known identity ([2, Equation (20)], or [12, Identity 14])

∑_{k=0}^{n} \binom{n}{k}H_{k} = 2^{n}(H_{n} − ∑_{k=1}^{n} \frac{1}{k2^{k}} ).

Corollary 3. For n ≥ 1, it is true that

∑_{k=0}^{n} \binom{n}{k}2^{k}H_{k} = (3^{n} − 1)H_{n} − 2 ∑_{k=0}^{n−1} 3^{k}H_{n−1−k}.  (7)

Proof. Set (a; b) = (2; 1) in (3).

Corollary 4. For n ≥ 1, we have the identity

∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}2^{k−1}H_{k} = \begin{cases}∑_{k=0}^{n−1}(−1)^{k}H_{n−1−k}, & if n is even; \\ H_{n} − ∑_{k=0}^{n−1}(−1)^{k}H_{n−1−k}, & if n is odd. \end{cases}  (8)

Proof. Setting (a; b) = (2; −1) in (3) yields

∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}2^{k}H_{k} = (1 − (−1)^{n})H_{n} − 2 ∑_{k=0}^{n−1}(−1)^{n−1−k}H_{n−1−k},

from which the result is deduced easily.

Theorem 1 also allows to establish some identities involving harmonic numbers and Fibonacci (Lucas) numbers. Recall, that Fibonacci numbers (F_{n})_{n≥0} and Lucas numbers (L_{n})_{n≥} are defined by F_{0} = 0, F_{1} = 1, L_{0} = 2, L_{1} = 1, and for n ≥ 2 we have the recurrences F_{n} = F_{n−1} + F_{n−2} and L_{n} = L_{n−1} + L_{n−2}.

Corollary 5. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers, respectively. Then, we have the relations

∑_{k=0}^{n} \binom{n}{k}F_{k}H_{k} = F_{2n}H_{n} − ∑_{k=0}^{n−1} F_{2k+1}H_{n−1−k},  (9)

and

∑_{k=0}^{n} \binom{n}{k}L_{k}H_{k} = (L_{2n} − 2)H_{n} − ∑_{k=0}^{n−1} L_{2k+1}H_{n−1−k}.  (10)

Proof. Evaluate (3) at (a; b) = (α; 1) and (a; b) = (β; 1), respectively, where α = (1 + \sqrt{5})/2 and β = −1/α. This gives

S_{n}(α, 1) = (α^{2n} − 1)H_{n} − ∑_{k=0}^{n−1} α^{2k+1}H_{n−1−k}

and

S_{n}(β, 1) = (β^{2n} − 1)H_{n} − ∑_{k=0}^{n−1} β^{2k+1}H_{n−1−k},

where we have used the relations α^{2} = α+1 and β^{2} = β+1. Now, calculate S_{n}(α, 1)±S_{n}(β, 1) and use the Binet forms for F_{n} and L_{n}, respectively.

Corollary 6. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers, respectively. Then, the following identities hold

∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}F_{2k}H_{k} = F_{n}H_{n} − ∑_{k=0}^{n−1}(−1)^{n−1−k}F_{k+2}H_{n−1−k},  (11)

and

∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}L_{2k}H_{k} = (L_{n} − 2(−1)^{n})H_{n} − ∑_{k=0}^{n−1}(−1)^{n−1−k}L_{k+2}H_{n−1−k}.  (12)

Proof. Evaluate (3) at (a; b) = (α^{2}; −1) and (a; b) = (β^{2}; −1), respectively. Combine the results as in the previous proof.

Corollary 7. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers, respectively. Then

∑_{k=0}^{n} \binom{n}{k}2^{k}(−1)^{n−k}F_{n−k}H_{k} = −(F_{2n} + (−1)^{n}F_{n})H_{n} + ∑_{k=0}^{n−1} F_{3k+1−n}H_{n−1−k},  (13)

and

∑_{k=0}^{n} \binom{n}{k}2^{k}(−1)^{n−k}L_{n−k}H_{k} = (L_{2n} − (−1)^{n}L_{n})H_{n} − ∑_{k=0}^{n−1} L_{3k+1−n}H_{n−1−k}.  (14)

Proof. Evaluate (3) at (a; b) = (2; −α) and (a; b) = (2; −β), respectively. Simplify using 2 − α = α^{−2} and 2 − β = α^{2}. Finally, calculate S_{n}(2, −α) ± S_{n}(2, −β) and keep in mind that F_{−n} = (−1)^{n+1}F_{n} and L_{−n} = (−1)^{n}L_{n}, respectively.

3 Harmonic sums with integer powers

Boyadzhiev [2, Proposition 10] obtained a representation for the combinatorial sum

S_{n}(a, 1, m) = S_{n}(a, m) = ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k}H_{k}, m ≥ 1,  (15)

in terms of Stirling numbers of the second kind S(m, k). The theorem below contains an alternative expression for the sum.

Theorem 8. For all n ≥ 1 we have

S_{n}(a, m) = ( ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k})H_{n} − ∑_{j=0}^{n−1} ∑_{k=j}^{n−1} \binom{k}{j}(j + 1)^{m}a^{j+1}H_{n−1−k}.  (16)

Proof. Apply the differential operator (a\frac{d}{da})^{m} to both sides of (3) with S_{n}(a, b) = S_{n}(a, 1). Note that

(a\frac{d}{da})^{m}(a + 1)^{n} = ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k}.

Remark 9. The Stirling numbers of the second kind S(n, k) are defined by

x^{n} = ∑_{k=0}^{n} S(n, k)(x)_{k},

with x_{0} = 1 and (x)_{n} = x(x − 1) ⋯ (x − n + 1), n ≥ 1, being the falling factorial. From the equation

(x + 1)_{n} = (x)_{n} + n(x)_{n−1},

it becomes clear, that we can restate the above result using Stirling numbers.

Corollary 10. For n ≥ 1, we have the following expression

∑_{k=0}^{n} \binom{n}{k}kH_{k} = n2^{n−1}H_{n} − ∑_{k=0}^{n−1}(k + 2)2^{k−1}H_{n−1−k}.  (17)

Proof. Use (a; m) = (1; 1) in (16) as well as the obvious identities

∑_{k=0}^{n} \binom{n}{k} = 2^{n} and  ∑_{k=0}^{n} \binom{n}{k}k = n2^{n−1}.

Corollary 11. For n ≥ 1, we have the identity

∑_{k=0}^{n} \binom{n}{k}(−1)^{k}kH_{k} = \begin{cases}−1, & if n = 1; \\ \frac{1}{n−1}, & if n ≥ 2. \end{cases}  (18)

Proof. Evaluate (16) at (a; m) = (−1; 1) and simplify.

Corollary 12. For n ≥ 1, we have the identity

∑_{k=j}^{n−1} \binom{k}{j}H_{n−1−k} = \binom{n}{j + 1}(H_{n} − H_{j+1}).  (19)

Proof. We have

S_{n}(a, m) = ( ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k})H_{n} − ∑_{j=0}^{n} \binom{n}{j}j^{m}a^{j}(H_{n} − H_{j})
= ( ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k})H_{n} − ∑_{j=1}^{n} \binom{n}{j}j^{m}a^{j}(H_{n} − H_{j})
= ( ∑_{k=0}^{n} \binom{n}{k}k^{m}a^{k})H_{n} − ∑_{j=0}^{n−1} \binom{n}{j + 1}(j + 1)^{m}a^{j+1}(H_{n} − H_{j+1}).

Comparing with (16) gives the result. Corollary 13. Let F_{n} and L_{n} be the Fibonacci and Lucas numbers, respectively. Then

∑_{k=0}^{n} \binom{n}{k}kF_{k}H_{k} = nF_{2n−1}H_{n} − ∑_{k=0}^{n−1}(kF_{2k} + F_{2k+1})H_{n−1−k},  (20)

and

∑_{k=0}^{n} \binom{n}{k}kL_{k}H_{k} = nL_{2n−1}H_{n} − ∑_{k=0}^{n−1}(kL_{2k} + L_{2k+1})H_{n−1−k}.  (21)

Proof. Evaluate (16) at (a; m) = (α; 1) and (a; m) = (β; 1), respectively, using

∑_{k=1}^{n} \binom{n}{k}kx^{k} = nx(1 + x)^{n−1},

in combination with

∑_{k=0}^{n} \binom{n}{k}α^{k} = α^{2n}, and ∑_{k=0}^{n} \binom{n}{k}β^{k} = β^{2n}.

Combine the sums as in the previous proofs.

Using similar elementary arguments we can prove the following identities for m = 2: Corollary 14. For n ≥ 1, it is true that

∑_{k=0}^{n} \binom{n}{k}k^{2}H_{k} = n(n + 1)2^{n−2}H_{n} − ∑_{k=0}^{n−1}(k + 1)(k + 4)2^{k−2}H_{n−1−k}.  (22)

Corollary 15. For n ≥ 1, we have the identity

∑_{k=0}^{n} \binom{n}{k}(−1)^{k}k^{2}H_{k} = \begin{cases}−1, & if n = 1; \\ 4, & if n = 2; \\ −\frac{n}{(n−2)(n−1)}, & if n ≥ 3. \end{cases}  (23)

Corollary 16. For n ≥ 1, the following relations hold:

∑_{k=0}^{n} \binom{n}{k}k^{2}F_{k}H_{k} = (nF_{2n−3} + n^{2}F_{2n−2})H_{n}

− ∑_{k=0}^{n−1} (kF_{2k−2} + k^{2}F_{2k−1} + 2kF_{2k} + F_{2k+1})H_{n−1−k},  (24)

and

∑_{k=0}^{n} \binom{n}{k}k^{2}L_{k}H_{k} = (nL_{2n−3} + n^{2}L_{2n−2})H_{n}

− ∑_{k=0}^{n−1} (kL_{2k−2} + k^{2}L_{2k−1} + 2kL_{2k} + L_{2k+1})H_{n−1−k}.  (25)

4 Concluding comments

Using the main results of this paper, it is possible to derive more identities involving harmonic numbers and Fibonacci (Lucas) numbers. In addition, we mention that from Theorems 1 and 8 relations between harmonic numbers and other important number sequences, such as Mersenne numbers or Pell numbers, are deducible.

5 Acknowledgments

The author is thankful to K. N. Boyadzhiev and T. Goy for their remarks on the first draft of the manuscript. He is also grateful to the anonymous referee and the editor-in-chief for valuable suggestions which improved the quality of the presentation.

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2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39, 11B65, 05A15. Keywords: harmonic number, binomial sum, Fibonacci number, generating function.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A001008}, and \underline{A002805}.)

Received October 20 2019; revised version received January 21 2020. Published in Journal of Integer Sequences, February 23 2020.

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