Harmonic Number Identities Via Euler’s Transform

Khristo N. Boyadzhiev Department of Mathematics Ohio Northern University Ada, Ohio 45810 USA

                     k-boyadzhiev@onu.edu

Abstract We evaluate several binomial transforms by using Euler’s transform for power series. In this way we obtain various binomial identities involving power sums with harmonic numbers.

1 Introduction and prerequisites

Given a sequence {a_{k}}, its binomial transform {b_{k}} is the sequence defined by

b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}, with inversion a_{n} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{n−k}b_{k},

or, in the symmetric version

b_{n} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{k+1}a_{k} with inversion a_{n} = ∑_{k=0}^{n} \binom{n}{k}(−1)^{k+1}b_{k}

(see [7, 12, 14]). The binomial transform is related to the Euler transform of series defined in the following lemma. Euler’s transform is used sometimes for improving the convergence of certain series [1, 8, 12, 13]. Lemma 1. Given a function analytical on the unit disk

f(t) = ∑_{n=0}^{∞} a_{n}t^{n},  (1)

then the following representation is true

\frac{1}{1 − t} f(\frac{t}{1 − t}) = ∑_{n=0}^{∞} t^{n} (∑_{k=0}^{n} \binom{n}{k}a_{k}) .  (2)

(Proof can be found in the Appendix.)

If we have a convergent series

s = ∑_{n=0}^{∞} a_{n},  (3)

we can define the function

f(t) = ∑_{n=0}^{∞} a_{n}t^{n}, |t| < 1.  (4)

Then, with t = \frac{1}{2} in (2) we obtain

s = ∑_{n=0}^{∞} (∑_{k=0}^{n} \binom{n}{k}a_{k}) \frac{1}{2^{n+1}} .  (5)

This formula is a classical version of Euler’s series transformation. Sometimes the new series converges faster, sometimes not – see the examples in [10].

We shall use Euler’s transform for the evaluation of several interesting binomial transformations, thus obtaining binomial identities of combinatorial and analytical character. Evaluating a binomial transform is reduced to finding the Taylor coefficients of the function on the left hand side of (2). In Section 2 we obtain several identities with harmonic numbers. In Section 3 we prove Dilcher’s formula via Euler’s transform.

This paper is close in spirit to the classical article [7] of Henry Gould.

Remark 2. The representation (2) can be put in a more flexible equivalent form

\frac{1}{1 − λt} f(\frac{μt}{1 − λt}) = ∑_{n=0}^{∞} t^{n} (∑_{k=0}^{n} \binom{n}{k}μ^{k}λ^{n−k}a_{k}) ,  (6)

where λ, μ are appropriate parameters.

To show the equivalence of (2) and (6) we first write

f(\frac{μt}{λ}) = ∑_{n=0}^{∞} a_{n} (\frac{μ}{λ})^{n} t^{n},  (7)

and then apply (2) to the function g(t) = f(\frac{μ}{λ}t). This provides

\frac{1}{1 − t} f(\frac{μ}{λ} \frac{t}{1 − t}) = ∑_{n=0}^{∞} t^{n} (∑_{k=0}^{n} \binom{n}{k} (\frac{μ}{λ})^{k} a_{k}) .  (8)

Replacing here t by λt yields (6).

With λ = 1 and μ = −1 we have

\frac{1}{t − 1} f(\frac{t}{t − 1}) = ∑_{n=0}^{∞} t^{n} (∑_{k=0}^{n} \binom{n}{k}(−1)^{k+1}a_{k}) ,  (9)

corresponding to the symmetrical binomial transform. Lemma 3. Given a formal power series

g(t) = ∑_{n=0}^{∞} b_{n}t^{n},  (10)

we have

\frac{g(t)}{1 − t} = ∑_{n=0}^{∞} (∑_{k=0}^{n} b_{k}) t^{n}.  (11)

This is a well-known property. To prove it we just need to multiply both sides of (11) by 1 − t and simplify the right hand side.

2 Identities with harmonic numbers

Proposition 4. The following expansion holds in a neighborhood of zero

\frac{log(1 − αt)}{1 − βt} = − ∑_{n=1}^{∞} (αβ^{n−1} + \frac{1}{2}α^{2}β^{n−2} + ⋯ + \frac{1}{n}α^{n}) t^{n}  (12)

where α, β are appropriate parameters. Proof. It is sufficient to prove (12) when β = 1 and then rescale the variable t, i.e. we only need

\frac{log(1 − αt)}{1 − t} = − ∑_{n=1}^{∞} (α + \frac{1}{2}α^{2} + ⋯ + \frac{1}{n}α^{n}) t^{n}.  (13)

This follows immediately from Lemma 3. Corollary 5. With α = 1 in (13) we obtain the generating function of the harmonic numbers

−\frac{log(1 − t)}{1 − t} = ∑_{n=0}^{∞} H_{n}t^{n},  H_{n} = 1 + \frac{1}{2} + ⋯ + \frac{1}{n}.  (14)

The next proposition is one of our main results Proposition 6. For every positive integer n and every two complex numbersλ, μ,

∑_{k=1}^{n} \binom{n}{k}H_{k}λ^{n−k}μ^{k} = H_{n}(λ + μ)^{n} − (λ(λ + μ)^{n−1} + \frac{λ^{2}}{2} (λ + μ)^{n−2} + ⋯ + \frac{λ^{n}}{n} ) . (15)

Proof. We apply (6) to the function

f(t) = −\frac{log(1 − t)}{1 − t} = ∑_{n=0}^{∞} H_{n}t^{n}.  (16)

On the left hand side we obtain

\frac{−1}{1 − λt} \frac{log(1 − \frac{μt}{1−λt})}{1 − \frac{μt}{1−λt}} = −\frac{log(1 − (λ + μ)t)}{1 − (λ + μ)t} + \frac{log(1 − λt)}{1 − (λ + μ)t},  (17)

which equals, according to Corollary 5 and Proposition 4,

∑_{n=1}^{∞} H_{n}(λ + μ)^{n}t^{n} − ∑_{n=1}^{∞} (λ(λ + μ)^{n−1} + \frac{λ^{2}}{2} (λ + μ)^{n−2} + ⋯ + \frac{λ^{n}}{n} ) t^{n}.  (18)

At the same time, by Euler’s transform the right hand side is

∑_{n=1}^{∞} t^{n} (∑_{k=1}^{n} \binom{n}{k}H_{n}λ^{n−k}μ^{k}) .  (19)

Comparing coefficients in (18) and (19) we obtain the desired result. Corollary 7. Setting λ = μ = 1 in (15) yields the well-known identity (see, for instance, [6, 14]):

∑_{k=1}^{n} \binom{n}{k}H_{k} = 2^{n} (H_{n} − ∑_{k=1}^{n} \frac{1}{k2^{k}} ) .  (20)

Corollary 8. Setting λ = 1 in (15) reduces it to

∑_{k=1}^{n} \binom{n}{k}H_{k}μ^{k} = H_{n}(1 + μ)^{n} − ((1 + μ)^{n−1} + \frac{(1 + μ)^{n−2}}{2} + ⋯ + \frac{1 + μ}{n − 1} + \frac{1}{n}) . (21)

We shall use this last identity to obtain a representation for the combinatorial sum

∑_{k=1}^{n} \binom{n}{k}H_{k}k^{m}μ^{k},  (22)

by applying the operator (μ\frac{d}{dμ})^{m} to both sides in (21). First, however, we need the following lemma. Lemma 9. For every positive integer m define the quantities

a(m, n, μ) = (μ\frac{d}{dμ})^{m} (1 + μ)^{n} = ∑_{k=0}^{n} \binom{n}{k}k^{m}μ^{k}.  (23)

Then

a(m, n, μ) = ∑_{k=0}^{n} \binom{n}{k}k!S(m, k)μ^{k}(1 + μ)^{n−k}.  (24)

This is a known identity that can be found, for example, in [6].

From Lemma 9 we obtain another of our main results.

Proposition 10. For every two positive integers m and n,

∑_{k=1}^{n} \binom{n}{k}H_{k}k^{m}μ^{k} = a(m, n, μ)H_{n} − ∑_{p=1}^{n−1} \frac{1}{n − p}a(m, p, μ).  (25)

Proof. Apply (μ\frac{d}{dμ})^{m} to both sides of (21) and note that (μ\frac{d}{dμ})^{m}μ^{k} = k^{m}μ^{k}.

The sums (22) were recently studied by M. Coffey [3] by using a different method (a recursive formula) and a representation was given in terms of the hypergeometric function

3 Stirling functions of a negative argument. Dilcher’s formula

Some time ago Karl Dilcher obtained the nice identity

∑_{k=1}^{n} \binom{n}{k}\frac{(−1)^{k−1}}{k^{m}} = ∑ \frac{1}{j_{1}j_{2} ⋯ j_{m}} , 1 ≤ j_{1} ≤ j_{2} ≤ ⋯ ≤ j_{m} ≤ n,  (26)

as a corollary from a certain multiple series representation [4, Corollary 3]; see also a similar result in [5]. As this is one binomial transform, it is good to have a direct proof by Euler’s transform method. Before giving such a proof, however, we want to point out one interesting interpretation of the sum on the left hand side in (26).

Let S(m, n) be the Stirling numbers of the second kind [9]. Butzer et al. [2] defined an extension S(α, n) for any complex number α ≠ 0. The functions S(α, n) of the complex variable α are called Stirling functions of the second kind. The extension is given by the formula

S(α, n) = \frac{1}{n!} ∑_{k=1}^{n} \binom{n}{k}(−1)^{n−k}k^{α},  (27)

with S(α, 0) = 0. Thus, for m, n ≥ 1,

(−1)^{n−1}n!S(−m, n) = ∑_{k=1}^{n} \binom{n}{k}\frac{(−1)^{k−1}}{k^{m}} .  (28)

For the next proposition we shall need the polylogarithmic function [11]

Li_{m}(t) = ∑_{n=1}^{∞} \frac{t^{n}}{n^{m}} .  (29)

Proposition 11. For any integer m ≥ 1 we have

(−1)^{n−1}n!S(−m, n) = ∑ \frac{1}{j_{1}j_{2} ⋯ j_{m}} , 1 ≤ j_{1} ≤ j_{2} ≤ ⋯ ≤ j_{m} ≤ n.  (30)

Proof. The proof is based on the representation

Li_{m}(\frac{−t}{1 − t}) = − ∑ \frac{t^{j_{m}}}{j_{1}j_{2} ⋯ j_{m}} , 1 ≤ j_{1} ≤ j_{2} ≤ ⋯ ≤ j_{m},  (31)

(see [15]) from which, in view of Lemma 2,

\frac{−1}{1 − t} Li_{m}(\frac{−t}{1 − t}) = ∑_{n=1}^{∞} A_{n}t^{n},  (32)

with coefficients

A_{n} = ∑ \frac{1}{j_{1}j_{2} ⋯ j_{m}} , 1 ≤ j_{1} ≤ j_{2} ≤ ⋯ ≤ j_{m} ≤ n.  (33)

The assertion now follows from (9).

In conclusion, many thanks to the referee for a correction and for some interesting comments.

4 Appendix

We prove Euler’s transform representation (2) by using Cauchy’s integral formula, both for the Taylor coefficients of a holomorphic function and for the function itself. Thus, given a holomorphic function f as in (1), we have

a_{k} = \frac{1}{2πi} ∮_{L} \frac{1}{λ^{k}} \frac{f(λ)}{λ}dλ,  (34)

for an appropriate closed curve L around the origin. Multiplying both sides by \binom{n}{k} and summing for k we find

∑_{k=0}^{n} \binom{n}{k}a_{k} = \frac{1}{2πi} ∮_{L} (∑_{k=0}^{n} \binom{n}{k}\frac{1}{λ^{k}} ) \frac{f(λ)}{λ}dλ = \frac{1}{2πi} ∮_{L} (1 + \frac{1}{λ})^{n} \frac{f(λ)}{λ}dλ.  (35)

Multiplying this by t^{n} (with t small enough) and summing for n we arrive at the desired representation (2), because

∑_{n=0}^{∞} t^{n} (1 + \frac{1}{λ})^{n} = \frac{1}{1 − t(1 + \frac{1}{λ})} = \frac{1}{1 − t} \frac{λ}{λ − \frac{t}{1−t}} ,  (36)

and therefore,

∑_{n=0}^{∞} t^{n} (∑_{k=0}^{n} \binom{n}{k}a_{k}) = \frac{1}{1 − t} \frac{1}{2πi} ∮_{L} \frac{f(λ)}{λ − \frac{t}{1−t}} dλ = \frac{1}{1 − t} f(\frac{t}{1 − t}) .  (37)

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2000 Mathematics Subject Classification: Primary 05A20; Secondary 11B73. Keywords: Harmonic numbers, binomial transform, Euler transform, Stirling number of the second kind, Stirling function, combinatorial identity, polylogarithm, Dilcher’s formula.

Received June 12 2009; revised version received July 25 2009. Published in Journal of Integer Sequences, August 30 2009.

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