A generalized alternating harmonic series

1.   Introduction

Riemann's Rearrangement Theorem says that if an infinite series of real numbers is conditionally convergent, then its terms can be rearranged in in such a way that the resulting series converges to any real sum ^[3]. It is well known that the alternating harmonic series converges to $\log 2$, that is,

As the prototypical conditionally convergent series, rearrangements of the alternating harmonic series have been studied extensively ^[1]. However, Riemann's Rearrangement Theorem is non-constructive; there is no general method to find the sum of a re-arrangement. It was shown in ^[8] that assigning plus or minus signs randomly produces sums that converge almost surely.

In this article, we consider a generalized version of the alternating harmonic series, one with the assignment of plus or minus signs, as follows. For each positive integer $k$, we consider the series

We will find the infinite sum, $S_k$, of the infinite series in two different formats and examine the interesting relationship between this sum and the harmonic number,

which appears often throughout mathematics ^[2].

Definition 1.1. We take $H_0 = 0$ and define the terms of $S_k$ by

for each positive integer $n$ and $k$. Thus,

Clearly, for a fixed $k$, $a_n(k) > 0$ and decreases to 0 as $n$ increases. Hence, by the alternating series test, $S_k$ is convergent for all positive integers $k$.

2.   Sum of generalized alternating harmonic series

Our first summation formula is given in integral form, as follows.

Theorem 2.1. If $k$ is a positive integer, then

Proof. Since

Equation (2.1) holds for $k = 1.$ For $k\ge 2,$ we first note that the harmonic number has an integral expression ^[7] as

Hence when $k\ge 2,$

It follows that for $k\ge 2,$

proving the theorem.

As an application of Theorem 2.1, we have

Example 2.2.

Our second summation formula is given in terms of trigonometric functions, as follows.

Corollary 2.3. If If $k$ is a positive integer, then

Proof. By Theorem 2.1,

where in Eq (2.3), for the second summation in the parenthesis, we changed the index from $m$ to $m' = k-m$. We simplify Eq (2.4) using the following well known result, see page 323 of ^[5].

This yields the desired result,

and this proves the Theorem.

Example 2.4.

3.   The relationship between $S_k$ and $H_k$

We now set about finding the relationship between $S_k$ and $H_k.$ Since on $[0, 1], \, \, 1\le 1+x^k \le 2$ for all integer $k\ge 1$, we have that,

Hence by Theorem 2.1, we have

We will now calculate and simplify the difference between $H_k$ and $S_k.$ Notice that

We make the following definition.

Definition 3.1. Let

Note that for each integer $k\ge 1$, $D_k$ is a convergent alternating series, which easily follows from the properties of $a_n(k)$. More interestingly, $D_k$ itself forms a convergent sequence, as given in the following.

Theorem 3.2.

Proof. Let $\varepsilon$ be an arbitrarily fixed positive real number. By the Wallis product formula ^[6], we have

Since the natural logarithm is a continuous function, there exist a positive integer $N_1$ such that when $l\ge N_1,$

Let $m$ be any positive integer and let $k \ge 1$. By (2.2), we have

Hence, there exists a positive integer $N_2$ such than when $m \ge N_2,$

for each and every $k \ge 1$.

Let $N = \max\{N_1, N_2\}.$ Recall the Euler-Mascheroni constant ^[4] is defined as

Since $\{ H_n-\log n\}$ converges, it is a Cauchy sequence. We therefore may choose a positive integer $K$ such that when $j, m \ge K,$

For convenience, we denote

Clearly, $D_k = D^{2N}(k)+T_{2N}(k)$ and by (3.2), for any $k\ge 1$, there is

Also notice that, for any integer $k\ge 1$, we have

Let

Henceforth, let $k\ge K$, where $K$ is defined by Eq (3.3). Using the representation of $D^{2N}(k)$ in Eq (3.5), we have

where "$\le$" in (3.6) follows from (3.3) because $k\ge K$, which makes all of $2nk$, $(2n-1)k$, and $(2n+1)k$ greater than $K$.

Finally, for $k \geq K,$ we have

where the last step follows from (3.1), (3.4), and (3.6). Therefore, by the arbitrariness of $\varepsilon$, we have

and this completes the proof.

Acknowledgments

The authors would like to extend their sincere gratitude to the anonymous reviewers for their constructive suggestions and comments, which have greatly helped improve the quality of this paper. We are grateful to Prof. Les Reid for stimulating discussions during the preparation of the manuscript. S. Zheng was supported by a Summer Faculty Fellowship from Missouri State University.

Conflict of interest

The authors declare that there is no conflicts of interest.