On the Summation of Certain Iterated Series

Ovidiu Furdui Campia Turzii 405100 Cluj Romania

                       ofurdui@yahoo.com

Tiberiu Trif Department of Mathematics Babeş-Bolyai University Cluj-Napoca Romania

                     ttrif@math.ubbcluj.ro

Abstract The paper gives a unified treatment of the summation of certain iterated series of the form ∑_{n=1}^{∞} ∑_{m=1}^{∞} a_{n+m}, where (a_{n})_{n∈ℕ} is a sequence of real numbers. We prove that, under certain conditions, the double iterated series equals the difference of two single series.

1 Introduction

The goal of this paper is to provide a unified treatment of the summation of a special class of double series that have appeared recently. Below, we collect some problems that led to the motivation of this article.

Problem 1 ([6]) Find the sum

∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{(−1)^{n+m}}{(⌊\sqrt{n + m}⌋)^{3}} ,

where ⌊a⌋ denotes the greatest integer less than or equal to a.

Problem 2 ([5]) Find

∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{(−1)^{n+m}}{n + m} .  (1)

Problem 3 ([4]) Compute the sum

∑_{n=0}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{ln(n + m)}{n + m} .  (2)

Problem 4 ([7]) Find

∑_{n=1}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{H_{n+m}}{n + m} ,  (3)

where H_{n} denotes the nth harmonic number.

Although the problems can be solved by various techniques, in this paper we give a general method for summing the iterated series

∑_{n=1}^{∞} ∑_{m=1}^{∞} a_{n+m},  (4)

where (a_{n})_{n∈ℕ} is a sequence of real numbers, and we show that, under certain conditions, the double iterated series equals the difference of two single series. The organization of the paper is as follows: in Section 2 we give the first two main results of the paper, which are about the evaluation of the double iterated series (4), and in Section 3 we give the closed form evaluation of multiple series of the form ∑_{n_{1}=1}^{∞} ⋯ ∑_{n_{k}=1}^{∞} a_{n_{1}+n_{2}+···+n_{k}}.

2 Double iterated series

The first main result of this section is the following theorem.

Theorem 1. Suppose that both series

∑_{k=1}^{∞} a_{k}  and  ∑_{k=1}^{∞} ka_{k}  (5)

converge and let σ and σ̃ denote their sums, respectively. Then the iterated series (4) converges and its sum s equals σ̃ − σ.

Proof. For positive integers ν and n we let

A_{ν} = ∑_{m=1}^{∞} a_{ν+m} and s_{n} = ∑_{ν=1}^{n} A_{ν}.

Likewise, for every positive integer n we let

σ_{n} = ∑_{k=1}^{n} a_{k} and σ̃_{n} = ∑_{k=1}^{n} ka_{k}.

Since A_{ν} = σ − σ_{ν}, it follows that

s_{n} = nσ − ∑_{ν=1}^{n} σ_{ν} = nσ − ∑_{k=1}^{n}(n + 1 − k)a_{k} = nσ + σ̃_{n} − (n + 1)σ_{n} = \frac{u_{n}}{v_{n}} ,  (6)

where

u_{n} = σ − σ_{n} + \frac{σ̃_{n} − σ_{n}}{n} and v_{n} = \frac{1}{n} .

On the other hand, it is straight-forward to show that

\frac{u_{n+1} − u_{n}}{v_{n+1} − v_{n}} = σ̃_{n+1} − σ_{n+1}.

Since

lim_{n→∞} \frac{u_{n+1} − u_{n}}{v_{n+1} − v_{n}} = σ̃ − σ,

an application of Cesáro-Stolz theorem, the (\frac{0}{0}) case, implies that

s = lim_{n→∞} s_{n} = lim_{n→∞} \frac{u_{n}}{v_{n}} = σ̃ − σ,

and the theorem is proved.

Solution to Problem 1. Let a_{k} = (−1)^{k}/(⌊\sqrt{k}⌋)^{3}, and note that both series (5)

∑_{k=1}^{∞} a_{k} = ∑_{k=1}^{∞}(−1)^{k} \frac{1}{(⌊\sqrt{k}⌋)^{3}}  and  ∑_{k=1}^{∞} ka_{k} = ∑_{k=1}^{∞}(−1)^{k} \frac{k}{(⌊\sqrt{k}⌋)^{3}} ,

converge based on the Leibniz test. Moreover,

σ̃ = ∑_{k=1}^{∞} ka_{k} = ∑_{k=1}^{∞} \frac{(−1)^{k}k}{(⌊\sqrt{k}⌋)^{3}} = ∑_{N=1}^{∞} ( ∑_{k=N^{2}}^{N^{2}+2N} \frac{(−1)^{k}k}{(⌊\sqrt{k}⌋)^{3}} ) = ∑_{N=1}^{∞} \frac{1}{N^{3}} ( ∑_{k=N^{2}}^{N^{2}+2N}(−1)^{k}k)

= ∑_{N=1}^{∞} \frac{(−1)^{N}(N^{2} + N)}{N^{3}} = ∑_{N=1}^{∞} \frac{(−1)^{N}}{N} + ∑_{N=1}^{∞} \frac{(−1)^{N}}{N^{2}}

= − ln 2 − \frac{π^{2}}{12}.

On the other hand,

σ = ∑_{k=1}^{∞} a_{k} = ∑_{N=1}^{∞} ( ∑_{k=N^{2}}^{N^{2}+2N} \frac{(−1)^{k}}{(⌊\sqrt{k}⌋)^{3}} ) = ∑_{N=1}^{∞} \frac{1}{N^{3}} ( ∑_{k=N^{2}}^{N^{2}+2N}(−1)^{k}) = ∑_{N=1}^{∞} \frac{(−1)^{N}}{N^{3}} = −\frac{3}{4} ζ(3),

where ζ(3) = ∑_{k=1}^{∞} 1/k^{3} denotes the Apéry constant. It follows, based on Theorem 1, that

∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{(−1)^{n+m}}{(⌊\sqrt{n + m}⌋)^{3}} = \frac{3}{4} ζ(3) − ln 2 − \frac{π^{2}}{12},

and the problem is solved.

Remark 2. It is worth mentioning that if the series ∑_{k=2}^{∞}(k − 1)a_{k}, would be absolutely convergent, the conclusion of Theorem 1 would follow by rearranging the terms of the iterated series (4) as follows

∑_{k=2}^{∞}(k − 1)a_{k} = a_{1+1} + (a_{1+2} + a_{2+1}) + ⋯ + (a_{1+(k−1)} + a_{2+(k−2)} + ⋯ + a_{(k−1)+1}) + · · · .

Unfortunately, when a_{k} = (−1)^{k}/⌊\sqrt{k}⌋^{3}, the absolute convergence of the series ∑_{k=2}^{∞}(k−1)a_{k} fails to hold. Also, the convergence of series (5) is sufficient but not necessary for the convergence of the iterated series (4). This is shown by the series (1), (2), and (3). Clearly, Theorem 1 does not apply to these series. However, the next theorem, which can be applied for summing the series (1), (2) and (3), is the second new result of this section.

Theorem 3. With the same notation as in the statement and the proof of Theorem 1, let us suppose that the first series in (5) converges, that σ̃_{2n} = o(n), and that the limit

ℓ = lim_{n→∞} (na_{2n} + σ̃_{2n−1})  (7)

exists in ℝ. Then the iterated series (4) converges and its sum s is given by s = ℓ − σ.

Proof. Let w_{n} = na_{2n} + σ̃_{2n−1}. With the same notation as in the statement and the proof of Theorem 1, we have, based on (6), that

σ + s_{2n} = (2n + 1)(σ − σ_{2n}) + σ̃_{2n} = \frac{x_{n}}{y_{n}} ,

where x_{n} = σ − σ_{2n} + \frac{1}{2n+1} σ̃_{2n} and y_{n} = \frac{1}{2n+1}. On the other hand,

\frac{x_{n+1} − x_{n}}{y_{n+1} − y_{n}} = \frac{2n + 1}{2} a_{2n+2} + σ̃_{2n+1} = w_{n+1} − \frac{1}{2} a_{2n+2} .

Since a_{n} → 0 as n → ∞, we have, based on the Cesáro-Stolz theorem, the (\frac{0}{0}) case, and (7), that lim_{n→∞}(σ + s_{2n}) = ℓ, and it follows that lim_{n→∞} s_{2n} = ℓ − σ. On the other hand, since

s_{2n+1} = s_{2n} + A_{2n+1} = s_{2n} + σ − σ_{2n+1},

we get that lim_{n→∞} s_{2n+1} = lim_{n→∞} s_{2n} = ℓ − σ. Thus, lim_{n→∞} s_{n} = ℓ − σ, and the theorem is proved.

Solution to Problem 2. Let a_{k} = (−1)^{k}/k and we note that

σ = ∑_{k=1}^{∞} a_{k} = ∑_{k=1}^{∞} \frac{(−1)^{k}}{k} = − ln 2.

On the other hand, σ̃_{2n} = ∑_{k=1}^{2n}(−1)^{k} = 0, and

w_{n} = na_{2n} + σ̃_{2n−1} = \frac{1}{2} + ∑_{k=1}^{2n−1} k · \frac{(−1)^{k}}{k} = −\frac{1}{2}, for all n ∈ ℕ.

Thus, ℓ = −\frac{1}{2}. An application of Theorem 3 shows that

∑_{n=1}^{∞} ∑_{m=1}^{∞} \frac{(−1)^{n+m}}{n + m} = ln 2 − \frac{1}{2} ,

and the problem is solved.

Before we give the solution to Problem 3 we need to introduce a notation. Recall that, the double factorial of a positive integer n is defined by the following formula

n!! = \{ \begin{matrix}n(n − 2) ⋯ 5 · 3 · 1, if n is odd; \\ n(n − 2) ⋯ 6 · 4 · 2, if n is even. \end{matrix}

Solution to Problem 3. Let a_{k} = (−1)^{k} \frac{ln k}{k}. It can be proved, see [9], that

σ = ∑_{k=1}^{∞}(−1)^{k} \frac{ln k}{k} = γ ln 2 − \frac{1}{2} ln^{2} 2,

where γ denotes the Euler-Mascheroni constant. On the other hand,

σ̃_{2n} = ∑_{k=1}^{2n}(−1)^{k} ln k = ln \frac{(2n)!!}{(2n − 1)!!} = \frac{1}{2} ln \frac{((2n)!!)^{2}}{((2n − 1)!!)^{2}(2n + 1)} + \frac{ln(2n + 1)}{2} ,

and

w_{n} = na_{2n} + σ̃_{2n−1} = \frac{1}{2} ln(2n) + ∑_{k=1}^{2n−1}(−1)^{k} ln k = \frac{1}{2} ln(2n) + ln \frac{(2n − 2)!!}{(2n − 1)!!}
= \frac{1}{2} ln \frac{((2n)!!)^{2}}{((2n − 1)!!)^{2}(2n + 1)} + \frac{1}{2} ln \frac{2n + 1}{2n} .

An application of the Wallis formula

lim_{n→∞} \frac{((2n)!!)^{2}}{((2n − 1)!!)^{2}(2n + 1)} = \frac{π}{2} ,

shows that σ̃_{2n} = o(n) and ℓ = lim_{n→∞} w_{n} = \frac{1}{2} ln \frac{π}{2} . Theorem 3 implies that

∑_{n=1}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{ln(n + m)}{n + m} = \frac{1}{2} ln \frac{π}{2} − σ,

and hence

∑_{n=0}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{ln(n + m)}{n + m} = σ + ∑_{n=1}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{ln(n + m)}{n + m} = \frac{1}{2} ln \frac{π}{2} ,

and the problem is solved.

Before we give the solution to Problem 4 we need to mention the properties of a special function. Recall that, the dilogarithm function, (see [11, 12]), denoted by Li_{2}(z), is the special function defined by

Li_{2}(z) = ∑_{n=1}^{∞} \frac{z^{n}}{n^{2}} = − ∫_{0}^{z} \frac{ln(1 − t)}{t}dt, |z| ≤ 1.

A special Landen identity involving this function, [12, p. 107], is given by

Li_{2}(z) + Li_{2}(1 − z) = \frac{π^{2}}{6} − ln(z) ln(1 − z).

This implies, when z = 1/2, that

Li_{2} (\frac{1}{2}) = \frac{π^{2}}{12} − \frac{ln^{2} 2}{2}.  (8)

Now we are ready to give the solution to Problem 4.

Solution to Problem 4. Let a_{k} = (−1)^{k} \frac{H_{k}}{k}. To calculate

σ = ∑_{k=1}^{∞}(−1)^{k} \frac{H_{k}}{k} ,

we will be using a general result established by Kantor [8]: given a sequence (α_{k})_{k≥1} of real numbers, we consider the power series

f(x) = ∑_{k=1}^{∞} α_{k}x^{k} and g(x) = ∑_{k=1}^{∞} α_{k}H_{k}x^{k}.  (9)

By the ratio test, the two power series have the same radius of convergence. Kantor [8] proved that for every real number x for which the first series in (9) converges one has

g(x) = ∫_{0}^{1} \frac{f(x) − f(tx)}{1 − t} dt.

Let α_{k} = (−1)^{k}/k, and note that

f(x) = ∑_{k=1}^{∞}(−1)^{k} \frac{x^{k}}{k} = − ln(1 + x), x ∈ (−1, 1].

It follows that

σ = g(1) = ∫_{0}^{1} \frac{ln(1 + t) − ln 2}{1 − t} dt.

The substitution x = 1 − t implies that

σ = ∫_{0}^{1} \frac{1}{x} ln (1 − \frac{x}{2}) dx = ∫_{0}^{1/2} \frac{ln(1 − y)}{y} dy = −Li_{2} (\frac{1}{2}) = \frac{ln^{2} 2}{2} − \frac{π^{2}}{12}.

On the other hand,

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

and

w_{n} = na_{2n} + σ̃_{2n−1} = \frac{1}{2} H_{2n} + σ̃_{2n} − H_{2n} = \frac{1}{2} (H_{n} − H_{2n}).

Since H_{n} = ln n + γ + o(1), we have that all the hypotheses of Theorem 3 are satisfied and ℓ = lim_{n→∞} w_{n} = −\frac{1}{2} ln 2. Thus,

∑_{n=1}^{∞} ∑_{m=1}^{∞}(−1)^{n+m} \frac{H_{n+m}}{n + m} = l − σ = −\frac{ln 2}{2} − \frac{ln^{2} 2}{2} + \frac{π^{2}}{12},

and the problem is solved.

3 Multiple series of a special form

In this section we study the multiple series of the form

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} a_{n_{1}+n_{2}+···+n_{k}},  (10)

where (a_{n})_{n∈ℕ} is a sequence of real numbers that verifies a special condition. Before we give the main theorem of this section we collect some results we need in our analysis. Recall that, the Stirling numbers of the first kind, denoted by s(n, k), are the special numbers defined by the generating function

z(z − 1)(z − 2) ⋯ (z − n + 1) = ∑_{k=0}^{n} s(n, k)z^{k}.

For recurrence relations as well as interesting properties satisfied by these numbers the reader is referred to the book by Srivastava and Choi [12]. The main result of this section is the following theorem.

Theorem 4. Let k ≥ 1 be a natural number and let (a_{n})_{n∈ℕ} be a sequence of real numbers such that ∑_{n=1}^{∞} a_{n}n^{k−1} converges absolutely. Then the multiple series (10) converges and

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} a_{n_{1}+n_{2}+···+n_{k}} = \frac{1}{(k − 1)!} ∑_{i=1}^{k} s(k, i) (∑_{p=k}^{∞} a_{p}p^{i−1}) ,

where s(k, i) are the Stirling numbers of the first kind.

Proof. We have

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} a_{n_{1}+n_{2}+···+n_{k}} = ∑_{p=k}^{∞} (  ∑_{n_{1}+n_{2}+···+n_{k}=p} a_{n_{1}+n_{2}+···+n_{k}})

= ∑_{p=k}^{∞} a_{p} ( ∑_{n_{1}+n_{2}+···+n_{k}=p} 1)

= ∑_{p=k}^{∞} a_{p}\binom{p − 1}{k − 1}

= \frac{1}{(k − 1)!} ∑_{p=k}^{∞} \frac{a_{p}}{p} (p(p − 1)(p − 2) ⋯ (p − k + 1))

= \frac{1}{(k − 1)!} ∑_{p=k}^{∞} \frac{a_{p}}{p} (∑_{i=0}^{k} s(k, i)p^{i})

= \frac{1}{(k − 1)!} ∑_{i=0}^{k} s(k, i) (∑_{p=k}^{∞} a_{p}p^{i−1})

= \frac{1}{(k − 1)!} ∑_{i=1}^{k} s(k, i) (∑_{p=k}^{∞} a_{p}p^{i−1}) ,

since s(k, 0) = 0 for k ∈ ℕ. The theorem is proved.

The next result, which is a consequence of Theorem 4, is about the calculation of another multiple series of a special form.

Corollary 5. Let 1 ≤ i ≤ k be fixed natural numbers and let (a_{n})_{n∈ℕ} be a sequence of positive numbers such that ∑_{n=1}^{∞} a_{n}n^{k} converges. Then

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} n_{i} · a_{n_{1}+n_{2}+···+n_{k}} = \frac{1}{k!} ∑_{i=1}^{k} s(k, i) (∑_{p=k}^{∞} a_{p}p^{i}) ,

where s(k, i) are the Stirling numbers of the first kind.

Proof. We have, based on symmetry reasons, that for all i, j = 1, …, k, one has that

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} n_{i} · a_{n_{1}+n_{2}+···+n_{k}} = ∑_{n_{1},n_{2},...,n_{k}=1}^{∞} n_{j} · a_{n_{1}+n_{2}+···+n_{k}},

and hence

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} n_{i} · a_{n_{1}+n_{2}+···+n_{k}} = \frac{1}{k} ∑_{n_{1},n_{2},...,n_{k}=1}^{∞}(n_{1} + n_{2} + ⋯ + n_{k}) · a_{n_{1}+n_{2}+···+n_{k}},

and the result follows based on Theorem 4 applied to the sequence (na_{n})_{n∈ℕ}.

Corollary 6. a) Let k be a fixed positive integer and let m > k. Then

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{1}{(n_{1} + n_{2} + ⋯ + n_{k})^{m}}

= \frac{1}{(k − 1)!} ∑_{i=1}^{k} s(k, i) (ζ(m + 1 − i) − 1 − \frac{1}{2^{m+1−i}} − ⋯ − \frac{1}{(k − 1)^{m+1−i}} ) ,

where the parenthesis contains only the term ζ(m + 1 − i) when k = 1.

b) Let k and i be fixed positive integers such that 1 ≤ i ≤ k, and let m be such that m − k > 1. Then

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{n_{i}}{(n_{1} + n_{2} + ⋯ + n_{k})^{m}}

= \frac{1}{k!} ∑_{i=1}^{k} s(k, i) (ζ(m − i) − 1 − \frac{1}{2^{m−i}} − ⋯ − \frac{1}{(k − 1)^{m−i}} ) ,

where the parenthesis contains only the term ζ(m − i) when k = 1.

The next result specializes to the case when the sequence a_{n} = 1/n!. Recall that, if k ∈ ℕ, the series S_{k} = ∑_{n=1}^{∞} \frac{n^{k}}{n!} , also known as the Wolstenholme series, equals an integral multiple of e, i.e., S_{k} = B_{k}e, (see [1, p. 197]). The integer B_{k} is known in the mathematical literature as the kth Bell number [2] and the equality S_{k} = B_{k}e is known as Dobinski’s formula [3]. For example, B_{1} = 1, B_{2} = 2, B_{3} = 5, B_{4} = 15, B_{5} = 52. One can prove that the sequence (B_{k})_{k∈ℕ} verifies the recurrence formula

B_{k} = ∑_{j=1}^{k−1} \binom{k − 1}{j}B_{j} + 1.

The next corollary gives a recurrence relation involving the Stirling numbers of the first kind and the Bell numbers. The formula is obtained by calculating, by two different methods, a multiple series involving a factorial term.

Corollary 7. Let k ≥ 2 and let i be fixed positive integers such that 1 ≤ i ≤ k. Then

a) ∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{n_{i}}{(n_{1} + n_{2} + ⋯ + n_{k})!} = \frac{e}{k!}.

b) ∑_{i=1}^{k} s(k, i) (B_{i}e − ∑_{p=1}^{k−1} \frac{p^{i}}{p!}) = e.

Proof. We have, based on Corollary 5, that

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{n_{i}}{(n_{1} + n_{2} + ⋯ + n_{k})!} = \frac{1}{k!} ∑_{i=1}^{k} s(k, i) (B_{i}e − ∑_{p=1}^{k−1} \frac{p^{i}}{p!}) .

On the other hand,

∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{n_{i}}{(n_{1} + n_{2} + ⋯ + n_{k})!} = \frac{1}{k} ∑_{n_{1},n_{2},...,n_{k}=1}^{∞} \frac{1}{(n_{1} + n_{2} + ⋯ + n_{k} − 1)!}

= ∑_{p=k}^{∞} ( ∑_{n_{1}+···+n_{k}=p} \frac{1}{(n_{1} + n_{2} + ⋯ + n_{k} − 1)!})

= \frac{1}{k} ∑_{p=k}^{∞} \frac{1}{(p − 1)!}\binom{p − 1}{k − 1}

= \frac{1}{k!} ∑_{p=k}^{∞} \frac{1}{(p − k)!}

= \frac{e}{k!},

and the corollary is proved.

The next result refers to the calculation of multiple Wolstenholme series. Corollary 8. Let m and k be natural numbers. Then

∑_{n_{1},...,n_{k}=1}^{∞} \frac{(n_{1} + n_{2} + ⋯ + n_{k})^{m}}{(n_{1} + n_{2} + ⋯ + n_{k})!} = \frac{1}{(k − 1)!} ∑_{i=1}^{k} s(k, i) (B_{m+i−1}e − ∑_{p=1}^{k−1} \frac{p^{m+i−1}}{p!} ) ,

where the parenthesis contains only the term B_{m+i−1}e when k = 1.

References

[1] T. J. J’A. Bromwich, An Introduction to the Theory of Infinite Series, AMS Chelsea Publishing, Third Edition, 1991.

[2] L. Comtet, Advanced Combinatorics, D. Reidel, 1974.

[3] G. Dobiński, Summirung der Reihe ∑ n^{m}/n! für m = 1, 2, 3, 4, 5, …, Arch. für Mat. und Physik 61 (1877), 333–336.

[4] C. Efthimiou, Problem 1838, Math. Mag. 83 (2010), 65.

[5] O. Furdui, Problem 5013, School Science and Mathematics 108 March (2008), 2.

[6] O. Furdui, Problem 1849, Math. Mag. 83 (2010), 227.

[7] O. Furdui, Problem 11519, Amer. Math. Monthly 117 (2010), 649.

[8] W. H. M. Kantor, Solution to problem 4946, Amer. Math. Monthly 69 (1962), 239.

[9] M. S. Klamkin, Problem 4592, Amer. Math. Monthly 61 (1954), 350.

[10] M. S. Klamkin, Problem 4946, Amer. Math. Monthly 68 (1961), 67.

[11] L. Lewin, Polylogarithms and Associated Functions, Elsevier, 1981.

[12] H. M. Srivastava and J. Choi, Series Associated with the Zeta and Related Functions, Kluwer, 2001.

2010 Mathematics Subject Classification: Primary 40A05; Secondary 40B05. Keywords: Bell numbers, iterated series, double alternating harmonic series, Stirling numbers, Wolstenholme series.

(Concerned with sequences \underline{A000110} and \underline{A094638}.)

Received February 10 2011; revised versions received May 9 2011. Published in Journal of Integer Sequences, May 17 2011.

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