Evaluations of Some Variant Euler Sums

Hongwei Chen Department of Mathematics Christopher Newport University Newport News, VA 23606 hchen@cnu.edu

Abstract In this note we present some elementary methods for the summation of certain

Euler sums with terms involving 1 + 1/3 + 1/5 + ⋯ + 1/(2k − 1).

1 Introduction

In the last decade, based on extensive experimentation with computer algebraic systems, a large class of Euler sums have been explicitly evaluated in terms of the Riemann zeta function ζ(k). For example, let

H_{k} = 1 + \frac{1}{2} + ⋯ + \frac{1}{k}.

Then

∑_{k=1}^{∞} \frac{1}{k^{2}} H_{k} = 2 ζ(3),
∑_{k=1}^{∞} \frac{1}{2^{k} k^{2}} H_{k} = ζ(3) − \frac{π^{2}}{12} ln 2,
∑_{k=1}^{∞} \frac{1}{k^{2}} H_{k}^{2} = \frac{17}{4} ζ(4),

and

∑_{k=1}^{∞} \frac{(−1)^{k−1}}{k^{2}} H_{k} = \frac{5}{8} ζ(3).

More details can be found in [1, 2, 3, 4] In particular, Borwein and Bradley [3] collected 32 beautiful proofs of the first sum above.

Motivated by the above results, in this note, replacing H_{k} by

h_{k} = H_{2k} − \frac{1}{2} H_{k} = 1 + \frac{1}{3} + ⋯ + \frac{1}{2k − 1},  (1)

we study the following variant Euler sums

∑_{k=1}^{∞} a_{k} h_{k}

where the a_{k} are relatively simple function of k.

2 The Main Results

We begin to derive some series involving h_{k}. Since

− ln(1 − x) = ∫_{0}^{x} \frac{dt}{1 − t} = ∑_{k=1}^{∞} \frac{x^{k}}{k} ,

replacing x by −x gives

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

Averaging these two series gives us

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

In term of the Cauchy product and partial fractions, we have

\frac{1}{4} ln^{2} (\frac{1 + x}{1 − x}) = ∑_{k=1}^{∞} (\frac{1}{(2k − 1) · 1} + \frac{1}{(2k − 3) · 3} + ⋯ + \frac{1}{1 · (2k − 1)}) x^{2k}

= ∑_{k=1}^{∞} \frac{1}{2k} [(\frac{1}{2k − 1} + \frac{1}{1}) + (\frac{1}{2k − 3} + \frac{1}{3}) + ⋯ + (\frac{1}{1} + \frac{1}{2k − 1})] x^{2k}

= ∑_{k=1}^{∞} (1 + \frac{1}{3} + ⋯ + \frac{1}{2k − 1}) \frac{x^{2k}}{k} .

Noting that h_{k} is given by (1), we have

∑_{k=1}^{∞} \frac{h_{k}}{k} x^{2k} = \frac{1}{4} ln^{2} (\frac{1 + x}{1 − x}) .  (3)

This enables us to evaluate a wide variety of interesting series via specialization, differentiation and integration.

First, setting x = 1/2, we find

∑_{k=1}^{∞} \frac{h_{k}}{2^{2k} k} = \frac{1}{4} ln^{2} 3.  (4)

For x = \sqrt{2}/2,

∑_{k=1}^{∞} \frac{h_{k}}{2^{k} k} = \frac{1}{4} ln^{2}(3 + 2\sqrt{2}).  (5)

Putting x = (\sqrt{5} − 1)/2 = ϕ, the golden ratio, we get

∑_{k=1}^{∞} \frac{h_{k}}{k} ϕ^{2k} = \frac{1}{4} ln^{2}(2 + \sqrt{5}).  (6)

Furthermore, for any α ≥ 2, putting x = (\sqrt{5}+1)/2α and x = (\sqrt{5}−1)/2α in (3) respectively, we get

∑_{k=1}^{∞} \frac{h_{k}}{α^{2k} k} (\frac{\sqrt{5} + 1}{2})^{2k} = \frac{1}{4} ln^{2} (\frac{(2α + 1) + \sqrt{5}}{(2α − 1) − \sqrt{5}})  (7)

and

∑_{k=1}^{∞} \frac{h_{k}}{α^{2k} k} (\frac{\sqrt{5} − 1}{2})^{2k} = \frac{1}{4} ln^{2} (\frac{(2α − 1) + \sqrt{5}}{(2α + 1) − \sqrt{5}}) .  (8)

Recalling the Fibonacci numbers which are defined by

F_{1} = 1, F_{2} = 1, F_{k} = F_{k−1} + F_{k−2} for k ≥ 2

and Binet’s formula

F_{k} = \frac{1}{\sqrt{5}} ((\frac{\sqrt{5} + 1}{2})^{k} − (\frac{1 − \sqrt{5}}{2})^{k}) ,

combining (7) and (8), we find

∑_{k=1}^{∞} \frac{h_{k}}{α^{2k} k} F_{2k} = \frac{\sqrt{5}}{20} ln (\frac{α^{2} + α − 1}{α^{2} − α − 1}) ln (\frac{α^{2} + α\sqrt{5} + 1}{α^{2} − α\sqrt{5} + 1}) .  (9)

In particular, for α = 2

∑_{k=1}^{∞} \frac{h_{k}}{2^{2k} k} F_{2k} = \frac{\sqrt{5}}{4} ln 5 ln(9 + 4\sqrt{5}).  (10)

Another step along this path is to change variables. Setting x = cos θ in (3) leads to

∑_{k=1}^{∞} \frac{h_{k}}{k} cos^{2k} θ = ln^{2} (cot(x/2)) .  (11)

Integrating both sides from 0 to π, and using

∫_{0}^{π} cos^{2k} θ dθ = \frac{π}{2^{2k}} ( \begin{matrix}2k \\ k \end{matrix} )

and

∫_{0}^{π} ln^{2} (cot(x/2)) dθ = \frac{π^{3}}{4} ,

we find

∑_{k=1}^{∞} \frac{h_{k}}{2^{2k} k} ( \begin{matrix}2k \\ k \end{matrix} ) = \frac{π^{2}}{4} .  (12)

This adds another interesting series to Lehmer’s list [6].

Next, for 0 < x < 1, differentiating (3), then multiplying both sides by x, we obtain

∑_{k=1}^{∞} h_{k}x^{2k} = \frac{x}{2(1 − x^{2})} ln (\frac{1 + x}{1 − x}) .  (13)

Setting x = 1/2, we get

∑_{k=1}^{∞} \frac{h_{k}}{2^{2k}} = \frac{1}{3} ln 3.  (14)

For x = \sqrt{2}/2,

∑_{k=1}^{∞} \frac{h_{k}}{2^{k}} = \frac{\sqrt{2}}{2} ln(3 + 2\sqrt{2}).  (15)

Similar to (10), we have

∑_{k=1}^{∞} \frac{h_{k}}{2^{2k}} F_{2k} = \frac{\sqrt{5}}{50} (10 ln(5 + 2\sqrt{5}) + 3\sqrt{5} ln 5 − 5 ln 5).  (16)

Finally, for 0 < x ≤ 1, dividing both sides of (3) by x and integrating from 0 to x, we obtain

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} x^{2k} = \frac{1}{2} ∫_{0}^{x} \frac{1}{t} ln^{2} (\frac{1 + t}{1 − t}) dt.  (17)

Using the substitution u = (1 − x)/(1 + x) and integration by parts, we get

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} x^{2k} = ∫_{(1−x)/(1+x)}^{1} \frac{ln^{2} u}{1 − u^{2}} du
= \frac{1}{2} ln x ln^{2} (\frac{1 − x}{1 + x}) + ∫_{(1−x)/(1+x)}^{1} \frac{ln u}{u} ln (\frac{1 − u}{1 + u}) du.

In view of (2), we have

∫_{(1−x)/(1+x)}^{1} \frac{ln u}{u} ln (\frac{1 − u}{1 + u}) du = −2 ∑_{k=0}^{∞} \frac{1}{2k + 1} ∫_{(1−x)/(1+x)}^{1} u^{2k} ln u du.

Since

∫ u^{2k} ln u du = \frac{1}{2k + 1} u^{2k+1} ln u − \frac{1}{(2k + 1)^{2}} u^{2k+1} + C,

we find

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} x^{2k} = \frac{1}{2} ln x ln^{2} (\frac{1 − x}{1 + x}) + 2 ln (\frac{1 − x}{1 + x}) ∑_{k=0}^{∞} \frac{1}{(2k + 1)^{2}} (\frac{1 − x}{1 + x})^{2k+1}

+2 ∑_{k=0}^{∞} \frac{1}{(2k + 1)^{3}} − 2 ∑_{k=0}^{∞} \frac{1}{(2k + 1)^{3}} (\frac{1 − x}{1 + x})^{2k+1} .  (18)

In terms of the polylogarithm function [5]

Li_{n}(x) = ∑_{k=1}^{∞} \frac{x^{n}}{k^{n}} ,

and noting that

∑_{k=0}^{∞} \frac{x^{n}}{(2k + 1)^{n}} = \frac{1}{2} (Li_{n}(x) − Li_{n}(−x))

and

∑_{k=0}^{∞} \frac{1}{(2k + 1)^{3}} = ∑_{k=0}^{∞} \frac{1}{k^{3}} − ∑_{k=0}^{∞} \frac{1}{(2k)^{3}} = \frac{7}{8} ζ(3),

we finally obtain

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} x^{2k} = \frac{7}{4} ζ(3) + \frac{1}{2} ln x ln^{2} (\frac{1 − x}{1 + x})

+ ln (\frac{1 − x}{1 + x}) (Li_{2} (\frac{1 − x}{1 + x}) − Li_{2} (\frac{x − 1}{1 + x})) − (Li_{3} (\frac{1 − x}{1 + x}) − Li_{3} (\frac{x − 1}{1 + x})) .

Setting x = 1, we get

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} = \frac{7}{4} ζ(3).  (19)

For x = 1/3,

∑_{k=1}^{∞} \frac{h_{k}}{3^{2k} k^{2}} = \frac{7}{8} ζ(3) − \frac{1}{2} ln 3 ln^{3} 2
+ \frac{1}{3} ln^{3} 2 + ln 2 Li_{2}(−1/2) + Li_{3}(−1/2),

where we have used

Li_{2}(1/2) = \frac{π^{2}}{12} − \frac{1}{2} ln^{2} 2;
Li_{3}(1/2) = \frac{7}{8} ζ(3) + \frac{1}{6} ln^{3} 2 − \frac{π^{2}}{12} ln 2.

Moreover, noting that

h_{k} = ∑_{i=1}^{k} ∫_{0}^{1} x^{2(i−1)} dt = ∫_{0}^{1} (∑_{i=1}^{k} x^{2(i−1)}) dt = ∫_{0}^{1} \frac{1 − x^{2k}}{1 − x^{2}} dx

and rewriting (8) as

∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} (1 − x^{2k}) = \frac{1}{2} ∫_{x}^{1} \frac{1}{t} ln^{2} (\frac{1 + t}{1 − t}) dt,

we have

∑_{k=1}^{∞} \frac{h_{k}^{2}}{k^{2}} = ∑_{k=1}^{∞} \frac{h_{k}}{k^{2}} ∫_{0}^{1} \frac{1 − x^{2k}}{1 − x^{2}} dx

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

Exchanging the order of the integration, we get

∑_{k=1}^{∞} \frac{h_{k}^{2}}{k^{2}} = \frac{1}{2} ∫_{0}^{1} (\frac{1}{t} ln^{2} (\frac{1 + t}{1 − t}) ∫_{0}^{t} \frac{1}{1 − x^{2}} dx) dt.

= \frac{1}{4} ∫_{0}^{1} \frac{1}{t} ln^{3} (\frac{1 + t}{1 − t}) dt.

Using the substitution x = (1 − t)/(1 + t) and the well-known fact that

∫_{0}^{1} x^{k} ln^{3} x dx = − \frac{6}{(k + 1)^{3}} ,

we find

∑_{k=1}^{∞} \frac{h_{k}^{2}}{k^{2}} = −\frac{1}{2} ∫_{0}^{1} \frac{ln^{3} x}{1 − x^{2}} dx

= −\frac{1}{2} ∑_{k=0}^{∞} ∫_{0}^{1} x^{2k} ln^{3} x dx = 3 ∑_{k=0}^{∞} \frac{1}{(2k + 1)^{4}} = \frac{45}{16} ζ(4)  (20)

Another path out of (3) is to bring in complex variables. Since

\frac{1}{i} tan^{−1}(iz) = tanh^{−1} z = \frac{1}{2} ln (\frac{1 + z}{1 − z} )

Replacing x by ix in (3), we obtain

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{k} x^{2k} = (tan^{−1} x)^{2}.  (21)

This series may be evaluated at values such as x = 2 − \sqrt{3}, \sqrt{3}/3, 1 explicitly:

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}(2 − \sqrt{3})^{2k}}{k} = \frac{π^{2}}{144} = \frac{3}{72} ζ(2),  (22)

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{3^{k} k} = \frac{π^{2}}{36} = \frac{1}{6} ζ(2),  (23)

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{k} = \frac{π^{2}}{16} = \frac{3}{8} ζ(2).  (24)

Similarly, applying differentiation and integration to (21), we deduce the corresponding formulas

∑_{k=1}^{∞} (−1)^{k−1}h_{k} x^{2k} = \frac{x}{1 + x^{2}} tan^{−1} x,  (25)

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{k^{2}} x^{2k} = 2 ∫_{0}^{x} \frac{(tan^{−1} t)^{2}}{t} dt.  (26)

In particular, we find

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{3^{k}} = \frac{\sqrt{3}}{24} π,  (27)

∑_{k=1}^{∞} \frac{(−1)^{k−1}h_{k}}{k^{2}} = G π − \frac{7}{4} ζ(3),  (28)

where G is the Catalan’s constant which is defined by

G = ∑_{k=0}^{∞} \frac{(−1)^{k}}{(2k + 1)^{2}} .

Finally, following the excellent suggestion of an anonymous referee, recalling that

h_{k} = H_{2k} − \frac{1}{2} H_{k},  (29)

we find from (19)

∑_{k=1}^{∞} \frac{1}{k^{2}} H_{2k} = ∑_{k=1}^{∞} \frac{1}{k^{2}} h_{k} + \frac{1}{2} ∑_{k=1}^{∞} \frac{1}{k^{2}} H_{k} = \frac{11}{4} ζ(3).  (30)

Furthermore, in terms of the multiple series [7]

∑_{i=1}^{∞} ∑_{j=1}^{∞} \frac{1}{ij(i + j)} = 2ζ(3), ∑_{i=1}^{∞} ∑_{j=1}^{∞} \frac{(−1)^{i+j}}{ij(i + j)} = \frac{1}{4} ζ(3),

the difference gives

∑_{i,j>1,i+j=odd} \frac{1}{ij(i + j)} = \frac{7}{8} ζ(3).

Setting i + j = 2k + 1 and using partail fractions, we have

∑_{i,j>1,i+j=odd} \frac{1}{ij(i + j)} = ∑_{k=1}^{∞} ∑_{j=1}^{2k} \frac{1}{j(2k + 1 − j)(2k + 1)}
= ∑_{k=1}^{∞} \frac{1}{(2k + 1)^{2}} ∑_{j=1}^{2k} (\frac{1}{j} + \frac{1}{2k + 1 − j} )
= ∑_{k=1}^{∞} \frac{1}{(2k + 1)^{2}} 2H_{2k}.

Thus,

∑_{k=1}^{∞} \frac{1}{(2k + 1)^{2}} H_{2k} = \frac{7}{16} ζ(3).  (31)

Subsequently, we have

∑_{k=1}^{∞} \frac{1}{(2k − 1)^{2}} H_{2k} = ∑_{k=1}^{∞} \frac{1}{(2k + 1)^{2}} H_{2k}+

∑_{k=1}^{∞} \frac{1}{(2k − 1)^{3}} + ∑_{k=1}^{∞} \frac{1}{2k(2k − 1)^{2}} = \frac{21}{16} ζ(3) + \frac{1}{8} (π^{2} − 8 ln 2).  (32)

From this and the known result [1]

∑_{k=1}^{∞} \frac{1}{(2k − 1)^{2}} H_{k} = \frac{1}{4} (π^{2} − π^{2} ln 2 − 8 ln 2 + 7ζ(3)),

we finally get

∑_{k=1}^{∞} \frac{1}{(2k − 1)^{2}} h_{k} = \frac{7}{16} ζ(3) + \frac{3}{4} ζ(2) ln 2.  (33)

3 Acknowledgments

The author would like to thank the referee and the editor for their thoughtful comments and suggestions for improving the original version of the manuscript.

References

[1] D. Bailey, J. Borwein and R. Girgensohn, Experimental evaluation of Euler sums, Experimental Math., 3 (1994), 17–30.

[2] J. Borwein, D. Bailey and R. Girgensohn, Experimentation in Mathematics, A. K. Peters, 2004.

[3] J. Borwein and D. V. Bradley, Thirty-two Goldbach variations, preprint. Available at

   http://users.cs.dal.ca/~jborwein/32goldbach.pdf.

[4] J. Borwein and I. J. Zucker and J. Boersma, The evaluation of character Euler double

   sums, preprint. Available at http://users.cs.dal.ca/~jborwein/bzb7.pdf.

[5] L. Lewin, Polylogarithms and Associated Functions, Elsevier North Holland, New York-Amsterdam, 1981.

[6] D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92 (1985), 449–457.

[7] D. Zagier, Values of zeta functions and their applications, First European Congress of Mathematics, Vol. 2, Paris, Birkhauser, 1994, pp. 497–512.

2000 Mathematics Subject Classification: Primary 40C15; Secondary 40A25, 11M41, 11M06. Keywords: Euler sums, Riemann zeta function, closed forms.

(Concerned with sequences \underline{A000045}, \underline{A000984}, \underline{A001008}, and \underline{A005408}.)

Received February 9 2006; revised version received April 21 2006. Published in Journal of Integer Sequences, May 18 2006.

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