Evaluations of Some Variant Euler Sums

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

May 18, 2006

Abstract The purpose of this note is to present some elementary methods for the summation

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

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]. Especially, [3] collected thirty two 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.

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)

Acknowledgment. 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 Mathematics, 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, v. 2, Pairs, Birkhauser, 1994, 497-512.

2000 Mathematics Subject Classification: Primary 40C15, 40A25; Secondary 11M41, 11M06.

Keywords: Euler sums, Riemann zeta function, closed forms.

(Concerned with sequence A005408, A000984, A000045, A001008.)