1. Introduction

Throughout this chapter n will denote a non-negative integer. Let sbn denote the sum of the base b digits of n, and put un=−1s2n. We study infinite products of the form

fbc≔∏n=1∞n+bn+cun.E1

(We show in Section 2 that fbc converges for b,c∈C\−1−2−3…).

Plainly fbc=1/fcb and fbb=1. Up to these relations, it seems that the only known nontrivial value of f is f1/21=2, which is the famous Woods-Robbins identity [1, 2]:

∏n=0∞2n+12n+2un=12.E2

Several infinite products inspired by it were discovered afterwards (see, e.g., [3, 4]). But none of them involve the sequence un. Moreover, almost nothing is known (see, e.g., [5, 6]) about the similar product

∏n=1∞2n2n+1un=f012.E3

Our goal is to study these infinite products in detail. This will allow us to gain a deeper understanding of such products as well as evaluate more products like the Woods-Robbins identity.

The material in this chapter is based on the two papers [7, 8].

2. General properties of the function f

First we establish a general result from [7] on convergence.

Lemma 1.1 Let R∈CX be a rational function such that the values Rn are defined and nonzero for n≥1. Then, the infinite product ∏nRnun converges if and only if the numerator and the denominator of R have the same degree and same leading coefficient.

Proof. If the infinite product converges, then Rn must tend to 1 when n tends to infinity. Thus the numerator and the denominator of R have the same degree and the same leading coefficient.

Now suppose that the numerator and the denominator of R have the same leading coefficient and the same degree. Decomposing them in factors of degree 1, it suffices, for proving that the infinite product converges, to show that infinite products of the form

∏n=1∞n+bn+cunE4

converge for complex numbers b and c such that n+b and n+c do not vanish for any n≥1. Since the general factor of such a product tends to 1, it is equivalent, grouping the factors pairwise, to proving that the product

∏n=1∞2n+b2n+cu2n2n+1+b2n+1+cu2n+1E5

converges. Since u2n=un and u2n+1=−un, we only need to prove that the infinite product

∏n=1∞2n+b2n+1+c2n+c2n+1+bunE6

converges. Taking (the principal determination of) logarithms, we see that

log2n+b2n+1+c2n+c2n+1+b=O1n2E7

which gives the convergence result.

Hence fbc converges for any b,c∈C\−1−2−3…. Using the definition of un, it follows that for any b,c,d∈C\−1−2−3…,

1. fbb=1.

2. fbcfcd=fbd.

3. fbc=c+1b+1fb2c2fc+12b+12.

One can ask the natural question: is f the unique function satisfying these properties?

2.1 A new function

Properties 1 and 2 above give

fbcfde=fbcfcdfdefdcfcdfdc=fbefdcfcc=fbefdc.E8

Hence we can rewrite property 3 as

fbc=fb2b+12b+1/fc2c+12c+1.E9

Thus fbc can be computed using only the quantities hx=fx2x+12, via

fbc=c+1b+1⋅hbhc.E10

So understanding f is equivalent to understanding h, in the sense that each can be completely evaluated in terms of the other.

Taking c=b+12 in Eq. (10) gives the functional equation:

hb=b+1b+32hb+12h2b.E11

Similar questions can be asked for h: is it the unique solution to Eq. (11)? What about monotonic/continuous/smooth solutions?

3. Infinite products

3.1 Direct approach

Theorem 1.1 The following relations hold.

For b,c∈C\−1−2−3…,
∏n=1∞n+bn+b+12n+c2n+cn+c+12n+b2un=c+1b+1.E12
For b,c∈C\0−1−2…,
∏n=0∞n+bn+b+12n+c2n+cn+c+12n+b2un=1.E13
For b∈C\−1−2−3…,
∏n=1∞n+bn+b+14n+b+34n+b2un=b+32b+1.E14
For c∈C\−1−2−3…,
∏n=1∞n+12n+c2n+cn+c+12un=c+1.E15

Proof.

This follows immediately using properties 1–3 in Section 2.
As above.
Take c=b+1/2 in Eq. (12).
Take b=0 in Eq. (12).

Corollary 1.1 For any positive rational number q, there exist monic polynomials P,Q∈14ℤX, both at most cubic, such that

∏n=1∞PnQnun=q.E16

Furthermore, if q is an integer, then P and Q can be chosen to be at most quadratic.

We still do not know exactly which numbers are given by such infinite products.

3.2 Functional equation approach

Recall the functional Eq. (11):

hb=b+1b+32hb+12h2b.E17

Taking b=0 in Eq. (11) gives

h0=23h12h0,E18

i.e., h1/2=3/2. This shows that

∏n=0∞4n+34n+1un=2.E19

Next, taking b=1/2 in Eq. (11) gives

h12=34h12;E20

hence h1=2, and we recover the Woods-Robbins identity

∏n=0∞2n+22n+1un=2.E21

Similarly, taking b=−1/2 in Eq. (11) gives

h−12=12h0h−1=12f012f−120=12f−1212,E22

i.e.,

∏n=1∞4n−12n+14n+12n−1un=12.E23

Taking b=1 in Eq. (11) gives

h1=45h32h2;E24

hence h3/2h2=52/4, and this gives

∏n=0∞4n+32n+24n+52n+3un=12.E25

Taking b=3/2 in Eq. (11) and using the previous result gives

h22h3=32E26

which is equivalent to

∏n=0∞2n+2n+12n+3n+2un=12.E27

These identities can be also combined in pairs to obtain other identities.

4. Some analytical results

We saw in the previous section that some of the infinite products we evaluated were integers, some were rational, and some were quadratic irrational. In the hope of further understanding their nature, we now study the analytical behaviors of f and h.

Lemma 1.2 Let b,c∈−1∞.

If b=c, then fbc=1.
If b>c, then
c+1b+12<fbc<1.E28
If b<c, then
1<fbc<c+1b+12.E29

Proof. Using properties 1–3 from Section 2, it suffices to prove 2.

Let b>c>−1, and put

an=logn+bn+c,SN=∑n=1Nanun,UN=∑n=1Nun.E30

Note that an is positive and strictly decreasing to 0. Using s22n+1=1+s22n, it follows that Un∈−2−10 and Un≡nmod2, for each n. Using summation by parts,

SN=aN+1UN+∑n=1NUnan−an+1.E31

So −2a1<SN<0 for N≥2. Exponentiating and taking N→∞ gives the desired result.

Lemma 1.2 together with Eq. (10) implies the following results.

Theorem 1.2hx/x+1 is strictly decreasing on −1∞, and hxx+1 is strictly increasing on −1∞.

Proof. Let −1<b<c. By Eqs. (29) and (10),

1<c+1b+1⋅hbhc<c+1b+12E32

from which the result follows.

Theorem 1.3 For b,c∈−1∞, fbc is strictly decreasing in b and strictly increasing in c.

Proof. By Eq. (10),

hcfbcc+1=hbb+1E33

and

b+1fbchb=c+1hcE34

hence the result follows from Theorem 1.2.

Theorem 1.4 For x∈−2∞,

1<hx<x+3x+22.E35

Proof. This follows from taking b=x/2 and c=x+1/2 in Eq. (10), then using Eq. (29).

We now prove some results on differentiability.

Theorem 1.5hx is smooth on −2∞.

Proof. Take b=x/2 and c=x+1/2 in Eq. (30). Then the sequence Sn of smooth functions on −2∞ converges pointwise to logh.

Differentiating with respect to x gives

SN′=∑n=1Nun2n+x2n+1+x=∑n=1Nun12n+x−12n+1+x.E36

Hence

SN′−SM′≤∑n=M+1N12n+x−12n+1+x≤∑n=M+1N12n−1+x−12n+1+x=12M+1+x−12N+1+x<12M−1→0E37

as M→∞, for any x∈−2∞ and N>M. Thus Sn′ converges uniformly on −2∞, which shows that logh, hence h, is differentiable on −2∞.

Now suppose that derivatives of h up to order k exist for some k≥1. Note that

SNk+1=−1kk!∑n=1Nun12n+xk+1−12n+1+xk+1.E38

As before,

SNk+1−SMk+1≤k!∑n=M+1N12n+xk+1−12n+1+xk+1≤k!∑n=M+1N12n−1+xk+1−12n+1+xk+1=k!2M+1+xk+1−k!2N+1+xk+1<k!2M−1k+1→0E39

as M→∞, for any x∈−2∞ and N>M. Hence Snk+1 converges uniformly on −2∞, i.e., hk is differentiable on −2∞.

Therefore, by induction, h has derivatives of all orders on −2∞.

Theorem 1.6 Let a≥0. Then

loghx=logha+∑k=1∞−1k−1k∑n=2∞unn+akx−akE40

for x∈a−1a+1.

Proof. Let Hx=loghx. By Theorem 1.5,

Hk+1x=−1kk!∑n=2∞unn+xk+1.E41

Hence

∣Hk+1x∣≤k!∑n=2∞1n+xk+1≤k!∑n=2∞1n+a−1k+1E42

for x∈a−1a+1. So by Taylor’s inequality, the remainder for the Taylor polynomial for Hx of degree k is absolutely bounded above by

1k+1∑n=2∞1n+a−1k+1x−ak+1E43

which tends to 0 as k→∞, since a≥0 and ∣x−a∣≤1. Therefore Hx equals its Taylor expansion about a for x in the given range.

5. Further remarks on h0

As mentioned in Section 1, not much is known about the quantity h0≈1.62816. We give the following explanation as to why h0 might behave specially in a sense.

Note that the only way nontrivial cancelation occurs in Eq. (11) is when b=0. Likewise, nontrivial cancelation occurs in Eq. (10) or property 3 in Section 2 only for bc=01/2 and 1/20. That is, the victim of any such cancelation is always h0 or h0−1. So we must look for other ways to study h0.

Using the two known values h1/2=3/2 and h1=2, the following expressions for h0 are obtained from Theorem 1.6 by choosing various values for x and a.

x=0 and a=1:
h0=2exp−∑k=1∞1k∑n=2∞unn+1kE44
x=1 and a=0:
h0=2exp∑k=1∞−1kk∑n=2∞unnkE45
x=0 and a=1/2:
h0=32exp∑k=1∞1k∑n=2∞u2n+12n+1kE46
x=1/2 and a=0:
h0=32exp∑k=1∞−1kk∑n=2∞u2n2nkE47

The Dirichlet series appearing in the above expressions were studied in [9]. We think that these identities and the results from Section 4 might help in shedding some light on the nature of h0.

6. Conclusions and future developments

We evaluated infinite products involving the digit sum function sbn by splitting the product based on the congruence classes modulo b. We illustrated two approaches for doing so, one by direct computation and another using functional equations. For b=2 we proved some analytical results to aid us in understanding the behavior of these products. Many open questions still remain.

Although we only considered the base b=2, many of the results above easily generalize to other bases. One possible direction toward a generalization is to take un=−1sbn. Another is un=ωbsbn, where ωb is a primitive b-th root of unity. We leave these as work to be done in the future.

Abeer Alharbi

Chapters authored

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