Open access peer-reviewed chapter

# Digit Sums and Infinite Products

Written By

Samin Riasat

Submitted: November 23rd, 2019 Reviewed: April 3rd, 2020 Published: July 1st, 2020

DOI: 10.5772/intechopen.92365

From the Edited Volume

## Number Theory and Its Applications

Edited by Cheon Seoung Ryoo

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## Abstract

Consider the sequence un defined as follows: un=+1 if the sum of the base b digits of n is even, and un=−1 otherwise, where we take b=2. Recall that the Woods-Robbins infinite product involves a rational function in n and the sequence un. Although several generalizations of the Woods-Robbins product are known in the literature, no other infinite product involving a rational function in n and the sequence un was known in closed form until recently. In this chapter we introduce a systematic approach to these products, which may be generalized to other values of b. We illustrate the approach by evaluating a large class of similar infinite products.

### Keywords

• digit sums
• Prouhet-Thue-Morse sequence
• Woods and Robbins product
• closed formulas for infinite products

## 1. Introduction

Throughout this chapter nwill denote a non-negative integer. Let sbndenote the sum of the base bdigits of n, and put un=1s2n. We study infinite products of the form

fbcn=1n+bn+cun.E1

(We show in Section 2 that fbcconverges for b,cC\123).

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

n=02n+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=12n2n+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 1.1Let RCXbe a rational function such that the values Rnare defined and nonzero for n1. Then, the infinite product nRnunconverges if and only if the numerator and the denominator of Rhave the same degree and same leading coefficient.

Proof.If the infinite product converges, then Rnmust tend to 1when ntends to infinity. Thus the numerator and the denominator of Rhave the same degree and the same leading coefficient.

Now suppose that the numerator and the denominator of Rhave 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=1n+bn+cunE4

converge for complex numbers band csuch that n+band n+cdo not vanish for any n1. Since the general factor of such a product tends to 1, it is equivalent, grouping the factors pairwise, to proving that the product

n=12n+b2n+cu2n2n+1+b2n+1+cu2n+1E5

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

n=12n+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 fbcconverges for any b,cC\123. Using the definition of un, it follows that for any b,c,dC\123,

1. fbb=1.

2. fbcfcd=fbd.

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

One can ask the natural question: is fthe 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 fbccan be computed using only the quantities hx=fx2x+12, via

fbc=c+1b+1hbhc.E10

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

Taking c=b+12in 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 1.1The following relations hold.

1. For b,cC\123,

n=1n+bn+b+12n+c2n+cn+c+12n+b2un=c+1b+1.E12

2. For b,cC\012,

n=0n+bn+b+12n+c2n+cn+c+12n+b2un=1.E13

3. For bC\123,

n=1n+bn+b+14n+b+34n+b2un=b+32b+1.E14

4. For cC\123,

n=1n+12n+c2n+cn+c+12un=c+1.E15

Proof.

1. This follows immediately using properties 1–3 in Section 2.

2. As above.

3. Take c=b+1/2in Eq. (12).

4. Take b=0in Eq. (12).

Corollary 1.1For any positive rational number q, there exist monic polynomials P,Q14X, both at most cubic, such that

n=1PnQnun=q.E16

Furthermore, if qis an integer, then Pand Qcan 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=0in Eq. (11) gives

h0=23h12h0,E18

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

n=04n+34n+1un=2.E19

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

h12=34h12;E20

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

n=02n+22n+1un=2.E21

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

h12=12h0h1=12f012f120=12f1212,E22

i.e.,

n=14n12n+14n+12n1un=12.E23

Taking b=1in Eq. (11) gives

h1=45h32h2;E24

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

n=04n+32n+24n+52n+3un=12.E25

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

h22h3=32E26

which is equivalent to

n=02n+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 fand h.

Lemma 1.2Let b,c1.

1. If b=c, then fbc=1.

2. If b>c, then

c+1b+12<fbc<1.E28

3. 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 anis positive and strictly decreasing to 0. Using s22n+1=1+s22n, it follows that Un210and Unnmod2, for each n. Using summation by parts,

SN=aN+1UN+n=1NUnanan+1.E31

So 2a1<SN<0for N2. Exponentiating and taking Ngives the desired result.

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

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

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

1<c+1b+1hbhc<c+1b+12E32

from which the result follows.

Theorem 1.3For b,c1, fbcis strictly decreasing in band 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 1.4For x2,

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

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

We now prove some results on differentiability.

Theorem 1.5hxis smooth on 2.

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

Differentiating with respect to xgives

SN=n=1Nun2n+x2n+1+x=n=1Nun12n+x12n+1+x.E36

Hence

SNSMn=M+1N12n+x12n+1+xn=M+1N12n1+x12n+1+x=12M+1+x12N+1+x<12M10E37

as M, for any x2and N>M. Thus Snconverges uniformly on 2, which shows that logh, hence h, is differentiable on 2.

Now suppose that derivatives of hup to order kexist for some k1. Note that

SNk+1=1kk!n=1Nun12n+xk+112n+1+xk+1.E38

As before,

SNk+1SMk+1k!n=M+1N12n+xk+112n+1+xk+1k!n=M+1N12n1+xk+112n+1+xk+1=k!2M+1+xk+1k!2N+1+xk+1<k!2M1k+10E39

as M, for any x2and N>M. Hence Snk+1converges uniformly on 2, i.e., hkis differentiable on 2.

Therefore, by induction, hhas derivatives of all orders on 2.

Theorem 1.6Let a0. Then

loghx=logha+k=11k1kn=2unn+akxakE40

for xa1a+1.

Proof.Let Hx=loghx. By Theorem 1.5,

Hk+1x=1kk!n=2unn+xk+1.E41

Hence

Hk+1xk!n=21n+xk+1k!n=21n+a1k+1E42

for xa1a+1. So by Taylor’s inequality, the remainder for the Taylor polynomial for Hxof degree kis absolutely bounded above by

1k+1n=21n+a1k+1xak+1E43

which tends to 0as k, since a0and xa1. Therefore Hxequals its Taylor expansion about afor xin the given range.

## 5. Further remarks on h0

As mentioned in Section 1, not much is known about the quantity h01.62816. We give the following explanation as to why h0might 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=01/2and 1/20. That is, the victim of any such cancelation is always h0or h01. So we must look for other ways to study h0.

Using the two known values h1/2=3/2and h1=2, the following expressions for h0are obtained from Theorem 1.6 by choosing various values for xand a.

• x=0and a=1:

h0=2expk=11kn=2unn+1kE44

• x=1and a=0:

h0=2expk=11kkn=2unnkE45

• x=0and a=1/2:

h0=32expk=11kn=2u2n+12n+1kE46

• x=1/2and a=0:

h0=32expk=11kkn=2u2n2nkE47

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 sbnby 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=2we 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 ωbis a primitive b-th root of unity. We leave these as work to be done in the future.

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Written By

Samin Riasat

Submitted: November 23rd, 2019 Reviewed: April 3rd, 2020 Published: July 1st, 2020