Table Of ContentSums and Products
Sums and Products
Titu Andreescu
Marian Tetiva
Library of Congress Control Number: 2018900205
ISBN-10: 0—9993428—1-9
ISBN-13: 978-0-9993428-1-7
© 2018 XYZ Press, LLC
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Contents
Introduction
1 TelescopingSumsand ProductsinAlgebra 19
TelescopingSumsand ProductsinTrigonometry 53
Complex Numbersandde Moivre’s Formula 93
TheAbel Summation Formula 107
Mathematical Induction 121
Combinatorial Identitiesand Generating Functions 139
Sumsand Products in NumberTheory 159
Problems 171
1 Easy Problems .................... 171
2 Medium Problems .................. 177
3 Hard Problems .................... 184
Solutions 193
1 Solutions to Easy Problems ............. ....... 193
2 Solutions to Medium Problems ........... 231
3 Solutions to Hard Problems ............. 279
vi Contents
References 335
Other BooksfromXYZPress 337
Introduction
Let us start with a simple (?) question. In a chess tournament there are n
players who play eliminatory games until only one winner remains. Thus, in
the first round, the players are arbitrarily playing games (chosen by drawing
lots), and only the winners of these games go to the second round. If the
number of players is odd there is one player staying aside, but he/she (and
all winners) will take part to the drawing for the second round. The process
repeats in the second and all the following rounds until, as we said, finally,
only one winner remains (and he is declared the winner ofthe championship).
The question is: how many games are necessary in order to establish the
champion?
Well, you might need a moment of thinking, and we strongly advise you to
take it (or, maybe, you already got the answer, which is great). You will
immediately see that if n = 2'" is a power of 2, then in the first round there
are 2m_1 games (and 2m_1 winners from these games accede to the second
round), and the process goes on and on so that there will be 2m_j games in
the jth round. The total number ofgames will then be
2m—1+2m-2+...+2+1=2m—1=n—1.
Although we cannot quite use this reasoning in the general case, the answer
n — 1 is correct for each and every value of n, because in each game precisely
one player is eliminated and, in order to arrive to the situation when only
one player still stands, n — 1 players must be eliminated, so, 77. — 1 games are
needed to see who the champion is.
The problem is solved, but we won’t stop here. This is because the reasoning
in the particular case ofn = 2m furnishes a hint for the general case. Namely,
2 Introduction
in the first round the number of games is k ifn = 2k is an even number, and
it is also k if n = 2k: + 1 is odd (and a player is forced — also by drawing, to
make the competition fair — to stay aside). This number can be expressed (for
both cases of even and odd n) as [11/2] where [x] denotes the integer part of
x (or the floor function ofac) — the largest integer which is not greater than a:
(that is, [as] = p if and only ifp is the only integer such that p _<_ :I: < p+ 1).
Thus in the first round there are [71/2] games and in the second round
”2 -_ n - lanl
players participate. Then the same pattern repeats: Lug/2] games are played
in the second round, and m = m — [a
2
players enter the third round, and so on. Thus, ifwe define the function f by
SC
f("’)=”_l§l
the total number of games played is
”1 n2
l3l + l3l + ' "
where the sequence (nk)k21 is defined by n1 = n and the recurrence nk =
f(nk_1) for every k 2 2. One can easily see that, starting with every positive
integer n the terms of the sequence (710121 eventually become equal to 1,
because this is a sequence of positive integers that strictly decreases as long
as its terms are greater than 1 (and thus, at some moment, a term equals 1,
and then all the terms that follow are also equal to 1). Thus the above sum
is actually a finite one (as we have already seen in the particular case of n
being a power of 2), as the integer parts [mg/2] are 0 as soon as nk becomes
1. Actually, we can find a formula for 774,, namely
mic—[2,711,
_ ’l’l
where the ceiling function of the real number as is defined by [at] = q if and
only if q is the unique integer such that q — 1 < a: S q, and this formula
Introduction 3
allows us to see that 71.], becomes 1 as soon as 2"—1 2 n (and the first such k
is [log2n] + l) — but this is not our point of interest here. Nevertheless, we
are interested in the fact that
_m _n2J = _ 1
i 2 J + l 2 + n ’
since we know now that the total number ofgames is n—1. For instance, with
m = n = 7, we have
n2=n1— [%J =7—3=4,
n3=n2—— [%J =4—2=2,
n
n4:n3— n —2—1=1,
and, consequently, m, = 1 for all k _ 4. Thus the number of games measured
as a sum is
EJ+l§J+l§l+lél+~= lél+l§J+El
Of course, this is 3 + 2 + 1 = 6 and corresponds to the result 7 — 1 given by
the (let us call it) global reasoning from the beginning.
To summarize: we defined a function
a:
f(x)=x‘l§l
and a sequence (”10121 starting with some arbitrary positive integer 77.1 = n
and satisfyingtherecurrence relationm, = f(nk_1) for k 2 2, and we obtained
l—Jl—J
This may be a somehow unexpected equation (although it is very clear in the
particular case n = 2m) and illustrates a simple principle in mathematics: do
something in two different ways (in this case “do something” is “count”), then
equate the two results (they must be equal, because, in the end, they represent
the same thing — in our case the total number ofgames). You will be amazed
4 Introduction
of what can be obtained by using this very simple (because fundamental)
principle.
Anyway, it is not ourpurposehereto examine such reasonings. Instead, we are
dealing throughout this bookwith sums and products (as the title says). Sums
and products are everywhere in mathematics. Probably the first matters that
a man (usually a child) learns in mathematics are addition and multiplication
(of natural numbers, then of integers, and so on) — operations for which the
results are called sum and product respectively. Clearly, we do not intend
to take it all from zero, but rather we will try to calculate as many sums as
possible of the form
n
Zak=al+a2+‘”+an
16:1
and products ofthe form
'n,
Halc = a1a2-..an
k=1
(that is, sums and products of an arbitrary number of terms, respectively
factors). Also, we will enter a little in the more advanced topic of infinite
sums and products, defined as limits ofcorresponding finite partial sums and
products, respectively:
k=1 =
and
00 n
H a], = lim H ak
n—>oo
k=1 k=1
We assume the reader to be familiar with the basic concepts of limit (and to
have the knowledge ofelementary limits), and (more rarely) ofderivative and
Riemann integral. There are, however, only few examples ofthis kind, and the
reader who is not familiar with these concepts can skip them without losing
the rest of the book. This clearly means that we expect the reader to know
basic arithmetic, algebra, and trigonometry (complex numbers in algebraic
