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Summations By Evan Chen
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SummationsEvan Chen
October 13, 2016
Mathematicians just love sigma notation for two reasons. First, it provides a convenient
way to express a long or even infinite series. But even more important, it looks really
cool and scary, which frightens nonmathematicians into revering mathematicians and
paying them more money.
— Calculus II for Dummies
Acknowledgments
Thanks to Zack Chroman, Michael Diao, Steven Hao, Ryan Kim, Kevin Qian, Colin
Tang, Michael Tang, Tyler Zhu, for helpful suggestions and comments.
Contents
1 Introduction 2
2 Algebraic manipulation 2
2.1 Telescoping and partial fractions . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Swapping the order of summation . . . . . . . . . . . . . . . . . . . . . . . . 3
2.3 Finite Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Sums modulo a prime 8
3.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4 Multiplicative number theory 10
4.1 Multiplicative functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4.2 Dirichlet convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.3 Möbius inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5 Generating functions 13
5.1 Toy applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.2 Linear recurrences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5.3 Common generating functions . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5.4 Snake Oil method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
6 Author’s Pick 19
7 Hints 21
1
,Evan Chen (October 13, 2016) Summations
§1 Introduction
This is a handout about how to deal with complicated sums. Broadly, sums on olympiad
contests fall into a few different categories:
• Purely algebraic sums, like n n(n+1) 1
P
, which are mostly exercise in algebraic
manipulation. These are the kind you would get in a lecture named “sums and
products”.
• Sums which are combinatorial, involving expressions like nk . You would see this
in lectures named “combinatorial sums” and “counting in two ways”, or perhaps
“generating functions”.
• Sums which are Pnumber theoretic, involving functions like ϕ, µ, or which involve
expressions like d|n , or which ask you to take a sum modulo p. Often appears in
“multiplicative number theory”.
I think it is preferable to avoid separating the learning of these apparently disjoint topics,
hence the unified handout. There are some connected themes which underlie all three; in
particular, one of the big ideas which runs through the entire handout is the concept of
swapping the order of summation.
Theorem 1.1 (Swapping the order of summation)
Let f (a, b) be a function. Then
XX XX
f= f.
a∈A b∈B b∈B a∈A
This seemingly obvious fact is a nontrivial step in more than 50% of summation problems
(in the same way that cyclic quadrilaterals is used in more than 50% of geometry problems).
We will see this key idea again and again in the examples that follow. Thus any time
you see a double sum, you should always consider P computing the sum in the reversed
order. In fact, even if you only have a single you should consider rewriting the
expression as a double sum (e.g. a generating function) so that the above theorem applies.
You might also need to change the variables before this step; for example we also have
XX X X
f= f.
a≥0 b≥0 k≥0 a,b
a+b=k
§2 Algebraic manipulation
These are the most low-tech problems, and often appear on short-answer contests as a
result. These techniques apply to all sums, and especially to those which don’t have N or
C flavor.
Examples of things you can do:
• Look at telescoping sequences. The classic example that everyone knows of is
100
X 1
.
n(n + 1)
n=1
2
, Evan Chen (October 13, 2016) Summations
• Generalizing the above point, looking at partial fractions is very often a good
thing to do if your expression has polynomial denominators.
• Try to factor the expression. In particular, one can factor through double sums:
! !
XX X X
a2 (b + 1) = a2 (b + 1) .
a∈A b∈B a∈A b∈B
• Try swapping the order of summation. (This is so important we’re saying it
again.)
§2.1 Telescoping and partial fractions
I won’t provide too many examples here (though there are plenty in the practice problems!)
since this is a topic that often appears at the AIME level anyways. So here is my token
example.
Example 2.1 (Stanford 2011)
Evaluate the sum
X 7n + 32 3 n
· .
n(n + 2) 4
n≥1
Solution. Again, since we have a polynomial denominator, our first reflex is to decompose
its partial fractions. This gives us
7n + 32 16 9
= − .
n(n + 2) n n+2
Suddenly we’re done, because the sum telescopes according to the fact that 9 = (3/4)2 · 16:
X 16 3 n 16
n+2
3
− .
n 4 n+2 4
n≥1
We see the trailing terms in the sum tend to zero as n → ∞. So the answer is
16 3 16 9 9 33
1 · 4 + 2 · 16 = 12 + 2 = 2 .
§2.2 Swapping the order of summation
For a less contrived example, we now prove linearity of expectation, which is the key
result from [2].
The motivating example which I always present is that this allows one to compute the
expected number of fixed points in a random permutation of {1, . . . , n}:
Example 2.2
A random permutation of {1, . . . , n} has an average of one fixed point.
To see this, let’s look at the case n = 4:
3
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