Finite sums

We begin with finite sums and products. While we have already used these con-cepts and the corresponding notations (the ∑ and ∏ symbols), let us nevertheless quickly recall some of their basic properties. We refer to [Grinbe15, §1.4] for a much more thorough (although probably not very exciting) survey of the properties of finite sums and products with detailed proofs, and to [GrKnPa94, Chapter 2] for a hands-on introduction to the art of manipulating them. Also, [AndTet18] is an entire book devoted to (mostly finite) sums and products.

This section will be a long yet (most likely) fast read, as much of it just will be spent stating rules and definitions that you have likely already encountered in your mathematical life history. Furthermore, almost all of these rules are just formalizing common sense about sums and products, so you won’t be surprised by them even if you see them for the first time. The rules will be illustrated by exercises, and some of the latter might actually be surprising. Thus, skim this section, but do not skip the exercises!

4.1.1. The ∑ sign

The most well-known type of finite sums are those of the form ∑^v

i=u

a_i (for example,

10∑

i=2

i²). But let us first define finite sums of a more general shape (from which we shall later obtain the ∑^v

i=u

a_i kind as a special case):⁶⁰

Definition 4.1.1. If S is a finite set, and if as is a number for each s ∈ S, then

s∑∈S

as denotes the sum of all of these numbers as. Formally, this sum is defined by recursion on |S|, as follows:

• If |S| = 0, then ∑

s∈S

as is defined to be 0. (In this case, ∑

s∈S

as is called an empty sum.)

• Let n ∈ N. Assume that we have defined ∑

s∈S

as for every finite set S with

|S| = n (and every choice of numbers as). Now, if S is a finite set with

|S| = n+1 (and if a number as is chosen for each s ∈ S), then ∑

s∈S

as is defined by picking any t∈ S and setting

s

∑

∈S

as =at+

∑

s∈S\{t}

as. (57)

It is not immediately clear why this definition is legitimate; in fact, the right hand side of (57) is defined using a choice of t, but theoretically one can imagine that different choices of t would lead to different results. Nevertheless, one can prove (see, e.g., [Grinbe15, Theorem 2.118] for this proof) that this definition is indeed legitimate (i.e., the right hand side of (57) does not depend ont). This is essentially saying that if we sum the numbers as for all s ∈ _S by starting with 0 and adding these numbers one by one, then the result does not depend on the order in which the numbers are being added. (Note that this is not true of multiplying matrices, or of putting on clothes, or of chemical reactions. So it’s not an obvious claim!)

Expressions of the form ∑

s∈S

as for finite sets S are called finite sums. Here are some examples:

60Throughout Section 4.1, “number” means an integer or a rational number or a real number or a complex number. More generally, the concept of a finite sum can be applied to any sort of object that has an addition operation with reasonable properties (such as commutativity and associativity); e.g., in linear algebra, one defines finite sums of vectors in a vector space, and in abstract algebra, one generalizes this further to finite sums of elements of an abelian group (or even monoid).

Example 4.1.2. (a) If S = {2, 6, 9} and as = 3^s for each s ∈ S, then ∑ theas with sranging overS” or “sum of theaswith srunning through all elements of S”. The letter “s” in the sum is called the “summation index”, and its exact choice is immaterial (for example, you can rewrite ∑

s∈S

a♠), as long as it does not already have a different meaning outside of the sum.

The sign ∑ itself is called “the summation sign” or “the ∑ sign”. The numbers as

are called the addends(or summandsorterms) of the sum ∑

s∈S

as. More precisely, for any givent ∈ S, we can refer to the numberat as the “addend corresponding to the index t” (or as the “addend fors =t”, or as the “addend fort”) of the sum ∑

s∈S

as. The summation index in a finite sum does not always have to be a single letter.

For instance, ifSis a set of pairs, then we can write ∑

(x,y)∈S

a(x,y) (meaning the same as ∑

s∈S

as). Here is an example of this notation:

∑

(here, we are using the notation ∑

(x,y)∈S

a_(x,y) with S={1, 2, 3}² and a_(x,y) = ^x y).

Warning 4.1.3. There is no agreement on the operator precedence of the ∑ sign versus the +sign. By this I mean the following question: Does ∑

s∈S

as+_b(where bis some other number) mean ∑

s∈S interpretation (i.e., I will read it as

s∑∈S

as

+b), but I have seen both used in the literature (although I believe that the second is more popular).

However, the · sign definitely has higher precedence than the∑ sign. That is, an expression of the form ∑

s∈S

basc is always understood to mean ∑

s∈S

(basc). As I mentioned above, the most common finite sums are of a special kind:

Definition 4.1.4. Let u and v be integers. We agree to understand the set {u,u+1, . . . ,v}to be empty if u>v.

Let as be a number for each s ∈ {_u,u+1, . . . ,v}. Then, the finite sum

s∈{u,u∑+1,...,v}

as will also be denoted by ∑^v

s=uas or by au+a_u+1+· · ·+av. Thus, finite sums of the form ∑^v

s=uas satisfy

∑

v s=u

as =0 for v<u, and (58)

∑

v s=u

as =au+

∑

v s=u+₁

as =

v−1 s

∑

=u

as+av forv≥u. (59) The equality (58) says that a sum that “ends before it begins” is 0. For example,

∑2 i=5

i =0. This is not completely uncontroversial⁶¹, but it shall be so in this course.

In a sum of the form

∑v

s=uas, the integeru is called thelower limit (orlower bound) of the sum, whereas the integer v is called the upper limit (or upper bound) of the sum.⁶² The sum is said to start(orbegin) at uand endatv.

Another way to use∑ signs is in describing sums that sum over all elements of a given set that satisfy a given statement. This is done as follows:

Definition 4.1.5. Let Sbe a finite set, and letA(s) be a logical statement defined for every s ∈ S. (For example, if S is a set of integers, then A(s) can be the statement “s is even”.) For each s∈ S satisfyingA(s), let as be a number. Then, the sum ∑

s∈S;

A(s)

as is defined by

s

∑

∈S;

A(s)

as =

∑

s∈{t∈S | A(t)}

as.

In other words, ∑

s∈S;

A(s)

as is the sum of the as for all s∈ Swhich satisfy A(s).

61I have seen authors who would “sum backwards” whenv <u, thus interpreting ∑²

i=5

ias−4−3.

This has some advantages (because why let a notation go to waste?), but it would mean that

∑v

s=uas cannot be rewritten as ∑

s∈{u,u+1,...,v}

as without checking the order relation betweenu and v, and I don’t want this extra headache.

62This has nothing to do with the notions of “bounds” and “limits” in analysis.

Example 4.1.6. If S = {1, 2, 3, 4, 5}, then ∑

s∈S;

sis even

as = a2+a₄ and ∑

s∈S;

sis odd

as = a₁+a3+a5.

4.1.2. The simplest rules: factoring out and splitting addends

Having defined the notations, let us recall some rules for manipulating finite sums.

We shall give no proofs (see [Grinbe15, §1.4] for them, or treat them as induction exercises), but we shall illustrate them with examples whenever reasonable.

We begin with four very simple rules. The first one ([Grinbe15, (6)]) says that if all addends in a sum are equal to some numbera, then the sum equals the number of addends timesa:

Theorem 4.1.7. Let Sbe a finite set. Let a be a number. Then,

s

∑

∈S

a=|S| ·a. (60)

This is trivial to prove by induction and needs no further explanation. Applying (60) toa =0 yields ∑

s∈S

0 =|S| ·0=0.

The next rule ([Grinbe15, (9)]) is just a distributive law for finite sums; it says that a factor common to all addends of a sum can be factored out:

Theorem 4.1.8. LetSbe a finite set. For every s∈ S, let as be a number. Also, let λbe a number. Then,

s

∑

∈S

λas =λ

∑

s∈S

as. (61)

Example 4.1.9. Let u and v be two integers. Let as be a number for each s ∈ {u,u+1, . . . ,v}. Also, let λ be a number. Then, Theorem 4.1.8 (applied to S = {u,u+1, . . . ,v}) yields

∑

v s=u

λas =λ

∑

v s=u

as. (62)

Equivalently,

λau+λau+1+· · ·+λav =λ(au+au+1+· · ·+av). (63) For example, if n∈ N, then the sum of the first n even positive integers is 2+4+· · ·+2n =2·1+2·2+· · ·+2·n

=2(1+2+· · ·+n)

by (63), applied to λ=2 and u=_{1 and}v=n and as =s

=2· ⁿ(n+1)

2 (by (9))

=n(n+₁).

The next rule ([Grinbe15, (7)]) says that if each addend in a sum is itself a sum of two numbers, then we can split the sum up:

Theorem 4.1.10. Let S be a finite set. For every s ∈ S, let as and bs be numbers.

(as a consequence of (60))

(by (64), applied to S ={1, 2, . . . ,n} and as = fs and bs =₁)

= fn+2−1+n·1= fn+2−1+n.

The equality (64) is often used forwards (i.e., in order to split a sum of the form

s∑∈S

(as+bs) into ∑

s∈S

as+ _∑

s∈S

bs) and backwards (i.e., in order to combine two sums

s∑∈S that the two sums being combined are ranging over the same set!

An easy consequence of Theorem 4.1.10 is the following:

Exercise 4.1.1. Let S be a finite set. For every s ∈ S, let as and bs be numbers.

Solution to Exercise 4.1.1. Theorem 4.1.10 (applied to as−bs instead of as) yields

s

∑

∈S Solving this equation for ∑

s∈S

4.1.3. Substitution of the summation index

The next rule ([Grinbe15, (12)]) is intuitively simple, yet highly useful:

Theorem 4.1.12. Let S and T be two finite sets. Let f : S → T be a bijective map⁶³. Let at be a number for eacht ∈ T. Then,

t

∑

∈T

at =

∑

s∈S

a_f(s). (65)

Roughly speaking, (65) holds because the sum ∑

s∈S

a_f(s) contains the same ad-dends as the sum ∑

t∈T

at. We say that the sum ∑

s∈S

a_f(s) is obtained from the sum

t∑∈T

at by substituting f (s) for t. (Conversely, ∑

t∈T

at is obtained from ∑

s∈S

a_f(s) by substituting t for f (s).)

Example 4.1.13. For any n∈ _{N, we have}

t∈{1,2,...,n

∑

}

t³ =

∑

s∈{−n,−n+1,...,−1}

(−s)³.

Indeed, this follows from (65), applied to S = {−n,−n+1, . . . ,−1}, T = {1, 2, . . . ,n}, f(s) =−s, and at =t³.

When substituting the index in a sum, it is perfectly okay to re-use the same letter for the new index. Thus, (65) can be rewritten as

s

∑

∈T

as =

∑

s∈S

a_f(s).

Theorem 4.1.12 has several well-known and oft-used consequences. The first one says that we can shift the index of a sum by any integerk:

Corollary 4.1.14. Let u, v and k be integers. Let at be a number for each t ∈ {u+k,u+k+_{1, . . . ,}v+k}. Then,

v+k t=

∑

u+k

at =

∑

v s=u

as+k.

Proof of Corollary 4.1.14. Let f denote the map

{u,u+1, . . . ,v} → {u+k,u+k+1, . . . ,v+k}, s 7→s+k.

63Recall: A map is said to be bijectiveif it is injective (i.e., one-to-one) and surjective (i.e., onto).

Bijective maps are also known as bijections or one-to-one correspondences. See, e.g., the detailed Wikipedia page for more about these kinds of maps.

This map f is easily seen to be well-defined and bijective⁶⁴. Thus, Theorem 4.1.12 (applied to S={u,u+1, . . . ,v} and T ={u+k,u+k+1, . . . ,v+k}) yields

t∈{u+k,u+

∑

k+1,...,v+k}

at =

∑

s∈{u,u+1,...,v}

a_f(s)

=|{z}a_s+k (since f(s)=s+k)

=

∑

s∈{u,u+1,...,v}

a_s+k.

In other words,

v+_k t=

∑

u+k

at =

∑

v s=u

as+k. This proves Corollary 4.1.14.

The next corollary says that sums can be turned around: i.e., we have

au+a_u+1+· · ·+av =av+a_v−1+· · ·+au. (66) Restated using ∑ signs, and generalized by shifting the index, this takes the fol-lowing form:

Corollary 4.1.15. Let u, v and k be integers. Let at be a number for each t ∈ {k−v,k−v+_{1, . . . ,}k−u}. Then,

k−u t=

∑

k−v

at =

∑

v s=u

a_k−s. (67)

Proof of Corollary 4.1.15. Let f denote the map

{u,u+1, . . . ,v} → {k−v,k−v+1, . . . ,k−u}, s7→ k−s.

This map f is easily seen to be well-defined and bijective⁶⁵. Thus, Theorem 4.1.12 (applied to S={u,u+1, . . . ,v} and T ={k−v,k−v+1, . . . ,k−u}) yields

t∈{k−v,k−

∑

v+1,...,k−u}

at =

∑

s∈{u,u+1,...,v}

a_f(s)

=|{z}a_k−s (since f(s)=k−s)

=

∑

s∈{u,u+1,...,v}

a_k−s.

In other words,

k−u t=

∑

k−v

at =

∑

v s=u

a_k−s. This proves Corollary 4.1.15.

Here is an example for the use of substitution in sums. Recall the formula (9), which is easily proved by induction. We shall now prove it in a more elegant way:

64Indeed, it has an inverse map, which sends eacht∈ {u+k,u+k+_{1, . . . ,}v+k}tot−k.

65Indeed, it has an inverse map, which sends eacht∈ {k−v,k−v+1, . . . ,k−u}tok−t.

Exercise 4.1.2. Let n∈ N. Prove the equality (9) without using induction.

Solution to Exercise 4.1.2. First of all, we notice that

∑

n

Thus, we shall focus on computing ∑ⁿ

i=0

(If you are too lazy to think about bijections, you can also derive this immediately by applying Corollary 4.1.15 to u = 0, v = n, k = n and at = t. This is the main purpose of Corollary 4.1.15: avoid the use of bijections.)

Now comes Gauss’s “doubling trick”: Recall that 2q = q+q for every q ∈ _Q.

Hence,

(here, we have used (64) backwards)

=

2 , which is precisely the equality (9). Hence, Exercise 4.1.2 is solved.

Let us remark how the computation (69) could be rewritten without the use of∑

signs:

2·(0+1+· · ·+n)

= (0+1+· · ·+n) + (0+1+· · ·+n)

= (0+1+· · ·+n) + (n+ (n−1) +· · ·+0)

(here, we have turned the second sum around using (66))

= (₀+_n)

| {z }

=n

+ (₁+ (_n−₁))

| {z }

=n

+· · ·+ (_n+₀)

| {z }

=n

(here, we have used (64) backwards)

=n+n+· · ·+n

| {z }

n+1 times

= (n+₁)n=n(n+₁).

66 This arguably looks simpler than (69), but it is clear that with more complicated sums it will become progressively harder to avoid the use of∑signs.

Gauss’s “doubling trick” used in the above solution is not a one-trick pony. Here is a more advanced use:

Exercise 4.1.3. Let n∈ _{N. Let} d be an odd positive integer. Prove that 1+2+· · ·+n |1^d+2^d+· · ·+n^d.

Solution to Exercise 4.1.3 (sketched). This is solved in detail in [19s, Exercise 2.10.8];

thus, we only give the skeleton of the argument. In view of (9), the claim that we

66We can also visualize this computation as a “picture proof”. For example, here is the picture for the casen=5:

5

1 4

2 3

3 2

4 1

5

n+1=6 n=₅

.

The addends of the sum 0+1+· · ·+nare drawn as purple rectangles, whereas the addends of the sumn+ (n−1) +· · ·+0 are drawn as red rectangles. (Rectangles of area 0, corresponding to 0 addends, are degenerate and thus invisible.) The rectangles are matched in such a way that the i-th column (counted from left, started with 0) has a purple rectangle of area i and a red rectangle of area n−i. This matching corresponds precisely to the way we matched our addends: (0+n)

| {z }

=n

+ (1+ (n−1))

| {z }

=n

+· · ·+ (n+0)

| {z }

=n

.

must prove rewrites as

n(n+1)

2 | 1^d+2^d+· · ·+n^d. This is equivalent to

n(n+1)| 2

1^d+2^d+· · ·+n^d

(71) (by Proposition 3.1.5). Hence, it suffices to prove (71).

In order to prove (71), it suffices to show that n| 2

1^d+2^d+· · ·+n^d

and (72)

n+₁| ₂₁^d+₂^d+· · ·+n^d

. (73)

Indeed, Example 3.5.2 (a) (applied to a = n) yields that n is coprime to n+_{1; in} other words, n⊥n+1. Hence, if we can prove (72) and (73), then we will get (71) by applying Theorem 3.5.7.

We shall prove (73) first:

[Proof of (73): We have 2

1^d+2^d+· · ·+n^d

=1^d+2^d+· · ·+n^d

+1^d+2^d+· · ·+n^d

=₁^d+₂^d+· · ·+n^d

+n^d+ (n−₁)^d+· · ·+₁^d

=

∑

n k=1

k^d+

∑

n k=1

(n+1−k)^d

| {z }

≡(−k)^dmodn+1

(by taking the congruencen+1−k≡−kmodn+1 to thed-th power)

≡

∑

n k=1

k^d+

∑

n k=1

(−k)^d =

∑

n k=1

k^d+ (−k)^d

| {z }

=0

(sincedis odd, so that(−k)^d=−k^d)

=

∑

n k=1

0=0 modn+1.

In other words,n+1| 2 1^d+2^d+· · ·+n^d

. Thus, (73) is proven.]

[Proof of (72):Ifn=0, then (72) boils down to 0| 2·0, which is obvious. Thus, for the rest of this proof, we WLOG assume that n6=_{0. Hence,}n−₁∈ N. Therefore, we can apply (73) to n−1 instead ofn(since we have already proven (73) for each n∈ N). We thus obtain

n |2

1^d+2^d+· · ·+ (n−1)^d.

In other words, 2

1^d+2^d+· · ·+ (n−1)^d ≡ 0 modn. Also, d > 0 (since d is odd), so thatn |n^d and thusn^d ≡0 modn. Now,

2

1^d+2^d+· · ·+n^d

=2



1^d+2^d+· · ·+ (n−1)^d+ n^d

≡0 mod|{z}n





≡2

1^d+2^d+· · ·+ (n−1)^d≡0 modn.

That is,n| 2 1^d+2^d+· · ·+n^d

. This proves (72).]

We have now proven both (72) and (73). As we have explained, this yields (71), which in turn solves Exercise 4.1.3.

4.1.4. The telescope principle

The next summation rule is the so-calledtelescope principle([Grinbe15, (16)]), which has a number of simple yet elegant applications:

Theorem 4.1.16. Let u and v be two integers such that u−1 ≤ v. Let as be a number for each s∈ {u−_1,u, . . . ,v}. Then,

∑

v s=u

(as−a_s−1) = av−a_u−1. (74)

Intuitively, the claim of Theorem 4.1.16 is obvious: The left hand side of (74) is

∑

v s=u

(as−a_s−1)

| {z }

=−as−1+as

=

∑

v s=u

(−a_s−1+as)

= (−au−1+au) + (−au+au+₁) + (−au+₁+au+2) +· · ·+ (−av−1+av). If we expand the right hand side of this equality, then it “contracts like a telescope”

(thus the name of Theorem 4.1.16): All addends except for the −au−1 and the av

cancel each other, and you are left with −au−1+av = av−au−1, which is exactly the right hand side of (74). (Be careful with this argument, though: It does not work for v = u−1. But this case is trivial anyway.) This intuitive argument can be formalized. Alternatively, Theorem 4.1.16 can be proved by induction onv (the proof is utterly straightforward), or derived from Exercise 4.1.1.

Note that Theorem 4.1.16 can be regarded as a discrete version of the Second Part of the Fundamental Theorem of Calculus. In fact, the latter fact says that Rv

u F⁰(x)dx = F(v)−F(u) for any differentiable function F on a (real) interval [u,v]. But the ∑ sign is a discrete analogue of theR

sign, whereas the consecutive differencesas−as−1 are discrete analogues of the values of the derivativeF⁰.

For all its simplicity, Theorem 4.1.16 is a surprisingly helpful tool for simplifying sums. The next four exercises are examples of this:

Exercise 4.1.4. Let n ∈ N. Simplify the sum ∑ⁿ

i=1

i·i! (that is, rewrite it without using the ∑sign).

Solution to Exercise 4.1.4. The trick is to realize that each i ∈_Nsatisfies

i·i! = (i+1)!−i! (75)

(because (i+1)! = (i+1)·i!=i·i!+i!). Thus,

∑

n i=₁

i·i!

|{z}

=(i+1)!−i!

(by (75))

=

∑

n i=₁

((i+1)!−i!) =

n+1 s

∑

=2







(s−1) +1

| {z }

=s



!−(s−1)!





(here, we have substituteds fori+1 in the sum)

=

n+1 s

∑

=₂

(s!−(s−1)!) = (n+1)!−(2−1)!

| {z }

=1!=1

by Theorem 4.1.16, applied tou =2, v =n+1 andas =s!

= (n+1)!−1. (76)

This solves Exercise 4.1.4.

Before we move on to the next example, let us observe what we have gained from the telescope principle. This equality (76), once it has been found, can easily be proved by induction on n (see [19s-hw0s, solution to Exercise 2 (b)] for such a proof); but the telescope principle has helped us find this equality in the first place (once we had the fancy to rewrite i·i! as (i+₁)!−i!). Thus, the telescope princi-ple turns the (often difficult) question of simplifying a sum into an (often simprinci-pler) question of rewriting its addends as differences.⁶⁷ Of course, it is not magic: Any proof using the telescope principle can easily be rewritten as an induction proof, because the telescope principle itself is easily proved by induction. The main ad-vantage of the principle is its convenience.

In future proofs, we shall be less detailed than in our above solution to Exercise 4.1.4, and simply say “by the telescope principle” instead of specifying what we are applying Theorem 4.1.16 to.

The next exercise is about generalizing the formula (29):

Exercise 4.1.5. Let pbe a positive integer, and let n∈ N. Simplify the sum

∑

n i=1

1

i(i+₁) (i+₂)· · ·(i+p)^.

67This is similar to computing an integral by finding an antiderivative of the function under the integral sign.

Solution to Exercise 4.1.5. We define

ai := ¹

(i+1) (i+2)· · ·(i+p) for eachi ∈N. Then, it is easy to see that

1

i(i+1) (i+2)· · ·(i+p) = ^ai−1 p −^ai

p (77)

for any positive integeri.

Indeed, ifiis a positive integer, then a_i−1

|{z}

= 1

i(i+1)· · ·(i+p−1)

(by the definition ofa_i−1)

− a_i

|{z}

= 1

(i+1) (i+2)· · ·(i+p)

(by the definition ofai)

= ¹

i(i+1)· · ·(i+p−1)

| {z }

= i+p

i(i+1)· · ·(i+p−1)·(i+p)

− ¹

(i+1) (i+2)· · ·(i+p)

| {z }

= i

i·(i+1) (i+2)· · ·(i+p)

= ⁱ+p

i(i+1)· · ·(i+p−1)·(i+p)− ⁱ

i·(i+1) (i+2)· · ·(i+p)

= ⁱ+p

i(i+1) (i+2)· · ·(i+p)− ⁱ

i(i+1) (i+2)· · ·(i+p)

= (i+p)−i

i(i+1) (i+2)· · ·(i+p) = ^p

i(i+1) (i+2)· · ·(i+p) ⁽⁷⁸⁾ and thus

a_i−1

p − ^ai p = ¹

p(a_i−1−a_i) = ¹

p · ^p

i(i+1) (i+2)· · ·(i+p) (by (78))

= ¹

i(i+1) (i+2)· · ·(i+p)^; this proves (77).

Now,

p (by the telescope principle)

= ¹

Note that Exercise 4.1.5 has no good answer for p = 0. Indeed, the sum ∑ⁿ

i=1

1 i cannot be simplified. (It is known as then-th harmonic number.)

Note also that, as an easy consequence of (79), we have

∑

∞ for any positive integer p. This also “holds” for p=0, in the sense that the infinite series ∑^∞

i=₁

1

i (known as the harmonic series) diverges (which is precisely what one would expect from seeing the 1

p = ¹

0 term on the right hand side of (80)). This is a well-known fact, with several proofs on the Wikipedia.

To facilitate its future application, let us restate Theorem 4.1.16 in a form that will allow us to apply it without substituting indices:

Corollary 4.1.17. Let u and v be two integers such that u−1 ≤ v. Let a_i be a (i.e., formally speaking, applied Corollary 4.1.14)

Corollary 4.1.17 is just Theorem 4.1.16 restated with a shifted index; thus, we shall still refer to it as the “telescope principle”.

Exercise 4.1.6. Let n∈ N. Simplify the sum ∑ⁿ

i=1

√ 1 i+√

i+1.

Solution to Exercise 4.1.6. If x and yare two distinct positive reals, then

√x−√

This formula might be well-known from high school (where it is used to rationalize denominators).

Now, for each positive reali, we have

√ 1

Hence,

(by the telescope principle⁶⁸). This simplifies further to √

n+1−1.

Finally, here is what might be the simplest application of the telescope principle;

we state it mainly because of its usefulness:

Exercise 4.1.7. Let a andb be any numbers. Letm ∈ _{N. Then,} (a−_b)

m−1 i

∑

=0

aⁱb^m−1−i =_a^m−_b^m_{. (82)}

Solution to Exercise 4.1.7. From (61), we obtain

(a−b)

by the telescope principle (Corollary 4.1.17, applied to u=_0, v=m−_{1 and}a_i =aⁱb^m−i)

!

=a^m−b^m.

We notice that Exercise 4.1.7 generalizes Exercise 2.1.1 and also yields a rather explicit new proof of Proposition 3.2.7 (check it!).

4.1.5. Splitting a sum into two

Another general rule for sums is the following ([Grinbe15, (3)]):

Theorem 4.1.18. Let Sbe a finite set. Let X andY be two subsets of Ssuch that X∩Y=_∅and X∪Y=S. (Equivalently,X andY are two subsets ofSsuch that

68specifically, by Corollary 4.1.17, applied tou=1,v=nanda_i =√ i

each element of S lies in exactly one of X and Y.) Let as be a number for each

Here, as we explained, ∑

s∈X Behind Theorem 4.1.18 stands the intuitively obvious fact that a sum ∑

s∈S

as can be computed by first sorting its addends into two “heaps” ∑

s∈X

as and ∑

s∈Y

as, then summing each heap separately, and finally adding the two heap sums together. For example, this says that

a₁+a2+a3+a₄+a5 = (a₁+a₄) + (a2+a3+a5).

(This is the particular case of (83) forS ={1, 2, 3, 4, 5},X ={1, 4}andY={2, 3, 5}.) Most commonly, Theorem 4.1.18 is used to split a sum of the form ∑^w

s=uas into two

Proof of Corollary 4.1.19. Recall the assumption u−1 ≤ v ≤ w. Thus, the sets {u,u+_{1, . . . ,}v} and {v+_1,v+_{2, . . . ,}w} are subsets of the set {u,u+_{1, . . . ,}w} Hence, Corollary 4.1.19 is proved.

Here are two examples of sums getting split:

Exercise 4.1.8. Let n∈ _N.

Solution to Exercise 4.1.8 (sketched). (a) Corollary 4.1.19 (applied to u = −n, v = 0, w=n and as =|s|) yields

(here, we have substitutedtfor−s in the sum, since the

map{1,2,...,n}→{−n,−n+1,...,−1}, s7→−s

here, we have renamed the summation index t assin the first sum be the set of all odd elements ofS. Then, each element ofSlies inexactlyone ofX andY(since each element of Sis either even or odd, but not both). In other words, X∩Y = _∅ and X∪Y = S. Therefore, we can apply Theorem 4.1.18 to as = s%2.

Let us now simplify the right hand side. Ifs ∈ X, thensis even (by the definition of X) and thus satisfies s%2 =0 (by Exercise 3.3.2(a)). Hence, ∑

s∈X

On the other hand, if s∈ Y, thens is odd (by the definition ofY) and thus satisfies s%2 =1 (by Exercise 3.3.2 (b)). Hence, ∑

69Strictly speaking, this needs to be proved. It can be easily proved by induction onn, or using Theorem 3.1.8.

Thus, |Y|=n. Hence, ∑

sign on the left hand side as

2n−1

s∑=₀. Thus, (86) rewrites as ²ⁿ

−1 s∑=₀

(s%2) = n.

The following is just a restatement of Theorem 4.1.18 using the notation from Definition 4.1.5:

(The right hand side is just a shorter way to rewrite the middle hand side; there is nothing profound happening here.)

4.1.6. Splitting a sum into several

Theorem 4.1.18 has a rather natural generalization ([Grinbe15, (26)]), in which a sum is split intoninstead of 2 smaller sums:

Theorem 4.1.21. Let S be a finite set. Let S1,S2, . . . ,Sn be finitely many subsets

Theorem 4.1.18 is the particular case of Theorem 4.1.21 obtained if we setn =2, S₁ =X and S₂=Y.

As an example for the use of Theorem 4.1.21, here is a pretty obvious exercise:

Exercise 4.1.9. Let I be a finite set of integers. For each pair(i,j) ∈ I×I, leta_(i,j) be a number. Prove that

(i,j)∈

∑

I×I

a_(i,j) =

∑

i∈I

a_(i,i)+

∑

(i,j)∈I×I; i<j

a_(i,j)+a_(j,i) .

Solution to Exercise 4.1.9. Let S be the set I×I. Define three subsets S1, S2 and S3

ofSby

S1 ={(i,j) ∈ I×I | i < j}, S2 ={(i,j) ∈ I×I | i = j}, S3 ={(i,j) ∈ I×I | i > j}.

Then, every element of S lies in precisely one of the subsets S1,S2,S3. In other words, the three subsetsS₁,S₂,S₃are pairwise disjoint and their union isS. Hence, (88) (applied to n=3) yields

s

∑

∈S

as =

∑

s∈S₁

as+

∑

s∈S₂

as+

∑

s∈S₃

as.

Renaming the indexs as(i,j)in each sum of this equality, we obtain

(i,j

∑

)∈S

a_(i,j) =

∑

(i,j)∈S₁

a_(i,j)+

∑

(i,j)∈S2

a_(i,j)+

∑

(i,j)∈S3

a_(i,j). (89) We shall now take a closer look at the sums on the right hand side.

The setS2 = {(i,j) ∈ I×I | i= j} consists of the pairs (i,i) for all i ∈ I. More precisely, there is a bijection

I →S₂, i7→ (i,i).

Thus, we can substitute (i,i) for (i,j) in the sum ∑

(i,j)∈S₂

a_(i,j). This sum therefore rewrites as follows:

(i,j

∑

)∈S₂

a(i,j) =

∑

i∈I

a(i,i). (90)

Furthermore, there is a bijection between the two setsS1={(i,j) ∈ I×I | i <j} and S₃={(i,j)∈ I×I | i >j}. Namely, the map

S1 →S3, (i,j) 7→(j,i)

is a bijection (because if (i,j) ∈ I×I satisfies i < j, then j > i, and vice versa).

The following theorem ([Grinbe15, (22)]) provides a more elaborate way of split-ting a sum:

The idea behind the formula (92) is the following: The left hand side is the sum of all as for s ∈ S. The right hand side is what you get if you first subdivide the

The idea behind the formula (92) is the following: The left hand side is the sum of all as for s ∈ S. The right hand side is what you get if you first subdivide the

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