Discusión y conclusiones

Proof: We need to compare the quantities

AN ≡

We use the binomial theorem to expand BN : BN = 1 +N

54 CHAPTER 3. SERIES OF NUMBERS Notice that every summand that appears in this last equation is positive. Thus, for 0≤ M ≤ N, In this last inequality we hold M fixed and let N tend to infinity. The result is that

On the other hand, our expansion for BN allows us to observe that BN ≤ AN. Thus

lim sup

N →∞ BN ≤ e . (3.37.2)

Combining (3.37.1) and (3.37.2) we find that e≤ lim inf

N →∞ BN ≤ lim sup

N →∞

BN ≤ e

hence that lim_{N →∞}BN exists and equals e. This is the desired result.

Remark 3.38 The last proof illustrates the value of the concepts of lim inf and lim sup. For we do not know in advance that the limit of the expressions BN

exists, much less that the limit equals e. However, the lim inf and the lim sup always exist. So we estimate those instead, and find that they are equal and that they equal e.

The next result tells us how rapidly the partial sums AN of the series defin-ing e converge to e. This is of theoretical interest, but will also be applied to determine the irrationality of e.

Proposition 3.39 With AN as above, we have that 0 < e− A^N < 1

3.4. SOME SPECIAL SERIES 55 Now the expression in parentheses is a geometric series. It sums to (N + 1)/N . Since AN < e, we have

e− A^N =|e − A^N| hence

|e − A^N| < 1 N· N!, proving the result.

Next we prove that e is an irrational number.

Theorem 3.40 Euler’s number e is irrational.

Proof: Suppose to the contrary that e is rational. Then e = p/q for some positive integers p and q. By the preceding proposition,

0 < e− A^q < 1 q· q!

or

0 < q!· (e − A^q) <1

q. (3.40.1)

Now

e− A^q =p q −



1 + 1 + 1 2!+ 1

3!+· · · + 1 q!



hence

q!· (e − A^q)

is an integer. But then equation (3.40.1) says that this integer lies between 0 and 1/q. In particular, this integer lies strictly between 0 and 1. That, of course, is impossible. So e must be irrational.

It is a general principle of number theory that a real number that can be approximated too rapidly by rational numbers (the degree of rapidity being mea-sured in terms of powers of the denominators of the rational numbers) must be irrational. Under suitable conditions an even stronger conclusion holds: namely, the number in question turns out to be transcendental. A transcendental num-ber is one which is not the solution of any polynomial equation with integer coefficients.

The subject of transcendental numbers is explored in the exercises. The exercises also contain a sketch of a proof that e is transcendental.

In Exercise 5 of Section 2.4, we briefly discussed Euler’s number γ. Both this special number and also the more commonly encountered number π arise in many contexts in mathematics. It is unknown whether γ is rational or irrational.

The number π is known to be transcendental, but it is unknown whether π + e (where e is Euler’s number) is transcendental.

56 CHAPTER 3. SERIES OF NUMBERS In recent years, questions about the the irrationality and transcendence of various numbers have become a matter of practical interest. For these proper-ties prove to be useful in making and breaking secret codes, and in encrypting information so that it is accessible to some users but not to others.

In Appendix II we prove that

SN ≡ XN j=1

j =N· (N + 1)

2 .

We conclude this section with a method for summing higher powers of j.

Say that we wish to calculate

Sk,N ≡ XN j=1

j^k

for some positive integer k exceeding 1. We may proceed as follows: Write

(j + 1)^k+1− j^k+1 =

EXERCISES 57 We have succeed in expressing Sk,N in terms of S1,N, S2,N, . . . , S_k−1,N. Thus we may inductively obtain formulas for Sk,N, any k. It turns out that

S2,N = N (N + 1)(2N + 1) 6

S3,N = N²(N + 1)² 4

S4,N = (N + 1)N (2N + 1)(3N²+ 3N− 1) 30

These formulas are treated in further detail in the exercises.

Exercises

1. Use induction to prove the formulas provided in the text for the sum of the first N perfect squares, the first N perfect cubes, and the first N perfect fourth powers.

2. A real number s is called algebraic if it satisfies a polynomial equation of the form

a0+ a1x + a2x²+· · · + a^mx^m= 0

with the coefficients ajbeing integers and am6= 0. Prove that if we replace the word “integers” in this definition with “rational numbers” then the set of algebraic numbers remains the same. Prove that n^p/q is algebraic for any positive integers p, q, n.

* 3. Refer to Exercise 2 for terminology. A real number is called transcendental if it is not algebraic. Prove that the number of algebraic numbers is countable. Explain why this implies that the number of transcendental numbers is uncountable. Thus most real numbers are transcendental;

however, it is extremely difficult to verify that any particular real number is transcendental.

* 4. Refer to Exercises 2 and 3 for terminology. Provide the details of the following sketch of a proof that Euler’s number e is transcendental. [Note:

in this argument we use some simple ideas of calculus. These ideas will be treated in rigorous detail later in the book.] Seeking a contradiction, we suppose that the number e satisfies a polynomial equation of the form

a0+ a1x +· · · + a^mx^m= 0 with integer coefficients aj.

(a) We may assume that a06= 0.

58 CHAPTER 3. SERIES OF NUMBERS (b) Let p be an odd prime that will be specified later. Define

g(x) = x^p−1(x− 1)^p· · · (x − m)^p (p− 1)!

and

G(x) = g(x) + g⁽¹⁾(x) + g⁽²⁾(x) +· · · g^(mp+p−1)(x) . (Here parenthetical exponents denote derivatives.) Verify that

|g(x)| < m^mp+p−1 (p− 1)!

for a suitable range of x.

(c) Check that

d dx

e^−xG(x)

=−e^−xg(x) and thus that

aj

Z j 0

e^−xg(x)dx = ajG(0)− a^je^−jG(j) . (∗)

(d) Multiply the last equation by e^j, sum from j = 0 to j = m, and use the polynomial equation that e satisfies to obtain that

Xm j=0

aje^j Z j

0

e^−xg(x)dx =− Xm j=0

mp+p−1X

i=0

ajg⁽ⁱ⁾(j) . (∗∗)

(e) Check that g⁽ⁱ⁾(j) is an integer for all values of i and all j from 0 to m inclusive.

(f ) Referring to the last step, show that in fact g⁽ⁱ⁾(j) is an integer di-visible by p except in the case that j = 0 and i = p− 1.

(g) Check that

g^(p−1)(0) = (−1)^p(−2)^p· · · (−m)^p. Conclude that g^(p−1)(0) is not divisible by p if p > m.

(h) Check that if p >|a⁰| then the right side of equation (∗∗) consists of a sum of terms each of which is a multiple of p except for the term

−a⁰g^(p−1)(0). It follows that the sum on the right side of (∗∗) is a nonzero integer.