Week 2: Sequences and Series

QF0: Quantitative Finance August 29, 207 Week 2: Sequences and Series Facilitator: Christopher Ting AY 207/208 Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate. Leonhard Euler 2. What is a sequence?. Let A be a set of objects. A sequence {a i } i= of elements of A is a function from the set of natural numbers N to the set A, i.e. a correspondence that associates one and only one element of A to each natural number n N. In other words, a sequence of elements of A is an ordered list of elements of A, where the ordering is provided by the natural numbers. 2. Thus if {a i } i= is a sequence, a is the first element, a 2 is the second element, and so on. Example: Define a sequence {a i } i= by characterizing its n-th element a n as a n = n. So it is a sequence of rational numbers: a =, a 2 = /2,.... Example: Define a sequence {a i } i= where each a n is a closed interval of real numbers, a n = [ n +, ], n which is a set of real numbers that are bounded by [ and inclusive ] [ of /(n + ) and /n. The elements of this sequence are closed intervals: a = 2,, a 2 = 3, ], and so on. 2 Example: Fibonacci sequence The Fibonacci sequence comprises the numbers in the following integer sequence: 0,,, 2, 3, 5, 8, 3, 2, 34, 55, 89, This sequence is generated by the recurrence relation, with F 0 = 0 and F = : F n = F n + F n 2. 2-

Week 2: Sequences and Series 2-2 2.2 Countable and uncountable sets. Let A be a set of objects. A is a countable set if all its elements can be arranged into a sequence, i.e. if there exists a sequence {a i } i= such that a A, n N : a n = a. In other words, A is a countable set if there exists at least one sequence {a i } i= such that every element of A belongs to the sequence. A is an uncountable set if such a sequence does not exist. The most important example of an uncountable set is the set of real numbers R. Example: The set of rational numbers Q is countable. Example: The set of irrational numbers is uncountable. 2. An uncountable sets are so densely populated that make counting impossible. 3. In mathematics, the cardinality of a set A is a measure of the number of elements of the set, and it is denoted by A. The cardinality of the natural numbers is denoted by ℵ 0. 4. A set A is uncountable if the cardinality of A is neither finite nor equal to ℵ 0, i.e., A, and A ℵ 0. 2.3 Limit of a sequence. First, we define the distance between two real numbers. It is the absolute value of their difference. For example, if a R and a n is a term of a sequence {a n }, the distance between a n and a, denoted by d(a n, a) is d(a n, a) = a n a. 2. Let {a n } be a sequence of real numbers. Let n 0 N. Denote by {a n } n>n0 a subsequence (subset of a sequence) of {a n } obtained by dropping the first n 0 terms, i.e., {a n } n>n0 = { a n0 +, a n0 +2,..., }. 3. Let a be a real number. We say that a is a limit of a sequence {a n } of real numbers, if, by appropriately choosing n 0, the distance between a and any term of the subsequence {a n } n>n0 can be made as close to zero as we like.

Week 2: Sequences and Series 2-3 4. More formally, let a R. We say that a is a limit of a sequence {a n } of real numbers, if ɛ > 0, n 0 N : d(a n, a) < ɛ, n > n 0. Namely, for any arbitrarily small number ɛ, there exists a natural number n 0 such that the distance between a n and a is less than ɛ for all the terms a n with n > n 0. In other words, for any arbitrarily small number ɛ, one can find a subsequence {a n } n>n0 such that the distance between any term of the subsequence and a is less than ɛ. By dropping a sufficiently high number of initial terms of {a n }, one can make the remaining terms as close to a as one wishes. 5. If a is a limit of the sequence {a n }, we say that the sequence {a n } is a convergent sequence and that it converges to a. We indicate the fact that a is a limit of {a n } by a = lim n a n. Example: Define a sequence {a n } by characterizing its n-th element a n as follows: a n = n. Intuitively, the limit of the sequence should be 0: 0 = lim n a n. To prove it, choose any ɛ > 0 and find an n 0 N such that all terms of the subsequence {a n } n>n0 have distance from zero less than ɛ: d(a n, 0) < ɛ, n > n 0. Note first that the distance between a generic term of the sequence a n and 0 is since each a n is positive. d(a n, 0) = a n 0 = a n = a n, We need to find an n 0 N such that all terms of the subsequence {a n } n>n0 satisfy a n < ɛ, n > n 0. Since a n < a n0 for all n > n 0, this condition is satisfied if a n0 < ɛ, which is equivalent to n 0 < ɛ. Therefore, it suffices to pick any n 0 such that n 0 > ɛ to satisfy the condition d(a n, 0) < ɛ, n > n 0.

Week 2: Sequences and Series 2-4 Thus, we have just shown that, for any ɛ, we are able to find n 0 N such that all terms of the subsequence {a n } n>n0 have distance from zero less than ɛ. It then follows that 0 is the limit of the sequence {a n }. Example: Kepler observed that the ratio of consecutive Fibonacci numbers converges. The limit is a number known as the golden ratio ϕ: F n+ ϕ = lim. n F n It is intriguingly remarkable that the sequence {2/, 3/2, 5/3, 8/5, 3/8, 2/3, 34/2,...}, which are the ratios of Fibonacci numbers, converges to ϕ, whose value is.6803398874. 2.4 The limit of a sequence in general. We now deal with the more general case where the terms of the sequence are not necessarily real numbers. We need a function d : A A R that associates to any couple of elements of A a real number measuring how far these two elements are. For example, if a and a are two elements of A, d(a, a ) needs to be a real number measuring the distance between a and a. 2. A function d : A A R is considered a valid distance function (and it is called a metric on A) if it satisfies some properties. 3. Let A be a set of objects. Let d : A A R. d is considered a valid distance function (in which case it is called a metric on A) if, for any a, a and a belonging to A:. Non-negativity: d(a, a ) 0; 2. Identity of indiscernibles: d(a, a ) if and only if a = a ; 3. Symmetry: d(a, a ) = d(a, a); 4. Triangle inequality: d(a, a ) + d(a, a ) d(a, a ). 4. All four properties are very intuitive. Property () says that the distance between two points cannot be a negative number. Property (2) says that the distance between two points is zero if and only if the two points coincide. Property (3) says that the distance from a to a is the same as the distance from a to a.

Week 2: Sequences and Series 2-5 Property (4) says that the distance one covers when one goes from a to a directly is less than (or equal to) the distance one covers when one goes from a to a passing through a third point a. In other words, if a is not on the way from a to a, the distance to be covered increases. 5. Whenever we are faced with a sequence of objects and we want to assess whether it is convergent, we need to first define a distance function on the set of objects and verify that the proposed distance function satisfies all the properties of a proper distance function. 6. Having defined the concept of a metric, we are now ready to state the formal definition of a limit of a sequence. Definition: Let A be a set of objects. Let d : A A R be a metric on A. We say that a A is a limit of a sequence {a n } of objects belonging to A, if: ɛ > 0, n 0 N : d(a n, a) < ɛ, n > n 0. If a is a limit of the sequence {a n }, we say that the sequence is a convergent sequence and that it converges to a, and write a = lim n a n. 7. A convergent sequence has a unique limit, i.e., if {a n } has a limit a, then a is the only limit of {a n }. Proof: (by contradiction) Suppose that a and a are two limits of a sequence {a n } and a a. By combining property ) and property 2) of a metric, it must be that d(a, a ) = d > 0. Pick any term a n of the sequence. By property 4) of a metric (the triangle inequality), we have d(a, a n ) + d(a n, a ) d(a, a ). It follows that d(a, a n ) + d(a n, a ) d > 0. Now, take any ɛ < d. Since a is a limit of the sequence, we can find n 0 such that d(a, a n ) < ɛ, n > n 0, which means that ɛ + d(a n, a ) d(a, a n ) + d(a n, a ) d > 0, n > n 0, and d(a n, a ) d ɛ > 0, n > n 0.

Week 2: Sequences and Series 2-6 Therefore, d(a n, a ) cannot be made smaller than d ɛ and consequently, a cannot be a limit of the sequence. 8. In practice, it is usually difficult to assess the convergence of a sequence using the above definition. Instead, convergence can be assessed using the following criterion: Lemma (criterion for convergence) Let A be a set of objects. Let d : A A R be a metric on A. Let {a n } be a sequence of objects belonging to A and a A. Then {a n } converges to a if and only if Proof: lim d(a n, a) = 0. n This is easily proved by defining a sequence of real numbers {d n } whose generic term is d n = d(a n, a). Note that the definition of convergence of {a n } to a, which is ɛ > 0, n 0 N : d(a n, a) < ɛ, n > n 0. Equivalently, ɛ > 0, n 0 N : d n 0 < ɛ, n > n 0. This is none other than the definition of the convergence of {d n } to 0. 9. So, in practice, the problem of assessing the convergence of a generic sequence of objects is simplified as follows:. Find a metric d(a n, a) to measure the distance between the terms of the sequence a n and the candidate limit a; 2. Define a new sequence {d n }, where d n = d(a n, a); 3 Study the convergence of the sequence {d n }, which is a simpler problem because {d n } is a sequence of real numbers. 2.5 Convergence of a monotone sequence of real numbers. A sequence {a n } of numbers is said to be monotone when a n a n+ for all n.

Week 2: Sequences and Series 2-7 2. If {a n } is a monotone sequence of real numbers, then this sequence has a finite limit if and only if the sequence is bounded. We shall prove that if an increasing sequence {a n } is bounded above, then it is convergent and the limit is sup{a n }. n Since {a n } is non-empty and by assumption, it is bounded above, then, by the least upper bound property of real numbers, c = sup n {a n } exists and is finite. Now for every ε > 0, there exists a n0 such that a n0 > c ε, since otherwise c ε is an upper bound of {a n }, which contradicts to c being sup n {a n }. Then since {a n } is increasing, n > n 0, c a n = c a n c a N < ε, hence by definition, the limit of {a n } is sup n {a n }. 2.6 The Bolzano-Weierstrass theorem. The Bolzano-Weierstrass theorem is a fundamental result about convergence in R. The theorem states that each bounded sequence in R has a convergent subsequence. 2. Lemma: Every sequence {x n } in R has a monotone subsequence. Proof: Let us call a positive integer n a peak of the sequence if m > n implies x n > x m i.e., if x n is greater than every subsequent term in the sequence. Suppose first that the sequence has infinitely many peaks, n < n 2 < < n j < Then the subsequence {x nj } corresponding to peaks is monotonically decreasing, and a monotone subsequence is found. Suppose next that there are only finitely many peaks, let N be the last peak and n = N +. Then n is not a peak, since n > N, which implies the existence of an n 2 > n with x n2 x n. Again, n 2 > N is not a peak, hence there is n 3 > n 2 with x n3 x n2. Repeating this process leads to an infinite non-decreasing subsequence x n x n2 x n3. In this way, a monotone subsequence { } x ni is constructed. 3. The proof of Bolzano-Weierstrass theorem is to apply this lemma. Suppose we have a bounded sequence in R; by the Lemma there exists a monotone subsequence, necessarily bounded. It follows from the earlier result (a monotone and bounded sequence of real numbers has a finite limit) this subsequence of real numbers, being monotone and bounded, must converge.

Week 2: Sequences and Series 2-8 2.7 Bounded theorem. The boundedness theorem states that a continuous function f in the closed interval [a, b] is bounded on that interval. That is, there exist real numbers m and M such that: m f(x) M for all x [a, b]. Proof: (by contradiction) Suppose the function f is not bounded above on the interval [a, b]. Then, by the property of the real numbers, for every natural number n, there exists an x n [a, b] such that f(x n ) > n. This defines a sequence {x n }. Because [a, b] is bounded, the Bolzano-Weierstrass theorem implies that there exists a convergent subsequence {x nk } of {x n }. Denote its limit by x. As [a, b] is closed, it contains x. { Because f is continuous at x, we know that f ( ) } x nk converges to the real number f(x) (as f is sequentially continuous at x.) On the other hand, f ( x nk ) > nk k for every k, which implies that +, a contradiction. Therefore, f must be bounded above on [a, b]. The proof that f attains its lower bound is similar. { f ( x nk ) } diverges to 2.8 Extreme value theorem. The extreme value theorem states that if a real-valued function f is continuous in the closed and bounded interval [a, b], then f must attain its maximum and minimum value, each at least once. That is, there exist numbers c and d in [a, b] such that f(c) f(x) f(d) for all x [a, b]. Proof: We look at the proof for the upper bound and the maximum of f. By applying these results to the function f, the existence of the lower bound and the result for the minimum of f follows. Also note that everything in the proof is done within the context of the real numbers. By the boundedness theorem, f is bounded from above, and by the least-upper-bound property of real numbers, the least upper bound (supremum) M of f exists. It is necessary to find a d in [a, b] such that M = f(d). Let n be a natural number. As M is the least upper bound, M /n is not an upper bound for f. Therefore, there exists d n in [a, b] so that M /n < f ( ) d n. This defines a sequence {dn }.

Week 2: Sequences and Series 2-9 Since M is an upper bound for f, we have M /n < f(d n ) M for all n. Therefore, the sequence { f(d n ) } converges to M. The Bolzano-Weierstrass theorem tells us that there exists a subsequence { d nk }, which converges to some d and, as [a, b] is closed, d is in [a, b]. Since f is continuous at d, the sequence { f(dnk ) } converges to f(d). But { f ( d nk ) } is a subsequence of {f(d n )} that converges to M, so M = f(d). Therefore, f attains its supremum M at d. 2.9 What is a series?. A series is an infinite ordered set of terms combined together by the addition operator. The term infinite series is sometimes used to emphasize the fact that series contain an infinite number of terms. 2. The order of the terms in a series can matter. By a suitable rearrangement of terms, a so-called conditionally convergent series may be made to converge to any desired value, or to diverge. 3. If the difference between successive terms of a series is a constant, then the series is said to be an arithmetic series. A series for which the ratio of each two consecutive terms a k+ /a k is a constant function of the summation index k is called a geometric series. The more general case of the ratio is a rational function of k produces a series called a hypergeometric series. 4. A series may converge to a definite value, or may not, in which case it is called divergent. Let the terms in a series be denoted a i, let the k-th partial sum be given by S k = k a i, i= and let the sequence of partial sums be given by {S = a, S 2 = a + a 2, S 3 = a + a 2 + a 3,...}. If the sequence of partial sums converges to a definite value, the series is said to converge. On the other hand, if the sequence of partial sums does not converge to a limit (e.g., it oscillates or approaches ± ), the series is said to diverge. Example: The Maclaurin series for a geometric series. Namely, for any number x whose absolute value x < is x x = x n = + x + x 2 +. n=0

Week 2: Sequences and Series 2-0 Example: An example of a convergent series is the geometric series. n=0 ( ) n = 2. 2 This is a special case of the Maclaurin series with x = /2. Example: An example of a divergent series is the harmonic series. n =. Example: Interestingly, while the harmonic series diverges to infinity, the alternating harmonic series converges to the natural logarithm of 2, ( ) n n = ln 2. Example: The series representation of e is e = where n! is the factorial, which is a compact symbol to denote n (n ) 2. Example: Show that the series n 2 converges. n=0 n!. S n = 2 + 2 2 + 3 2 + + n 2 < + 2 + 3 + + n(n ) ( = + ) ( + 2 2 ) ( + + 3 n ) n = 2 n < 2. Since the series is bounded by 2 for any n, it is necessarily convergent. In fact, Euler showed that n 2 = π2 6. 5. A series of terms a n is said to be absolutely convergent if the series formed by taking the absolute values of the a n, i.e., n a n, converges. 6. An especially strong type of convergence is called uniform convergence, and series which are uniformly convergent have particularly nice properties. For example, the sum of a uniformly

Week 2: Sequences and Series 2- convergent series of continuous functions is continuous. A convergent series can be differentiated term by term, provided that the functions of the series have continuous derivatives and that the series of derivatives is uniformly convergent. Finally, a uniformly convergent series of continuous functions can be integrated term by term. 2.0 Exercises A. Suppose a n is a divergent series where each a n is strictly positive. Prove the following (a) The series (b) But the series B. Prove that the series a n + a n is also divergent. a n + n 2 a n converges. ( ) n converges. n C. Prove that if a decreasing sequence {a n } is bounded below, then it is convergent and the limit is inf n {a n}. D. Complete the extreme value theorem for the infimum m.