1 Sequences and Series

1 Sequences and Series We start with a review of some very basic notions. Definition 1.0.1. (a) A sequence fang of real numbers is a function N ! R defined by n 7! an. (b) fang is said to be non-decreasing " (non-increasing #) if a1 ≤ a2:::(a1 ≥ a2 ≥ a3::). In either case, fang is said to be a monotone sequence. (c) fang is said to be a bounded sequence if janj ≤ M < 1 for n ≥ 1. (d) A subsequence of fang is a restriction of the function to fnk : k ≥ 1; n1 < n2 < n3:::g. (e) We say that an ! l 2 R if given " > 0, there exists N such that jan − lj < " for n ≥ N and that an ! 1(−∞) if given M 2 R, there exists N such that an > M(an < −M) for n ≥ N. (f) A sequence fang is said to be a Cauchy sequence if given " > 0, there exists N such that jan − amj < " for n; m ≥ N. Exercises: 1. Show that (c) fails to hold if and only if there exists a subsequence of fang converging to ±∞. 1 n 2. Show that the sequence defined by an = (1 + n ) is monotone increasing and bounded. 1 1 1 3. Show that the sequence defined by an = n + n+1 + ::: + 2n−1 is # and hence convergent. Definition 1.0.2. If S ⊆ R, we say that a real number M is an upper(lower) bound of S if every x 2 S satisfies x ≤ M(x ≥ M) and M is said to be the least upper bound (lub) (greatest lower bound(glb)) if no number in (−∞;M)((M; 1)) is an upper bound (a lower bound). The completeness axiom Every S ⊂ R which is bounded above (below) has a lub in R. Exercise: A sequence which is " and bounded above converges to its lub. The following results are some of the crucial consequences of the completeness of R. Monotone Boundedness Theorem: Every bounded monotone sequence of real numbers converges to a limit in R. Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a monotone (and hence convergent) subsequence. Cauchy Criterion: Every Cauchy sequence of real numbers is convergent and conversely. Exercises: 1 n 1. Show that xn = (1 + n ) is convergent. Write: xn ! l. 1 n+1 Clearly, yn = (1 + n ) ! l; xn < yn and fyng #. Hence for every n; xn < l < yn. Use a calculator and a large value of n to estimate l. 1 1 2. Let tn = 2 + 2! + ::: + n! . Show that ftng " and tn < 3. Thus ftng is convergent. P1 1 Moreover, lim xn ≥ lim tn and thus xn ! n=0 n! . Fact: A function f :(a; b) ! R is continuous if and only if whenever an ! c 2 (a; b); f(an) ! f(c), provided (an); c ⊆ (a; b). 1 1.1 Sum of an infinite series Definition 1.1.1. 1. Given a sequence fang, let sn = a1 + ::: + an denote the sequence of partial P sums. If sn ! s 2 R, then we say that the series an converges to s. P 2. If tn = ja1j + +::: + janj converges to in R, then we say that an converges absolutely. P P 3. If an converges and tn ! 1, then we say that an converges conditionally. The simplest example of an conditionally convergent series is the alternating harmonic series P n 1 (−1) n . In modern calculus, divergence of harmonic series is shown by using the integral test and convergence of the alternating harmonic series is based on the alternating series test. Here’s a note of historical significance. 1. The simplest and most elegant proof of divergence of the harmonic series due to Orseme (Four- teenth century). 1 1 n n Observe that s2 > 2 and s2 − s2(n−1) > 2 . Thus s2 ! 1. 2. The alternating series test is due to Leibniz and perhaps its significant feature is that one can use it to estimate the error of approximation. Test: If u1 ≥ u2::: and un ! 0, then let an = s2n; bn = s2n+1. Also, fang "; fbng #. So bn ! β; an ! α and an − bn ! 0 implies α = β = s. Moreover, js − skj ≤ uk+1. 1 1 Exercise: How many terms do we need to add to estimate the sum of 1 − 2 + 3 ::: within an error of :01? P P Definition 1.1.2. Rearrangement of a convergent series an is a series aπ(n) where π : N ! N is a 1-1, onto function, or a permutation of N. The most crucial difference between a convergent and an absolutely convergent series is the following theorem, due to Riemann. Theorem: Any rearrangement of an absolutely convergent series converges to the same sum. A conditionally convergent series can be rearranged to converge to any number whatsoever. 1 1 1 1 1 1 1 1 Exercise: Show that 1 − 2 − 4 + 3 − 6 − 8 ::: = 2 [1 − 2 + 3 :::]. The following is used in discussing the convergence of series of positive terms. Exercises 1. If a ≥ 0, then lim(a )1=n exists if and only if lim an+1 exists and the two limits are equal. n n an 1 1 n 2. Find lim n (n!) as n ! 1. The following is an important extension of series of numbers. Weierstrass’s M-test: Suppose ffng is a sequence of functions defined on a common domain P P S ⊆ R. Asuume jfn(x)j ≤ Mn for x 2 S and n 2 N. If Mn < 1, then fn converges (uniformly) to a function f on S. 2 1.2 Euler’s number e 1 n 1 n+1 Let xn = (1 + n ) . Then xn+1 = (1 + n+1 ) . Pn n 1 Pn+1 n+1 1 By the binomial theorem, xn = k=0 k nk and xn+1 = k=0 k (n+1)k . 1 1 k 1 1 k The k-th term in xn is k! (1 − n ):::(1 − n ) and the k-th term in xn+1 is k! (1 − n+1 ):::(1 − n+1 ). k k k k Clearly, n+1 < n ) (1 − n+1 ) > (1 − n ). It follows that xn < xn+1 since xn+1 has one extra term. 1 1 Let tn = 2 + 2! + ::: + n! . Then tn+1 − tn = 1=(n + 1)! ) (tn) ". The following lemma is proved by simple induction on k. Lemma 1.2.1. For every k ≥ 1 we have: 2k−1 ≤ k!. Now, 1 1 t ≤ 2 + + ::: + n 2! n! 1 1 ≤ 2 + + ::: + 2 2n ≤ 3 Thus xn ≤ tn ≤ 3 and by monotone convergence theorem, both the sequences are convergent. Write l1 = lim xn and l2 = lim tn. We aim to show that l1 = l2. Observe that xn ≤ tn ) l1 ≤ l2. Fix m and let n ≥ m, then 1 1 1 1 m − 1 x ≥ 2 + (1 − ) + (1 − ):::(1 − ) n 2! n m! n n Hold m fixed and let n ! 1. We see that lim xn ≥ tm. i.e. l1 ≥ tm for every m. Now let m ! 1. We see that: l1 ≥ l2. We write: 1 1 1 e = lim (1 + )n = e = 2 + + + ::: n!1 n 2! 3! Consequently, 2 < e < 3 Claim: e is irrational. Proof. (By contradiction). Clearly, e is not an integer. p Suppose e = q where p; q 2 N and q ≥ 2. Then q!e 2 N. Pq 1 P1 1 Then q!e − k=0 k! = k=q+1 k! 2 N. P1 1 P1 1 But since q > 1, then k=q+1 k! < k=1 2k = 1. This is a contradiction since (0; 1) \ N = φ. 1 n+1 Exercise: Let yn = (1 + n ) . Show that : 1. (yn) # and that xn ≤ yn for every n. 2. Conclude that xn ≤ e ≤ yn for all n. 3. Pick a large value of n and estimate e. 3 2 Area under the curve y = xn Pn n(n+1) It is known that Archimedes knew how to add the first n natural numbers by k=1 k = 2 . At the end of the tenth century, Alhazen gave the following recursive relation. n n n p X X X X (n + 1) ik = ik+1 + ( ik) i=1 i=1 p=1 i=1 A formal proof can be written using induction. What matters is the following corollary. Exercise: Use induction on k to prove the following corollary. Corollary 2.0.1. For every pair of natural numbers fk; ng the following holds: n X nk+1 ik = + p (n) k + 1 k i=1 where pk(n) is a polynomial of degree k in n. One immediate consequence Wallis’s first result. Wallis’s Theorem 1: p 1 If p 2 N, area under y = x for x 2 [0; 1] is p+1 . The proof is achieved easily by dividing [0; 1] into n equal subintervals and computing (what we now call)the Riemann sum. Observe that y = xp and x = y1=p are inverse functions and thus the two areas add up to 1. Consequently, we have the following. Wallis’s Theorem 2: n 1 If either n or 1=n is a natural number, area under the the curve y = x ; 0 ≤ x ≤ 1 is n+1 . The following result should follow naturally, but Wallis was not able to prove it. n 1 1+n Wallis’s Conjecture: If n 2 Q, then area under the curve y = x for x 2 [0; a] is n+1 a . p Fermat’s proof: Write n = q and let a = 1. Fix r 2 (0; 1) and divide [0; 1] into subintervals as follows.

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