2 Sequences and seriesWe will first deal with sequences, and then study infinite series in terms ofthe associated sequence of partial sums.2.1 SequencesBy a sequence, we will mean a collection of numbersa1, a2, a3, . . . , an, an+1, . . . ,which is indexed by the set N of natural numbers. We will often denote itsimply as {an}.A simple example to keep in mind is given by an=1n, which appears todecrease towards zero as n gets larger and larger. In this case we would liketo have 0 declared as the limit of the sequence. A quick example of a sequencewhich does not tend to any limit is given by the sequence {1, −1, 1, −1, . . . },because it just oscillates between two values; for this sequence, an= (−1)n+1,which is certainly bounded.Definition 2.1 A sequence {an} is said to converge, i.e., have a limit A,iff for any ε > 0 there exists N = N(ε) > 0 s.t. for all n ≥ N we have|an− A| < ε. Notation: limn→∞an= A, or an→ A as n → ∞.Two Remarks:(i) It is immediate from the definition that for a sequence {an} to converge,it is necessary that it be bounded. However, it is not sufficient, i.e.,{an} could be bounded without being convergent. Indeed, look at theexample an= (−1)n+1considered above.(ii) It is only the tail of the sequence which matters for convergence. Oth-erwise put, we can throw away any number of the terms of the sequenceoccurring at the beginning without upsetting whether or not the laterterms bunch up near a limit point.Lemma 2.2 If limn→∞an= A and limn→∞bn= B, then11) limn→∞(an+ bn) = A + B2) limn→∞(can) = cA, for any c ∈ R3) limn→∞anbn= AB4) limn→∞an/bn= A/B, if B = 0.Proof of 1) For any ε > 0, choose N1and N2so that for n1≥ N1,n2≥ N2, |an1− A| < ε/2, |bn− B| < ε/2. Then for n ≥ max{N1, N2}, wehave|an+ bn− A − B| <ε2+ε2= ε.Hence A + B is the limit of an+ bnas n → ∞. 2Proofs of the remaining three assertions are similar. For the last one,note that since {bn} converges to a non-zero number B, eventually all theterms bnwill necessarily be non-zero, as they will be very close to B. So, inthe sequence an/bn, we will just throw away some of the initial terms whenbn= 0, which doesn’t affect the limit as the bnoccurring in the tail will allbe non-zero. 2Example: We have limn→∞1n= 0. Indeed, given ε > 0, we may choosean N ∈ N such that ∃N >1ε, because N is unbounded. This implies that forn ≥ N, we have |1n− 0| ≤1N< ε. Hence {1n} converges to 0. 2Definition A sequence is said to be monotone increasing, denoted an↗,if an+1≥ anfor all n ≥ 1, and monotone decreasing, denoted an↘, ifan+1≤ anfor n ≥ 1. We say {an} is monotone (or monotonic) if it is of oneof these two types.Theorem 2.3 A bounded, monotonic sequence converges.Proof. Assume bounded and an↗. Let A = sup{an}. (We write sup forsupremum, which is the same as the least upper bound, and inf for infimum,which is the greatest lower bound.) For any ε > 0, A − ε is not an upperbound ⇒ ∃aN> A − ε. But an≥ aNfor n ≥ N ⇒ −ε < an− A ≤ 0 for alln ≥ N. Hence {an} converges with limit A.If {an} is (bounded and) monotone decreasing, then look at {bn}, withbn= −an. Then this new sequence is monotone increasing and has a limitB, which is the sup of {bn}. Then A = −B is the inf of {an}, and an→ Aas n → ∞.22Example: limn→∞12n= 0. Indeed, the sequence is bounded and mono-tone decreasing, hence ∃A such that limn→∞1/2n= A. Note that if wemultiply the terms of the sequence by 2, then this new sequence also con-verges to A. On the other hand, by Lemma 1.2, part 2), this limit should be2A. Then 2A = A, implying that A = 0. 2This example is a special case of a more general phenomenon:Lemma 2.4 Let y be a positive real number. Then {yn} is unbounded ify > 1, whilelimn→∞yn= 0 if y < 1.Proof of Lemma. Suppose y > 1. Write y = 1 + t with t > 0. Thenby the binomial theorem (which can be proved by induction),yn= (1 + t)n=n∑k=0(nk)tk,which is ≥ 1 + nt (as t > 0). Since 1 + nt is unbounded, i.e, larger than anynumber for a big enough n, ynis also unbounded.Now let y < 1. Then y−1is > 1 and hence {y−n} is unbounded. Thisimplies that, for any ϵ > 0, ynis < ϵ for large enough n. Hence the sequenceynconverges to 0. 2As an exercise, try to extend this Lemma and prove that for any y ∈ Rwith |y| < 1, the sequence {yn} converges to 0.Here is an example. Define a sequence {sn} by puttingsn= 1 +11!+12!+ . . . +1(n − 1)!.It is not hard to see that this sequence is bounded. Try to give a proof.(In fact one can show that it is bounded by 3, but we do not need the bestpossible bound at this point.) Clearly, sn+1> sn, so the sequence is alsomonotone increasing. So we may apply Theorem 1.3 above and concludethat it converges to a limit e, say, in R. But it should be remarked that onecan show with more work that e is irrational. So there is a valuable lessonto be learned here. Even though {an} is a bounded, monotone sequence ofrational numbers, there is no limit in Q; one has to go to the enlarged numbersystem R.32.2 The Squeeze PrincipleAn efficient way to prove the convergence of a sequence is to see if it can besqueezed between two other sequences which converge to the same limit.Proposition 1 Suppose {an}, {bn}, {cn} are sequences of real numbers suchthatbn≤ an≤ cn, ∀ n ≥ 1.Suppose moreover that the sequences {bn} and {cn} are convergent with thesame limit L, say. Then the sequence {an} converges as well, withlimn→∞an= L.Example: For any fixed real number x, consider the sequence {an}, withan=sin(nx)n. Then, since sin(nx) takes values between −1 and 1, the hypothe-ses of the Proposition are satisfied if we take bn= −1/n and cn= 1/n, sinceanand bnboth converge to zero. We conclude that {sin(nx)n} also convergesto 0, regardless of what x is.Before giving a proof of Proposition 1, let us note the following usefulconsequence for sequences {an} with non-negative terms, by taking bn= 0for all n.Corollary 2.5 Suppose {an}, { cn} are sequences of non-negative real num-bers, with {cn} convergent, such thatan≤ cn, ∀ n ≥ 1,andlimn→∞cn= 0.Then the sequence {an} converges as well, withlimn→∞an= 0..4Proof of Proposition 1. Pick any ε > 0. Then, the convergence of{bn}, resp. {cn}, implies that we can find some N1> 0, resp. N2> 0, suchthat for all n ≥ N1, resp. n ≥ N2,|L − bn| < ε, resp. |L − cn| < ε.Put N = max(N1, N2). Then for all n ≥ N, the fact that anis