We begin by considering the following three sequences:

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1 STUDENT S COMPANIONS IN BASIC MATH: THE TWELFTH The Concept of Limits for Sequences and Series In calculus, the concept of limits is of paramount importance, in view of the fact that many basic objects of study such as derivatives, integrals, series etc. are defined by pushing some expressions to their limits. This should be motivating enough for a careful study of limits. But most students find the standard definition of limits is incomprehensive or hard to use. Here we introduce an ad hoc definition of limits designed especially for beginners. This definition still complies with the general standard of mathematical rigor. It is much easier to understand and often easier to work with. After learning this ad hoc definition, you will not find any difficulty in switching to the standard one, which will be given near the end. We begin by considering the following three sequences:, /2, /3, /4, /5, /6,... /2, /4, /8, /6, /32,... 0., 0.0, 0.00, 0.000, ,... Here we only write down the first few terms of each sequence. It is more economic and more precise to express a sequence by writing down a general expression of its nth term a n and then putting curly brackets around it to render the sequence into the form {a n }. For example, the nth term of the first sequence above is /n and hence this sequence is usually written as {/n}, or more precisely as {/n} n. Similarly the second sequence can be written as {/2 n } and the third sequence as {0 n }. All three sequences above have the following common feature: their terms become smaller and smaller and get closer and closer to zero. QUESTION. Are you convinced that these sequences have this common feature? Technically, becoming smaller and smaller is called decreasing and getting closer and closer to zero called tending to zero. Putting together we simply say decreasing to zero. Now we state the following formal definition to give the precise meaning. We say that a sequence {a n } is decreasing if a n a n+ for all n.

2 We say that a sequence {α n } of positive numbers is decreasing to zero if {α n } is decreasing and, for every ε > 0, there exists some n such that α n < ε; (in other words, there is no positive number ε such that α n ε for all n.) QUESTION 2. How do you use the definition to check that {/n} is decreasing to zero? Now we come to our ad hoc definition of limits of sequences. Roughly it says that a sequence {a n } converges to some A if the distance between a n and A can be controlled by some positive number α n which decreases to zero. Definition ( ad hoc ). We say that a sequence {a n } converges to A and write lim n a n = A or simply lim a n = A, if there is a sequence {α n } of positive numbers decreasing to zero such that a n A < α n for all n. In the above definition, a n A is the distance between a n and A, which is simply a n A when a n A, and is A a n when a n A. You cannot find this ad hoc definition in any textbook. You may say it is a result of my effort to improve the teaching of mathematical analysis, which is not an easy subject. EXAMPLE 3.. We are asked to guess the limit of the sequence { n n } (if it exists). Here n n will be rewritten as n /n. The problem is about the behavior of p n n /n for large n. So let us take a large concrete n to find out what n /n is, say n = 0 0 (0 0 is certainly a large number explicitly, it is followed by ten zeros): (0 0 ) /00 = (0 0 ) 0 0 = = 0 (0 9) = =. This suggests that n /n converges to. The question is, can we prove this rigorously? n EXAMPLE 4. Now we are going to give a rigorous proof of lim n n =. Let p n = n n. Notice that p n for all n. (Otherwise we would have p n <, which would lead to n = p n n < n =, contradicting n.) Since we anticipate that, for large n, p n is only slightly larger than, we have to estimate the difference d n between p n and, that is d n = p n. Now we use the following binomial estimation : n = p n n = ( + d n ) n = + nd n + n(n ) 2 d 2 n + n(n + ) 2 d 2 n. 2

3 (The reason why we only keep n(n+) 2 d 2 n and drop all other terms will be clear in a moment.) So we have d 2 n 2/(n + ), which gives 0 d n 2/(n + ), or 0 p n 2/(n + ). It is easy to see that the sequence { 2/(n + )} is decreasing to zero. Therefore n n = p n converges to as n. Why do we have to trouble ourselves with getting a rigorous definition of convergence? Well, without it, we cannot produce a correct argument. An incorrect argument is one of the most woeful things in mathematics, because it may lead us to erroneous conclusions with disastrous consequences (such as a rejected manuscript, a failed exam, a collapsing bridge or a sinking ship). Take a look of Gardener in Lewis Carroll s Sylvie and Bruno Concluded, who sang his verse about finding an error in his argument that would proved he was the Pope a dreadful fact that extinguished all hope, he lamented, with tears streaming down his cheeks. In the definition of limits of sequences, we have used the absolute value a n A of a n A for the distance between a n and A. You may say that a is a just a nice way to write down the awkward expression a 2. A proper definition of the absolute value of a real number a is given by a = { a if a 0; a if a < 0. Here are some elementary facts about absolute values that you should keep in mind: (A) Triangular inequality: a + b a + b (for all real a and b.) (A2) a b = b a, ab = a b. (A3) a a 0 ε ε a a 0 ε a 0 ε a a 0 + ε. (Here stands for if and only if.) These elementary facts are easy to verify. Indeed, accepting the usage of the square root function, (A2) just says (a b) 2 = (b a) 2 and (ab)2 = a 2 b 2. Also, (A) can be rewritten as (a + b) 2 a 2 + b 2, which can be checked by squaring both sides. EXERCISE 5a. Check (A), (A2) and (A3) above. Besides the brute force method suggested above for checking (A) to (A3), there is a conceptual one, described next, which is usually regarded as a more preferable approach. 3

4 Notice that every nonzero real number a can be uniquely written as σ a a, where σ a the sign of a, which is either + or. Also notice that a = σ a a and σ ab = σ a σ b. is EXERCISE 5b. Give the detail of proving (A) by this conceptual approach. QUESTION 6. How do you use (A) and (A3) to derive the inequality a b a b? The following observation will be routinely used in our development of the basic theory of limits of sequences. Observation. If {α n } and {β n } are sequences of positive numbers decreasing to zero, then so are the sequences {α n + β n } and {Kα n }, for any K > 0. EXERCISE 7. Verify this observation with great care. EXAMPLE 8. Now we use what we have so far to prove the following basic fact about limits: if {a n } and {b n } are sequences of real numbers converging to a and b respectively, then the sequence {a n + b n } converges to a + b and {a n b n } converges to ab. Indeed, by assumption, there are sequences {α n } and {β n } in of positive numbers decreasing to zero, such that a n a < α n and b n b < β n for all n. Hence (a n + b n ) (a + b) = (a n a) + (b n b) a n a + b n b < α n + β n for all n. Since α n + β n decreases to zero, we conclude that a n + b n tends to a + b as n. Next, notice that a n = (a n a) + a a n a + a α n + a α + a. So a n b n ab = a n b n a n b + a n b ab = a n (b n b) + b(a n a) a n b n b + b a n a ( a + α )β n + b α n. Our observation tells us that b α n +( a +α )β n to ab. The proof is complete. decreases to zero. Hence a n b n converges EXERCISE 9. Prove the following basic fact: if lim a n = a and if a n [c, d] for all n, then a [c, d]. In other words, if c a n d for all n and if lim n a n = a, then c a d. 4

5 EXERCISE 0. (This is a harder exercise.) Prove the following basic fact: if lim a n = lim n b n = A and if a n x n b n for all n, then lim x n = A. One may describe this fact picturesquely as follows: a and b are two police officers, x is a thief, and A is a police station. The condition a n x n b n means that, at any moment n, the officers are grabbing the thief from both sides. The condition lim n an a n = lim n b n = A means that both officers is marching resolutely to the police station. The conclusion lim n x n = A means that the thief is forced to approach to the police station, with no choice.) EXERCISE. Prove that, if lim n a n = a, then lim n a n = a. Hint: QUESTION 6. EXERCISE 2. Prove that, if a n 0 and if lim n a n = a, then lim n an = a. According to our definition, we have lim n a n = a if there is a sequence of positive numbers α n decreasing to zero such that a n a < α n for all n. We may relax the last requirement a bit: instead of asking a n a < α n holds for all n, we only require this inequality for n starting from certain stage. This is not a substantial improvement. However it gives us flexibility in some arguments. Observation 2. If the inequality a n a < α n holds for all n > n 0, where α n decreases to zero and n 0 is some positive integer, then lim n a n = a. QUESTION 3. How do you justify this observation? ( Justify here means verify.) EXAMPLE 4. Now we use this simple observation to show the following: If lim b n = b, lim a n = a, (a n 0, a 0), then lim b n /a n = b/a. In view of the basic fact proven in EXAMPLE 8 above, it is enough to show lim a n = a. By assumption, there is a sequence {α n } decreasing to zero such that a n a < α n for all n. Now a n a = a n a a n a = a n a a n a < α n a n a. ( ) [Aside: We cannot expect that the right hand side α n / a n a is decreasing to zero. The trouble is caused by a n in the denominator of this expression. In the next step we show that it stays away from zero for large n.] Now a = a a n +a n a a n + a n, or a n a a n a. Since α n decreases to zero and since a /2 is positive, there is a positive integer n 0 such that α n0 < a /2. Thus, for n > n 0, we have a n a < α n α n0 < a /2 and hence a n a a n a > a ( a /2) = a /2. 5 Returning to ( ), we conclude

6 a n a < Kα n for all n > n 0, where K = 2/ a 2. Since {Kα n } is a sequence decreasing to zero, Observation 2 tells us that lim a n = a. The proof is complete. EXERCISE 5. (Working out this exercise is great fun.) Prove that if {a n } is a sequence converges to a, then the sequence {b n } of arithmetic means of {a n } given by b n = a + a a n n also converges to a. Hint: First verify that, if {a n } decreases to zero, so does {b n }. In many practical problems, it is convenient to express a sequence in the form of a series. By a series of numbers we mean a formal infinite sum a + a 2 + a 3 +, which is often written as n= a n (or its variants, such as k=0 a k.) A sum of finitely many numbers is a number, but a formal sum is just an expression (not a number) of grouping some numbers by the addition sign without performing the actual addition. It is a bit hard to swallow, but you have to take it that way. Given a series n= a n, we may form the sequence of partial sums s, s 2, s 3,... which are given by s = a, s 2 = a + a 2, s 3 = a + a 2 + a 3, etc. In general, s n = a + a a n ( = n k= a k Conversely, given a sequence {s n } n, if we let a = s and a n = s n s n for n 2, then s n are partial sums of the series n= a n. (Notice that a + a a n = s + (s 2 s ) + + (s n s n ) = s n, ). which is a telescoping sum.) This shows that there is a one to one correspondence between sequences and series, which are related by taking partial sums and telescoping. Therefore we can transplant all concepts about sequences of numbers to series of numbers. For example, we say that the series n= a n converges to s (we also say n= a n is summable with s as its sum) if its partial sums s n n k= a n converge to s as n. Similarly we say that a series diverges if its corresponding sequence of partial sums diverges. EXAMPLE 6. Let a be a fixed number and consider the geometric series n=0 an. The nth partial sum of this series is s n = + a + a a n. [Note that the zeroth partial sum is s 0 = a 0 =.] We can find a closed form of this sum by the following well known 6

7 trick. Multiplying both sides of s n = + a + a 2 + a a n + a n by a, we obtain as n = a + a 2 + a 3 + a a n + a n+. Hence the difference between s n and as n is s n as n = a n+. Thus, in case a, we arrive at s n + a + a a n = an+. a Suppose a <. Then lim n a n = 0 and hence lim n s n = a. We conclude: If a <, then n=0 an converges to a. EXERCISE 7. Show that the series n(n + ) = n n + and telescope. n= n(n + ) converges to. Hint: Write EXERCISE 8. Find the sum of the series n= n(n + )(n + 2). EXERCISE 9. Prove by the method described below that, if 0 < a <, then lim n nan = 0. First, write x n = na n and r n = x n /x n. Show that lim n r n = a. Take any b with a < b <. Then there exists some positive integer N such that r n < b for all n N. Let n > N and write down the telescoping product na n = x n = x n r n = x n 2 r n r n = = x r 2 r 3 r n x r 2 r N b n N 0 as n. PROBLEM 20. (a) Fix a positive number b and write s k = + b + + b k = bk b. Compute the sum n k= s k in two different ways to deduce the identity n + (n )b + + 2b n 2 + b n = (n (n + )b + b n+ )/( b) 2. 7

8 (b) Letting b = /a, multiply both sides of the above identity by a n to get a closed expression for the sum n k= kak. Then let n to deduce that, for a with a <, the series k=0 kak converges to ( a) 2. Now we introduce the standard definition of limits of sequences. Before doing this, we introduce a special terminology to help us to understand this definition. Let S be a set of numbers and {a n } be a sequence. We say that {a n } is eventually in S if there is a positive integer N such that a n are in S for all n > N. To put in another way, that {a n } is eventually in S means that a n are in S except for finitely many n. Now suppose that {a n } is a sequence converging to a according to our ad hoc definition, that is a n a < α n for some sequence {α n } of positive numbers decreasing to zero. Let ε be an arbitrary positive number. Then there exists some positive integer N that α N < ε. Now, when n > N, we have such a n a < α n α N < ε. Note that a n a < ε means the same as a ε < a n < a + ε, that is, a n is in the open interval (a ε, a + ε). So we have seen that {a n } is eventually in (a ε, a + ε). Thus we know: for every ε > 0, the sequence {a n } is eventually in (a ε, a + ε). From the above discussion, we see that {a n } is eventually in (a ε, a + ε) means there exists some N such that a n a < ε for all n > N. We have rendered the ad hoc definition for lim n a n = a to the following statement: given ε > 0 there exists N such that whenever n > N, we have a n a < ε. The last statement is the standard definition for lim n a n = a. Finally, we briefly discuss a crucial issue: existence of limits. We let an example lead our way of discussion. EXAMPLE 2. Given a positive number a, consider a sequence of numbers x n (n 0) defined recursively by x 0 =, x n+ = x2 n + a 2x n for n 0. What is the limit of this sequence? Well, it is very easy to find, provided that the limit does exist. Indeed, we can rewrite the recursion relation as 2x n x n+ = x 2 n + a. Letting 8

9 n, this identity becomes 2L 2 = L 2 + a., where L is the limit of the sequence {x n }, which is assumed to exist. Thus we have L 2 = a. Hence L = a. QUESTION 22. Why is L above cannot be a? In the above example, we can find the limit if it exists. The problem is, does it really exist? The answer is yes. We can use the following fact to justify this answer: Fact. If {a n } is a decreasing sequence (recall that this means a n+ a n for all n) and if a n 0 for all n, then lim n a n exists. (More generally, any bounded monotone sequence of real numbers converges.) Now you are asked to verify that the sequence given in EXAMPLE 2 above satisfies the conditions of this fact: PROBLEM 23. Prove that x n+ x n in EXAMPLE 2. for all n for the sequence {x n } described QUESTION 24. Define a sequence of positive numbers a n (n ) recursively by putting a = and a n+ = a 2 n +. Let L be the limit of {a n }. Rewrite the recursive relation as a 2 n+ = a 2 n + and let n. Then L 2 = L 2 +, or 0 =. What goes wrong here? The fact mentioned above for checking existence of limits only deals with monotone sequences. For a general sequence, the following criterion (described in the ad hoc way ) is of paramount importance: Cauchy s Criterion. A sequence {a n } is convergent if and only if there is a sequence {α n } of positive numbers decreasing to zero such that a m a n α n for all m, n with m n. Here we only give an inadequate description of some basic facts about existence of limits, leaving their important consequences and examples to an advanced course in mathematical analysis. 9