From Calculus II: An infinite series is an expression of the form

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1 MATH 3333 INTERMEDIATE ANALYSIS BLECHER NOTES Infinite series of numbers From Calculus II: An infinite series is an expression of the form = a m + a m+ + a m+2 + ( ) Let us call this expression (*). The here are real (or complex) numbers. What does expression (*) mean? In fact we shall see shortly that the expression means two things. or Usually m = 0 or, that is, (*) usually is a 0 + a + a 2 + a + a 2 + a 3 +. We call the number the kth term in the series. Sometimes we will be sloppy and write k when we mean (*). The most important question about an infinite series, just as for an infinite sequence, is ) does the series converge? and 2) if it converges, what is its sum? We will explain these in a minute. In fact an expression like (*) has two meanings: Meaning # : A formal sum. That is, it is a way to indicate that we are thinking about adding up all these numbers in the expression (*), in the order given. It does not mean that these numbers do add up. Before we go to Meaning # 2, let me say how you add up all the numbers in an infinite series. To do this, we define the nth partial sum s n to be the sum of the first n terms in the series. In this way we get a sequence s, s 2, s 3, called the sequence of partial sums. For example, for the series k=0, we have s n = n k=0. We say the original series converges if the sequence {s n } converges. If it does not converge then we say it diverges. We call lim s n the sum of the n series if this limit exists. Meaning # 2: k = lim n s n if this limit exists. Cauchy test: k converges iff given ɛ > 0 there exists an N 0 such that a n+ + a n a m < ɛ whenever m > n N. [Proof: Since s m s n = a n+ + a n a m, this is just saying that the partial sums s n = n k= are a Cauchy sequence. And we know from Theorem 4.8 that a sequence converges iff it is a Cauchy sequence.]

2 76 MATH 3333 INTERMEDIATE ANALYSIS BLECHER NOTES In a sum like k= (k+)k, the k is a dummy index. That is, it is only used internally inside the sum, and we can feel free to change its name, to j= (j+)j, for example, In a series let us call m the starting index. Thus for example, the starting index of k=2 k k is 2. Any series can be renumbered so 2 that its starting index is 0. That is, any infinite series may be rewritten as k=0. For example,, which is the same as a m + a m+ + a m+2 +, can be relabelled by letting j = k m, or equivalently k = j + m. Then = a j+m. There is no reason of course why we chose 0 for the starting index. j=0 One can make all series begin with the starting index if you wanted to, by a similar trick (let j = k m + or equivalently k = j + m ). Geometric series: This is a series of form c + cx + cx 2 + cx 3 +, or k=0 cxk, for constants c and x. We call x the constant ratio of the geometric series. Note that if you divide any term in the series by the previous term, you get x. We assume c 0, otherwise this is the trivial series with sum 0. The MAIN FACT about geometric series, is that such a series converges if and only if x <, and in that case its sum is c x. FACT: If k=0 and k=0 b k both converge, and if c is a constant, then: k=0 ( + b k ) converges, with sum k=0 + k=0 b k; k=0 ( b k ) converges, with sum k=0 k=0 b k; k=0 (c) converges, with sum c k=0. [Proof: We just prove the first and third, the second is quite similar. The nth partial sum of k=0 ( + b k ) is n k=0 ( + b k ) = n k=0 + n k=0 b k. By a fact about sums of limits of sequences from Chapter 4 (Fact 9 ()), this converges, as n, to k=0 + k=0 b k. Similarly the nth partial sum of k=0 (c) is n k=0 (c) = c n k=0. By Fact 9 (4) in Chapter 4, this converges, as n, to c k=0.] For any positive integer m we can write k=0 = (a 0 +a + +a m )+. Indeed k=0 converges if and only if converges. If these series converge, then their sum also obeys the rule: k=0 = (a 0 + a + + a m ) +.

3 MATH 3333 INTERMEDIATE ANALYSIS BLECHER NOTES 77 From the last fact it follows that the first few terms of a series, do not affect whether the series converges or not. It will affect the sum though. The Divergence Test: If lim 0 then the series k diverges. A matching statement (the contrapositive): If k converges, then lim = 0. [Beware: If lim = 0 we cannot conclude that k converges. [Proof: Suppose that k=0 = s. If s n is the nth partial sum then s n s as n. Clearly s n+ s too, as n. Thus a n = s n+ s n s s = 0.] 9. Nonnegative Series, and tests for series convergence. A series k is called a nonnegative series if all the terms are 0. For a nonnegative series, the sequence {s n } of the partial sums is a nondecreasing (or increasing) sequence. Indeed if s n = a 0 + a + + a n say, then s n+ = a 0 + a + + a n + a n, so that s n+ s n = a n 0. Therefore, by a fact we saw in Theorem 4.4 in Chapter 4 for monotone sequences, the sum of the series equals the least upper bound of the sequence {s n } of partial sums. Thus the sum of the series always exists, but may be + if (s n ) is unbounded. In the latter case k = lim n s n = +. More importantly, a nonnegative series converges if and only if the {s n } sequence is bounded above. The latter happens if and only if the sum of the series is finite. Thus to indicate that a nonnegative series converges we often simply write k <. The Integral Test: If f(x) is a continuous decreasing positive function defined on [, ) [Picture drawn in class], then k= f(k) converges if and only if f(x) dx converges (i.e. is finite). p-series. An almost identical argument shows that k= and only if p >. These are called p-series. Basic Comparison Test: Suppose that 0 b k for all k. ) If k b k converges, then k converges. 2) If k diverges, then k b k diverges. k p converges if [Proof: We have n k= n k= b k. So for ), if ( n k= b k) is bounded above then ( n k= ) is bounded above. That is, by the third bullet in this section, if k b k converges, then k converges.

4 78 MATH 3333 INTERMEDIATE ANALYSIS BLECHER NOTES Note that 2) is the contrapositive to ).] Limit Comparison Test: Suppose that k and k b k are nonnegative series. If lim sup If lim inf b k b k <, and if k b k converges then k converges. > 0 and k b k diverges then k diverges. Root Test: Suppose that k is a nonnegative series with lim sup ( ) k = r. If 0 r < then k converges. If < r then k diverges. Ratio Test: Suppose that k is a nonnegative series with lim sup + R and lim inf then k diverges. + = = r. If 0 R < then k converges. If < r From Homework 2 Question 3 (a) it is easy to see that the root test is more powerful theoretically than the ratio test. That is if the ratio test works to prove convergence or divergence, then the root test would give the same conclusion. 0. Absolute and conditional convergence A series k is called absolutely convergent if k converges. Recall that here could be a complex number (you can view complex numbers as elements of R 2 here). Any absolutely convergent series is convergent. [Proof: By the Cauchy test above (the first and second page of this chapter), since k converges, given ɛ > 0 there exists an N 0 such that a n+ + a n a m a n+ + a n a m < ɛ, m > n N. By the Cauchy test again (but used in the other direction), k converges.] The converse is false, a series may be convergent, but not absolutely convergent. Such a series is called conditionally convergent. If k= converges absolutely then k= k=. [Proof. Note n k= n k=. Take the limit as n in this inequality, and use Fact 4 about sequences from Chapter 4 to see that k= k=.] The Alternating Series Test (a.k.a. Leibniz Test)/Alternating Series approximation: Suppose that a 0 a a 2, and that

5 MATH 3333 INTERMEDIATE ANALYSIS BLECHER NOTES 79 lim k = 0. Then a 0 a + a 2 a 3 + (which in sigma notation is k=0 ( )k ) converges, and moreover s n k=0 ( )k a n for all n, where s n is the nth partial sum n k=0 ( )k. The fundamental fact about power series (this was already mentioned in the last bullet before Theorem 4.4). Namely consider the power series k=0 c kx k (here x and the c k can be complex numbers if you wish). We set R = if limsup n c n n = 0, and R = 0 if limsup n c n n =. Otherwise set R =. This is the radius of convergence of the power series. limsup n c n n In the homework you are asked to show that in place of limsup n c n n you can use limsup n sn+ s n. Theorem 0.. Let k=0 c kx k be a power series, and define R as above. Then k=0 c kx k converges absolutely if x < R, and it diverges whenever x > R. If k= is a series, and f : N N is a bijection (that is, is one-toone and onto), then the series k= a f(k) is called a rearrangement of k=. It is not hard to see that a rearrangement of a convergent series need not converge. Theorem Any rearrangement of an absolutely convergent series is convergent and has the same sum. Dirichlet test: Let k be a series whose partial sums form a bounded sequence. Suppose (b n ) is a decreasing sequence with limit 0. Then k b k converges. Abel s test: Suppose that k converges and b n is a monotonic convergent sequence. Then k b k converges. Adding parentheses: Let k= be a series, and suppose that n < n 2 < are integers. Let b = n k=, b 2 = n 2 k=n a + k, b 3 = n 3 k=n a 2+ k,. We call k= b k a series obtained from k= by adding parentheses. Theorem 0.2. If k= converges, and b k, n k are as above, then k= b k converges and has sum k=. Also: (a)] If k= is a nonnegative series, then k= converges iff k= b k converges. (b) If....