Series. 1 Convergence and Divergence of Series. S. F. Ellermeyer. October 23, 2003

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1 Series S. F. Ellermeyer October 23, 2003 Convergence and Divergence of Series An infinite series (also simply called a series) is a sum of infinitely many terms a k = a + a 2 + a 3 + () The sequence a n is called the sequence of terms of the series () and the sequence S n = a + a a n is called the sequence of partial sums of the series. Example The series has sequence of terms k 2 a n = n 2 and sequence of partial sums S n = n 2. Example 2 The series k

2 has sequence of terms a n = n and sequence of partial sums S n = n. The concept of a series allows us to formalize the idea of adding up infinitely many numbers. In particular, if the sequence of partial sums of a series converges to a real number, L, then we take this to mean (by definition) that the sum of the infinitely many terms which make up the series is L. Definition 3 If P a k is a series with sequence of partial sums S n and if S n converges to the limit L, then we say that the series converges to L (or that the series has sum L) and we write a k = L If S n diverges, then we say that the series diverges. Example 4 (Geometric Series) Consider a geometric series r k (where r is a constant). The sequence of partial sums of this series is S n = r + r r n. By observing that we see that which means that rs n = r 2 + r r n+, S n rs n = r r n+ ( r) S n = r r n+ 2

3 and hence that S n = r rn+ r (assuming that r 6= ). If r <, then lim n r n+ =0and we obtain lim S r r n+ n = lim n n r and hence we conclude that if r <, then r k = = r 0 r = r r r r. If r, thenlim n S n = which means that r k = and we can say that the geometric series diverges properly to. If r, thens n diverges by oscillation and we say that the geometric series diverges by oscillation. Example 5 (A Telescoping Series) Consider the series k 2 +3k = In order to analyze the sequence of partial sums of this series we first perform a partial fraction decomposition of the sequence of terms: a n = n 2 +3n = n (n +3) = 3n 3(n +3) = 3 Now we examine the sequence of partial sums: µ 4 µ n n +3. S = a = 3 µ S 2 = a + a 2 = 3 3

4 = µ S 3 = µ = µ S 4 = µ µ S 5 = µ = µ Upon examination of this sequence of partial sums, we see that, in general, for all n 3 we have Thus Therefore S n = 3 lim S n = lim n 3 = 3 µ n + n +2. n +3 n µ µ = 8. k 2 +3k = 8. n + n +2 n +3 Example 6 (The Harmonic Series) The harmonic series is the series We will show that this series diverges by examining its sequence of partial sums. Note that S = 4 k.

5 In general S 2 = + 2 =.5 S 4 = S > =2 µ S 8 = S µ > =2.5 µ S 6 = S µ > = 3 S 2 n + 2 n. This shows that the sequence of partial sums is not bounded above and, hence, is divergent. We conclude that the harmonic series diverges properly to. 2 A Basic Criterion for Divergence of a Series In order for the series a k to converge, the terms a n must all be very small (in absolute value) for large n. In particular, the following is true: Theorem 7 If the series P a k converges, then lim n a n =0. Stated in another way, we have Theorem 8 (Basic Divergence Test) If a n does not converge to the limit 0, then the series P a k diverges. ProofoftheBasicDivergenceTest:Suppose that the series P a k converges to the sum L (where L is some real number). Then the sequence of partial sums of this series, S n,convergestol. Since S n = S n + a n for all n, 5

6 then Therefore, a n = S n S n for all n. lim n a n = lim n (S n S n )=L L =0. Example 9 The series ( ) k = + + diverges because its sequence of terms is a n =( ) n and this sequence does not converge to 0. In particular, a n diverges by oscillation. Example 0 The series nx k k + diverges because its sequence of terms is a n = n n + and this sequence does not converge to 0. In particular lim n n n + =. Caution: If it is true that lim n a n =0, we cannot automatically conclude that the series P n a k converges! An example of why this is true istheharmonicseries k for which we have a n =/n and lim a n = lim n n n =0 but the series still diverges. 6

7 3 Two Basic Questions About Series Given a series a k, there are two basic questions which can be asked:. Doestheseriesconverge? 2. If the series does converge, then what is its sum? There are many theorems which provide criteria to determine whether certain types of series converge or diverge. We will study some of these theorems. Once it has been determined that a given series converges, it is usually a much more difficult problem to actually determine the sum of the series. Nonetheless, there are certain classes of series whose sums we can compute. These include the geometric series and telescoping series, examples of which were given previously. 4 Convergence/Divergence Tests for Series with Positive Terms Here we consider some convergence/divergence tests that apply to series with positive terms. 4. The Integral Test Theorem (The Integral Test) Let P n a k be a series such that a n > 0 for all n and suppose that there is a function f such that f is decreasing on some interval [N, ) and f (n) =a n for all n N. Then the series P n a k converges if and only if the improper integral converges. Z N f (x) dx 7

8 Example 2 (p series) A p series is a series of the form where p is some fixed positive real number. We will use the Integral Test to show that the p series converges if p> and diverges if 0 <p. First, let us assume that p 6=. If we let f (x) =/x p, then f is decreasing on the interval [, ) and f (n) =/n p for n. Also, Z Z k p f (x) dx = x dx Z p t = lim t x dx Z p t = lim x p dx t t à = lim = lim t p + x p+ p + x=t x= t p p +! If p>, then lim t à p + t! = p p + p + which shows that, in this case, the improper integral converges. If 0 <p<, then lim t à Z f (x) dx p + t! = p p + which shows that, in this case, the improper integral Z f (x) dx 8

9 diverges. By the Integral Test, we conclude that the p series converges if p> and diverges if 0 <p<. If p =, then we have the harmonic series, P, which we have already k shown diverges. The Integral Test can also be used to show that the harmonic series diverges. 4.2 The Standard Comparison Test Theorem 3 (The Standard Comparison Test) Suppose that P a k and P b k are series such that 0 a n b n for all n and such that P b k converges. Then P a k converges. Remark The Standard Comparison Test can also be used to obtain divergence. If the series P a k in the above theorem diverges, then it must be the case that P b k diverges. Proof of the Standard Comparison Test: Let S n be the sequence of partial sums of the series P a k and let T n be the sequence of partial sums of the series P b k. If we know that the series P b k converges, then we know that the sequence T n converges to some limit L (a real number). Also, since 0 a n for all n, we know that the sequence S n is monotone increasing. Furthermore, since a n b n for all n, thens n T n for all n, and since lim n T n = L, thens n is bounded above. Since S n is monotone increasing and bounded above, it converges. Therefore, the series P a k converges. Example 4 We will use the Standard Comparison Test to show that the series k 2 + k +8 converges. Since 0 n 2 + n +8 n 2 for all n and since is a convergent p series, then k 2 k 2 + k +8 9

10 converges by the Standard Comparison Test. In applying the Standard Comparison Test, it is not actually necessary that 0 a n b n for all n. The test still applies if 0 a n b n for all n N (where N can be any positive integer not necessarily ). This is because the convergence or divergence of any series is not affected by the first finitely many terms of the series. This idea is illustrated in the next example, where we show a comparison, 0 a n b n, that holds for n 0. Example 5 We will use the Standard Comparison Test to show that the series 2k k diverges. If n 0, thenn Thus,ifn 0, then we have Note that and that Therefore 0 2n n 2 + n 2 2n n 2 + n 2 = n k =. 2n n k k =. 4.3 The Limit Comparison Test Theorem 6 (The Limit Comparison Test) Suppose that P a k and P b k are series such that a n > 0 and b n > 0 for all n, and suppose that a n lim = c n b n where c is a positive real number. Then the series P a k and P b k either both converge or both diverge. 0

11 Example 7 We will use the Limit Comparison Test to show that the series k 2 + k +8 converges. In order to do this, we must have a series that we already know is convergent which we can compare this series to. We take this to be the p series k. 2 The sequence of terms of the original series in question is a n = n 2 + n +8 and the sequence of terms of the p series is b n = n 2. Now we observe that a n n 2 lim = lim n b n n n 2 + n +8 =. Since this limit exists and is a positive real number, and since we know that the series P is convergent, then we can conclude that the series k 2 is also convergent. P k 2 +k+8 Example 8 We will use the Limit Comparison Test to show that the series 2k k diverges. The sequence of terms of this series is 2n a n = n We will compare this to the harmonic series, k,

12 whosesequenceoftermsis b n = n. Since a n 2n 2 lim = lim n b n n n =2, and since we know that the series P series P 2k k is also divergent. 5 Algebraic Properties of Series k is divergent, we conclude that the If we have a series, P a k, with sequence of partial sums S n,andifc is a constant, then the series P ca k has sequence of partial sums T n = cs n. If lim n S n = L (a real number), then lim n T n = cl. Therefore, if the series P a k converges to the sum L, then the series P ca k converges to the sum cl. Example 9 In an earlier example, we showed that k 2 +3k = 8. Therefore 4 k 2 +3k = If P a k and P b k are convergent series, with P a k = L and P b k = M, then the series P (a k + b k ) is also convergent and P (a k + b k )= L + M. Example 20 Since and then à µ 3 5 k = k 2 +3k + µ 3 5 k 2 +3k = ³ 3 5 k! 2 = 3 8, = = 7 72.

13 Finally,wenotethatif P a k is a convergent series and P b k is a divergent series, then the series P (a k + b k ) is divergent. Example 2 Since the series P is convergent and the series P k 2 k is divergent, then the series is divergent. µ k 2 + k If we have two series, P a k and P b k, both of which are divergent, then the series P (a k + b k ) mightbeconvergentoritmightbedivergent. As an exercise, make up examples of each possibility. 3