Convergence Tests We are now interested in developing tests to tell - PowerPoint PPT Presentation

Convergence Tests We are now interested in developing tests to tell whether or not a series converges, without having to guess at its sum. The first such test is entirely a negative result. Theorem . (the divergence test). ≠ ∑ lim uk 0 uk (a) If , then the series diverges. →∞ k = ∑ uk lim uk 0 (b) If , then the series may either converge →∞ k or diverge.

Proof . (a) It is clear by definition that u k = s k − s k -1 . If the series = = converges to s, then we have lim s lim s s . − →∞ →∞ 1 k k k k = − = − = − = Thus lim u lim ( s s ) lim s lim s s s 0. − − →∞ →∞ 1 →∞ →∞ 1 k k k k k k k k k (b) Follows by showing both a convergent series and a = divergent series for which lim The two series are uk 0. →∞ k 1 1 1 1 1 1 + + + + + � � + + + + + � � 1 and 1 2 k 2 4 2 3 k The divergence test is strictly a negative test. It can be used to show that a series diverges, but it cannot be used to show that a series converges.

Problem . ∞ 2 k 2 4 6 2 k = + + + + + � � ∑ (a) Show that the series + + k 2 3 4 5 k 2 = k 1 diverges. Solution . The n th term sequence has the limit 2 k 2 = = ≠ lim lim 2 0 + →∞ →∞ + 2 k 2 k k 1 k Therefore the series diverges.

Problem . ∞   1 3 5 9 1 + = + + + + + + � � (a) Show that the series ∑   1 1 k   k 2 4 8 = 2 2 k 1 diverges. Solution . The n th term sequence has the limit 1 + = ≠ lim 1 1 0 →∞ 2 k k Therefore the series diverges.

Theorem (a) If Σ u k and Σ v k are convergent series, then Σ ( u k + v k ) and Σ ( u k − v k ) are also convergent, and we have ∞ ∞ ∞ + = + ∑ ∑ ∑ ( ) u v u v k k k k = = = k 1 k 1 k 1 ∞ ∞ ∞ − = − ∑ ∑ ∑ ( u v ) u v k k k k = = = k 1 k 1 k 1 (b) If c is a nonzero constant, then the series Σ u k and Σ cu k both converge or both diverge. In the case of convergence we have ∞ ∞ = ∑ ∑ cu c u k k = = k 1 k 1

Example . Find the sum of the series ∞ 5 7 − ∑ − 1 k k 1 = 3 4 k The series breaks down into the difference of two series, ∞ ∞ ∞ 5 7 5 5 5 5 5 = + + + + � � ∑ ∑ ∑ The first series and . − k k 1 1 k k 3 9 27 = = = 3 4 3 3 k 1 k 1 k is geometric with a = 5/3 and r = 1/3. It therefore converges to ∞ 5 7 5 7 7 = = + + + 3 � is also ∑ The second series . 7 − 1 k 1 − 1 2 4 16 1 = 4 k 3 geometric with a = 7 and r = 1/4. It therefore converges to

7 7 28. = = Thus by (a) the original series converges to − 1 3 3 1 4 4 5/2 − 28/3 = (15 − 56)/6 = − 41/6. Example . Determine whether the following series converges or diverges. ∞ 7 7 7 7 = + + + + � ∑ 7 1 k 2 3 4 = k ∞   ∞ 7 1 = ∑   ∑ 7 . Solution . We showed before that the series 1 k k   = =  k 1  k ∞   1 ∑   diverges. This series is a constant multiple of a 1 k   =  k  divergent series and so diverges by part (b).

Theorem Convergence or divergence is not affected by deleting a finite number of terms from a series; in particular, for any positive K , the series ∞ = + + + � ∑ u 1 u u u k 2 3 = k 1 ∞ = + + + � ∑ u u u u and k + + K K 1 K 2 = k K both converge or both diverge. Warning . The above theorem says that convergence is unaffected by removal of a finite number of terms. However, the sum usually is changed.

Example . Determine whether the following series converge or diverge. ∞ 1 1 1 1 1 1 1 + − + + + + + � = + + + (a) (b) � 4 3 5 7 ∑ 8 k 2 4 8 8 9 10 = k ∞ 1 1 1 1 = + + + � ∑ Solution . (a) is just the divergent series 8 k 8 9 10 = k ∞   1 ∑   With the first 7 terms missing. By the theorem above 1 k   =  k  above, it must also diverge.

Example . Determine whether the following series converge or diverge. ∞ 1 1 1 1 1 1 1 + − + + + + + � = + + + (a) (b) � 4 3 5 7 ∑ 8 k 2 4 8 8 9 10 = k 1 1 1 + + + � Solution . (b) The series is a geometric series with 2 4 8 a = 1/2 and r = 1/2. It therefore converges. Thus by the previous theorem the sequence 1 1 1 + − + + + + + � 4 3 5 7 2 4 8 must also converge. Of course the sums are different. The geometric series formed when the first 4 terms are dropped has sum (1/2)/(1 − 1/2) = 1, and therefore the original series converges to 9 + 1 = 10.

The integral test ∞ ∞ 1 1 ∑ ∫ dx . Let us compare the series with the improper integral 2 2 = x 1 k k 1 Clearly, if the series converges, the sum of the series is greater than the integral, while if the integral diverges, so does the series. 1/4 1/9 1

On the other hand, the diagram below shows that if the integral ∞ 1 ∑ converges, so does the series and therefore so does the 2 = 2 k k original series. On the other hand, if the series diverges, then so must the integral. 1/16 1/4 1/9

This leads to the following theorem. Theorem (The Integral Test) Suppose that a series Σ u k = Σ f ( k ) has positive terms, and that f ( x ) is the formula resulting from replacing k by x in the expression for the terms of the series. If the function with formula f ( x ) is decreasing and continuous on ∞ the interval [ a , ∞ ), then the series and the integral ∑ uk = k 1 ∞ both converge or both diverge. ∫ f x dx ( ) a Note that the value of a does not matter.

Example . Use the integral test to determine whether the following series converge or diverge. ∞ ∞ ∞ 1 1 1 ∑ ∑ ∑ (a) (b) (c) , a positive number. p 2 p 1 k = = = 1 k k k k k 1 ∞ r dx dx = = =∞ Solution (a). ∫ lim ∫ lim ln( ) r . →∞ →∞ x x r r 1 1 Since the improper integral converges, so does the series.

Example . Use the integral test to determine whether the following series converge or diverge. ∞ ∞ ∞ 1 1 1 ∑ ∑ ∑ (a) (b) (c) , a positive number. p 2 p 1 k = = = 1 k k k k k 1 ∞ r r     dx dx 1 1 = = − = − = Solution (b). ∫ lim ∫ lim lim 1 1.     2 2 →∞ →∞ →∞ x r     r r 1 r x x 1 1 Since the improper integral converges, so does the series. Note. The series does not converge to 1. The theorem does not say that the series and the improper integral have the same value when they converge.

Example . Use the integral test to determine whether the following series converge or diverge. ∞ ∞ ∞ 1 1 1 ∑ ∑ ∑ (a) (b) (c) , a positive number. p 2 p 1 k = = = 1 k k k k k 1 Solution (c). We already know that this diverges if p = 1. If p is not 1, then we have −   ∞ 1 p r r − −   dx x 1 p 1 p = =   = − ∫ lim ∫ x dx lim lim r 1   − − p   →∞ →∞ →∞   1 p 1 p r r 1 r x   1 1  1 if > p 1  − = p 1  ∞ < if p 1 

Thus we have the following result of using the integral test ∞ 1 ∑ on series .It is called the p - series test . 1 p = k k Theorem (The p - series test) ∞ 1 1 1 1 = + + + + + � � ∑ 1 1 p p p p = k 2 3 k k converges if p > 1 and diverges if 0 < p ≤ 1.

Problem . Use any method to determine if the following series converges or diverges. ∞ 3 ∑ 1 k 5 = k Solution . The first thing we usually do is check the divergence test. Clearly the terms in this series do converge to 0, so the divergence test gives us no information. ∞   ∞ 3 1 3 = ∑   ∑ We can write the series as 1 k 5 k   = 5 =  k 1  k Since the series in the brackets diverges, so does the series ∞ 3 ∑ 1 k 5 = k

Problem . Use any method to determine if the following series converges or diverges. ∞ 3 ∑ 1 k 5 = k ∞ 1 ∑ Solution 2 . The series is a p - series, and so by the p - 1 k = k series test it diverges. ∞ 3 ∑ is a constant multiple of this p - series, so also diverges. 1 k 5 = k

Problem . Use any method to determine if the following series converges or diverges. ∞ 2 1 + k ∑ 2 3 + = k k 1 Solution . Again check the divergence test. In this case we have 1 + 1 2 + = 2 k 1 = ≠ k lim lim 1 0 →∞ →∞ 2 3 + k k + k 3 1 2 k ∞ 2 1 + k ∑ Thus by the divergence test, the series diverges. 2 3 + = k k 1

Problem . Use any method to determine if the following series converges or diverges. ∞ ln k ∑ k = k 3 ln x = Solution . Here the function is integrable and ( ) f x x satisfies the conditions of the integral test. Thus the series behaves exactly as does the improper integral ∞ r ln x ln x = ∫ dx lim ∫ dx →∞ x x r 3 3 ln xdx ∫ . We pause to evaluate Let u = ln x , du = dx / x . x