Convergence Tests. Theorem. (the divergence test)., then the series u k diverges. k k. (b) If lim u = 0, then the series u may either converge

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1 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). (a) If lim u k 0, then the series u k diverges. k (b) If lim u = 0, then the series u may either converge k k k or diverge.

2 Proof. (a) It is clear by definition that u k = s k s k-. If the series converges to s, then we have lim lim. k s k = k s k = s Thus lim u lim ( s s ) lim s lim s s s 0. k k = k k k = k k k k = = (b) Follows by showing both a convergent series and a divergent series for which lim u = 0. The two series are k k k + and 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.

3 Problem. (a) Show that the series diverges. k 4 6 k = k k+ Solution. The nth term sequence has the limit lim k = lim = 0 k k+ k + k Therefore the series diverges.

4 Problem. (a) Show that the series diverges k = k Solution. The nth term sequence has the limit lim + 0 k k = Therefore the series diverges.

5 Theorem (a) If Σu k and Σv k are convergent series, then Σ(u k + v k ) and Σ(u k v k ) are also convergent, and we have ( uk + vk) = uk + vk ( uk vk) = uk vk (b) If c is a nonzero constant, then the series Σu k and Σcu k both converge or both diverge. In the case of convergence we have cuk = c uk

6 Example. Find the sum of the series k k k = The series breaks down into the difference of two series, and. 3 k The first series k 3 k = k is geometric with a = 5/3 and r = /3. It therefore converges to =. The second series k 4 k = is also = geometric with a = 7 and r = /4. It therefore converges to

7 7 = 7 = 8. Thus by (a) the original series converges to / 8/3 = (5 56)/6 = 4/6. Example. Determine whether the following series converges or diverges k = Solution. 7. k k = k k We showed before that the series = = k k diverges. This series is a constant multiple of a = divergent series and so diverges by part (b).

8 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 uk = u + u + u + 3 and uk = u + u + u + K K+ 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.

9 Example. Determine whether the following series converge or diverge. (a) (b) 8 k = Solution. (a) is just the divergent series 8 k = k k With the first 7 terms missing. By the theorem above = above, it must also diverge.

10 Example. Determine whether the following series converge or diverge. (a) (b) 8 k = Solution. (b) The series is a geometric series with 4 8 a = / and r = /. It therefore converges. Thus by the previous theorem the sequence must also converge. Of course the sums are different. The geometric series formed when the first 4 terms are dropped has sum (/)/( /) =, and therefore the original series converges to 9 + = 0.

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

12 On the other hand, the diagram below shows that if the integral converges, so does the series and therefore so does the k = k original series. On the other hand, if the series diverges, then so must the integral. /4 /9 /6

13 This leads to the following theorem. Theorem (The Integral Test) Suppose that a series Σu k = Σf(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 a f() xdx u k both converge or both diverge. Note that the value of a does not matter. and the integral

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

15 Example. Use the integral test to determine whether the following series converge or diverge. (a) k (b) k (c) p k, p a positive number. Solution (b). dx r dx r = lim lim lim. = x = = x r x r r r Since the improper integral converges, so does the series. Note. The series does not converge to. The theorem does not say that the series and the improper integral have the same value when they converge.

16 Example. Use the integral test to determine whether the following series converge or diverge. (a) k (b) k (c) p k, p a positive number. Solution (c). We already know that this diverges if p =. If p is not, then we have dx r x p r lim x p dx lim lim r p = p = = p x r r r p if p> = p if p<

17 Thus we have the following result of using the integral test on series.it is called the p - series test. k p Theorem (The p - series test) p = p 3 p p k k converges if p > and diverges if 0 < p.

18 Problem. Use any method to determine if the following series converges or diverges. 3 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 3 We can write the series as k 5k = 5 k k = = Since the series in the brackets diverges, so does the series 3 5 k

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

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

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

22 Then ln x u dx= udu ln( x ) x = = Continuing with the previous equation, we have lnx r lnx r ln( ) ln(3) r dx= lim dx= lim [ ln( x) ] = lim = x x 3 3 r 3 r r Therefore the series diverges.

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

24 Then xe x dx e u = du= e u= e x Continuing with the previous equation, we have x r xe dx lim xe x dx lim e r e = = = r r e Therefore the series converges (but not to /e).

25 Problem. Use any method to determine if the following series converges or diverges. k = k+ Solution. This series is a p - series with p = /, with terms dropped off. Thus it diverges. Solution. We can use the integral test directly. dx r dx r = lim = lim x+ = lim r+ 3 = x+ x r + r r We see again that the series must diverge.