The Integral Test. P. Sam Johnson. September 29, P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

Size: px

Start display at page:

Download "The Integral Test. P. Sam Johnson. September 29, P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39"

Maximilian Davidson
5 years ago
Views:

1 The Integral Test P. Sam Johnson September 29, 207 P. Sam Johnson (NIT Karnataka) The Integral Test September 29, 207 / 39

2 Overview Given a series a n, we have two questions:. Does the series converge? 2. If it converges, what is its sum? In this lecture we answer that question by making a connection to the convergence of the improper integral f (x) dx. However, as a practical matter the second question is also important, and we will discuss it. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

3 Overview In this lecture, we study series that do not have negative terms. The reason for this restriction is that the partial sums of these series form nondecreasing sequences, and nondecreasing sequences that are bounded from above always converge. To show that a series of nonnegative terms converges, we need only show that its partial sums are bounded from above. It may at first seem to be a drawback that this approach establishes the fact of convergence without producing the sum of the series in question. Surely it would be better to compute sums of series directly from formulas for their partial sums. But in most cases such formulas are not available, and in their absence we have to turn instead to the two step procedure of first establishing convergence and then approximating the sum. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

4 Nondecreasing Partial Sums Suppose that is an infinite series with n= a n a n 0 for all n. Then each partial sum is greater than or equal to its predecessor because s n+ = s n + a n : s s 2 s 3 s n s n+. Since the partial sums form a nondecreasing sequence, the Nondecreasing Sequence Theorem tells us the series will converge if and only if the partial sums are bounded from above. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

5 Corollary of Nondecreasing Sequence Theorem Corollary. A series a n of nonnegative terms converges if and only if its partial n= sums are bounded from above. Example 2 (The Harmonic Series). The series n= is called the harmonic series. n = n + The harmonic series is divergent, but this doesnt follow from the nth-term Test. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

6 The Harmonic Series The nth term n does go to zero, but the series still diverges. The reason it diverges is because there is no upper bound for its partial sums. To see why, group the terms of the series in the following way: ( ) ( ) ( ) + 6 P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

7 The Harmonic Series The sum of the first two terms 3/2. The sum of the next two terms is /3 + /4, which is greater than /4 + /4 = /2. The sum of the next four terms is /5 + /6 + /7 + /8, which is greater than /8 + /8 + /8 + /8 = /2. The sum of the next eight terms is /9 + /0 + / + /2 + /3 + /4 + /5 + /6, which is greater than 8/6 = /2. The sum of the next 6 terms is greater than 6/32 = /2, and so on. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

8 The Harmonic Series In general, the sum of 2 n terms ending with /2 n+ is greater than 2 n /2 n+ = /2. The sequence of partial sums is not bounded from above: If n = 2 k, the partial sum s n is greater than k/2. Therefore the harmonic series n= { Note that the sequence n} converges. n diverges. Don t get cofused with the terms sequence and series. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

9 The Integral Test We introduce the Integral Test with a series that is related to the harmonic series, but whose nth term is /n 2 instead of /n. Example 3 (Does the following series converge?). n 2 = n 2 +. n= Solution : We determine the convergence of n= by comparing it with n2 x 2 dx. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

The Integral Test To carry out the comparison, we think of the terms of the series as values of the function f (x) = /x 2 and interpret these values as the areas of rectangles under the curve

10 The Integral Test To carry out the comparison, we think of the terms of the series as values of the function f (x) = /x 2 and interpret these values as the areas of rectangles under the curve y = /x 2. The sum of the areas of the rectangles under the graph of f (x) = /x 2 is less than the area under the graph. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

11 The Integral Test From the figure, we have Thus the partial sums of s n = n 2 = f () + f (2) + f (3) + + f (n) n < f () + x 2 dx < + < + = 2. n= x 2 dx are bounded from above (by 2) and the n2 series converges. The sum of the series is known to be π 2 / P. Sam Johnson (NIT Karnataka) The Integral Test September 29, 207 / 39

12 The Integral Test Theorem 4 (The Integral Test). Let {a n } be a sequence of positive terms. Suppose that a n = f (n), where f is a continuous, positive, decreasing function of x for all x N, where N is a positive integer. Then the series and the integral both converge or both diverge. N n=n a n f (x) dx P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

13 Proof of the Integral Test We establish the test for the case N =. The proof for general N is similar. We start with the assumption that f is a decreasing function with f (n) = a n for every n. This leads us to observe that the rectangles in the figure, which have areas a, a 2,, a n, collectively enclose more area than that under the curve y = f (x) from x = to x = n +. That is, n+ f (x) dx a + a a n. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

14 Proof of the Integral Test (contd...) In the figure, the rectangles have been faced to the left instead of to the right. If we don t consider the first rectangle, of area a, we see that a 2 + a a n If we include a, we have n a + a a n a + f (x) dx. n f (x) dx. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

15 Proof of the Integral Test (contd...) Combining these results gives n+ f (x) dx a + a a n a + n f (x) dx. These inequalities hold for each n, and continue to hold as n. If f (x) dx is finite, the right-hand inequality shows that a n is finite. If f (x) dx is infinite, the left-hand inequality shows that a n is infinite. Hence the series and the integral are both finite or both infinite. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

16 The p-series Example 5 (The p-series). Show that the p-series n= n p = p + 2 p + 3 p + + n p + (p is a real constant) converges if p >, and diverges if p. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

17 The p-series : Solution : If p >, then f (x) = /x p is a positive decreasing function of x. Since x p dx = = = [ x x p p+ dx = lim b p + ( ) p lim b b p (0 ) = p p, ] b the series converges by the Integral Test. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

18 The p-series : Solution (contd...) We emphasize that the sum of the p-series is not /(p ). The series converges, but we don t know the value it converges to. If p <, then p > 0 and x p dx = p lim b (b p ) =. The series diverges by the Integral Test. If p =, we have the (divergent) harmonic series n +. We have convergence for p > but divergence for every other value of p. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

19 The p-series Test The p-series with p = is the harmonic series. The p-series Test shows that the harmonic series is just barely divergent; if we increase p to , for instance, the series converges! The slowness with which the partial sums of the harmonic series approaches infinity is impressive. For instance, it takes about 78, 482, 30 terms of the harmonic series to move the partial sums beyond 20. It would take your calculator several weeks to compute a sum with this many terms. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

20 A Convergent Series Example 6. The series n= n 2 + converges by the Integral Test. The function f (x) = /(x 2 + ) is positive, continuous, and decreasing for x, and x 2 + dx = lim [arctan b x]b = lim [arctan b arctan ] b = π 2 π 4 = π 4. Again we emphasize that π/4 is not the sum of the series. The series converges, but we do not know the value of its sum. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

21 Conclusion Convergence of the series n= n 2 + can also be verified by comparison with the series n= n 2. Comparison test are studied in the next lecture. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

22 Exercise Exercise 7 (Applying the Integral Test). Use the Integral Test to determine if the following series converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied n= n= n=2 n 0.2 e 2n ln(n 2 ) n n= n=2 n 2 e n/3 n 4 n 2 2n + P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

23 Solution. f (x) = is positive, continuous and decreasing for x. By the x 0.2 Integral Test, the given series diverges. 2. The function is decreasing for x. By the Integral Test, the given series converges. 3. The function is decreasing for x 3. By the Integral Test, the given series diverges. 4. The function is decreasing for x 7. By the Integral Test, the given series converges. 5. The function is decreasing for x 8. By the Integral Test, the given series diverges. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

24 Exercise Exercise 8 (Determining Convergence or Divergence). Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series convergence or divergence.) n= n= n= 0 n n n + 2 n n n=2 n= n= ln n n 2 n n + n( n + ) P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

25 Solution. converges 2. diverges 3. converges 4. diverges 5. diverges 6. diverges P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

26 Exercise Exercise 9 (Determining Convergence or Divergence). Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series convergence or divergence.) (ln 3) n (/n) (ln n) ln 2 n n tan n n= n=3 n= n= n= 8 tan n + n 2 n n 2 + P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

27 Solution. diverges 2. converges 3. diverges 4. converges 5. converges P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

28 Theory and Examples Exercise 0.. For what values of a, if any, do the series converge? 2. Are there any values of x for which reasons for your answer. n= ( a n + 2 ) n + 4 (/(nx)) converges? Give n= P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

29 Solution. Solution will be discussed in tutorial session. 2. No. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

30 Theory and Examples Exercise. Is it true that if a n is a divergent series of positive numbers then there n= is also a divergent series n= of positive numbers with b n < a n for every n? (a) Is there a smallest divergent series of positive numbers? Give reasons for your answers. (b) Is there a largest convergent series of positive numbers? Explain. b n Solution will be discussed in tutorial session. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

31 Theory and Examples Exercise 2 (The Cauchy Condensation Test). Let {a n } be a non increasing sequence (a n a n+ for all n) of positive terms that converges to 0. Then prove that a n converges if and only if 2 n a 2 n converges. Proof will be discussed in tutorial session. Example 3. (/n) diverges because 2n (/2 n ) = diverges. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

32 The Cauchy condensation test Exercise 4.. Use the Cauchy condensation test to show that (a) n ln n diverges; n=2 (b) converges if p > and diverges if p. np n= Solution will be discussed in tutorial session. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

33 Logarithmic p-series Exercise 5.. Show that dx x(ln x) p (p is a positive constant) converges if and only if p > What implications does the fact in the above exercise have for the convergence of the series Give reaons for your answer. n=2 n(ln n) p? P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

34 Exercise Theorem 6 (Logarithmic p-series Test). Let p be a positive constant. The series converges if and only if p >. n=2 n(ln n) p Exercise 7 (Logarithmic p-series). Use Logarithmic p-series test determine which of the following series converge and which diverge. Support your answer in each case. (a) n=2 n(ln n).0 (b) n=2 n ln(n 3 ) P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

35 Euler s Constant The figures suggest that as n increases there is little change in the difference between the sum n and the integral ln n = n x dx. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

36 Euler s Constant By taking f (x) = /x in the proof of The Integral Test, show that ln(n + ) n + ln n or 0 < ln(n + ) ln n ln n. n Thus, the sequence a n = n ln n is bounded from below and from above. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

37 Exercise Exercise 8 (Euler s Constant). Show that n+ n + < n x dx = ln(n + ) ln n. and use this result to show that the sequence {a n } defined by a n = n ln n is decreasing. Since a decreasing sequence that is bounded from below converges, the numbers a n converge: ln n γ. n The number γ, whose value is , is called Euler s constant. In contrast to other special numbers like π and e, no other expression with a simple law of formulation has ever been found for γ. Solution will be discussed in tutorial session. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

38 Exercise Exercise 9. Use the integral test to show that converges. n=0 e n2 Solution will be discussed in tutorial session. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

39 References. M.D. Weir, J. Hass and F.R. Giordano, Thomas Calculus, th Edition, Pearson Publishers. 2. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translated by George Yankovsky). 3. S.C. Malik and Savitha Arora, Mathematical Analysis, New Age Publishers. 4. R. G. Bartle, D. R. Sherbert, Introduction to Real Analysis, Wiley Publishers. P. Sam Johnson (NIT Karnataka) The Integral Test September 29, / 39

Substitutions in Multiple Integrals

Substitutions in Multiple Integrals P. Sam Johnson April 10, 2017 P. Sam Johnson (NIT Karnataka) Substitutions in Multiple Integrals April 10, 2017 1 / 23 Overview In the lecture, we discuss how to evaluate