INFINITE SERIES. function,where is a set of natural numbers and R is a set of real numbers. A sequence can be expressed as
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1 CHAPTER 2 INFINITE SERIES 2.1 Sequences: A sequence of real numbers is defined as a function,where is a set of natural numbers and R is a set of real numbers. A sequence can be expressed as or For example is a sequence. Convergent sequence: A sequence converges to a number l, if for given, there exists a positive integer m depending on, such that. Then l is called the limit of the given sequence and we can write or 2.2 Definition of an Infinite Series An expression of the form infinite series of real numbers, where each denoted by is known as the is a real number. It is For example is an infinite series. Convergence of an infinite series Consider an infinite series Let us define and so on. Then the sequence so formed is known as the sequence of partial sums (S.O.P.S.) of the given series.
2 Convergent series: A series converges if the sequence of its partial sums converges i.e. if exists. Also if then is called as the sum of the given series. Divergent series: A series diverges if the sequence of its partial sums diverges i.e. if Example 1 Show that the Geometric series + +..,where, is convergent if and diverges if. Solution: Let us define Case 1: Consider = (As if ) Since is finite the sequence of partial sums i.e. converges and hence the given series converges. Case2: Consider ( As if ) Since diverges and hence the given series diverges. Case2: Consider
3 Since diverges and hence the given series diverges. Positive term series An infinite series whose all terms are positive is called a positive term series. p-series:an infinite series of the form + is called p-series. It converges if and diverges if. For example: 1. + converges 2. + converges 3. + converges Necessary condition for convergence: If an infinite series is convergent then. However, converse need not be true. Proof: Consider the sequence of partial sums of the series. We know that Now Taking limit, we get
4 ..(1) As is convergent sequence of its partial sums is also convergent. Let, then Substituting these values in equation (1), we get. To show that converse may not hold, let us consider the series. Here But is a divergent series (As Corollary:If, then cannot converge. Example 2Test the convergence of the series Solution: Here Hence the given series is not convergent. Example 3 Test the convergence of the series Solution: Here Hence the given series is not convergent.
5 2.3 Tests for the convergence of infinite series 1. Comparision Test: Let and be two positive term series such that (where is a positive number) Then (i) If converges then also converges. (ii) If diverges then also diverges. Example 4 Test the convergence of the following series (i) (ii) (iii) Solution: (i) Here We know that for Hence for Now is a geometric series ( whose common ratio is. Since is also convergent. is a convergent series. Thus by comparision test (ii) Here We know that for Hence for for Now is a divergent series (As p. Thus by comparision test is also divergent. (iii) Here. Clearly
6 Now is a geometric series ( whose common ratio is Since is also convergent. is a convergent series. Thus by comparision test Example 4 Test the convergence of the series Solution: Here Clearly + + Now is a convergent series (As p. Thus by comparision test is also convergent. 2. Limit Form Test: Let and be two positive term series such that (where is a finite and non zero number). Then and behave in the same manner i.e. either both converge or both diverge. Example 5Test the convergence of the series +.. Solution: Here Let. Now consider =
7 = = (which is a finite and non zero number) Hence by Limit form test, and behave similarly. Since converges (as also converges. Example 6Test the convergence of the series (i)..(ii) Solution:(i) Here Let. Now consider (which is a finite and non zero number) Hence by Limit form test, and behave similarly. Since diverges (as also diverges. (ii) Here = Let. Now consider
8 = 1 (which is a finite and non zero number) Hence by Limit form test, and behave similarly. Since converges (as also converges. Example 7 Test the convergence of the series (i) (ii ) Solution:(i) Here = Let ( which is a finite and non zero number) Since converges (as also converges (by Limit form test). (ii ) Here. Let. Then
9 =1 (which is a finite and non zero number) Since diverges (as also diverges (by Limit form test). Exercise 2A Test the convergence of the following series: 1. Ans. Convergent 2. Ans. Convergent 3. Ans. Convergent Ans. Divergent Ans. Convergent 6. Ans. Divergent 7. Ans. Divergent 8. Ans. Convergent Ans. Divergent 10. Ans. Divergent 3. D Alembert s Ratio Test Let be a positive term series such that Then converges if l< 1 diverges if l > 1 Test fails if l = 1
10 Example 8 Test the convergence of the following series: (i) +.. (ii) (iii) Solution: (i)here Then = = = 0 < 1 Hence by Ratio test,the given series converges. (ii) Here Then = = = 0 <1 Hence by Ratio test, the given series converges. (iii) Here Then = = = = < 1 Hence by Ratio test, the given series converges. Example 9 Test the convergence of the following series: (i) +.. (ii) +.. Solution: (i) Here
11 Then = = Hence by Ratio test, the given series diverges. (ii) Here Then = = Hence by Ratio test, the given series converges. Example 10 Test the convergence of the following series: (i) +.. (ii) ( Solution: (i) Here Then = Hence by Ratio test, the given series converges if if and diverges Test fails if. i.e. x When x, Let Now consider (which is a finite and non zero number) Since diverges (as also diverges for x=1 (by Limit form test).
12 the given series converges for and diverges for. (ii) Here Then x = x Hence by Ratio test, the given series converges if and diverges if Test fails if x When x, Let (which is a finite and non zero number) Since converges (as also converges for x=1 (by Limit form test). the given series converges for and diverges for. 3. Cauchy s n th Root Test Let be a positive term series such that Then converges if l< 1 Test fails if l = 1 diverges if l > 1
13 Example 11 Test the convergence of the following series: (i) +.. (ii) (iii) Solution: (i) Here = 0 <1 Hence by Cauchy s root test, the given series converges. (ii) Here = <1 (iii) Here Hence by Cauchy s root test, the given series converges. = Hence by Cauchy s root test, the given series converges. Example 12 Test the convergence of the following series: Solution: Here
14 Hence by Cauchy s root test, the given series converges. Exercise 2B Test the convergence of the following series: 1. Ans. Convergent 2. Ans. Divergent 3. Ans. Convergent 4. + Ans. Convergent Ans. Convergent if, divergent if 6. Ans. Convergent 7. Ans. Convergent 8. Ans. Convergent (p > 0) Ans. Convergent 10. Ans.Convergent if, divergent if
15 4. Raabe s Test Let be a positive term series such that Then converges if l> 1 diverges if l < 1 Test fails if l = 1 Example 13 Test the convergence of the following series: (i) +.. (ii) ( Solution:(i) Here Then = 1 Hence Ratio test fails. Now applying Raabe s test, we have < 1 Hence by Raabe s test, the given series diverges. (ii) Ignoring the first term, Then =
16 Hence by Ratio test, the given series converges if if and diverges Test fails if x When x, Hence by Raabe s test, the given series converges if the given series converges if 1 and diverges if 4. Logarithmic Test Let be a positive term series such that Then converges if l> 1 diverges if l < 1 Example 14 Test the convergence of the series Solution: Here Then
17 Hence by Ratio test, the given series converges if and diverges if Test fails if ex Since involves e applying logarithmic test. for = = + = < 1. Hence the given series converges for and diverges for. 5. Cauchy s Integral Test If is non-negative, integrable and monotonically decreasing function such that =, then if converges then the series also converges.
18 Example 15 Test the convergence of the following series (i) (ii) Solution:(i) Here. Let Clearly function. is non-negative, integrable and monotonically decreasing Consider = which is finite. Hence converges so also converges. (ii) Here. Let Clearly function. is non-negative, integrable and monotonically decreasing Consider Hence diverges.
19 Exercise 2C Test the convergence of the following series: 1. Ans. Convergent if, divergent if 2. Ans. Convergent if, divergent if 3. Ans. Convergent 4. Ans. Convergent if, divergent if Ans. Convergent if, divergent if Ans. Convergent if, divergent if. 7. Ans. Convergent 2.4 Alternating Series An infinite series of the form ( is called an infinite series. We write
20 Leibnitz s Test The alternating series following conditions: converges if it satisfies the (i) (ii) Example 16 Test the convergence of the following series (i) (ii) Solution: (i) The given series is. Here Since Also Hence by Leibnitz s test, the given series converges. (ii) The given series is. Here Since Also Hence by Leibnitz s test, the given series converges. Absolute Convergence A series is absolutely convergent if the series is convergent. For example as geometric series whose common ratio ). is absolutely convergent is a convergent series (Since it is a
21 Result: Every absolutely convergent series is convergent. But the converse may not be true. Conditional Convergence A series which is convergent but not absolutely convergent is called conditionally convergent series. Example 17 Test the convergence and absolute convergence of the following series: (i) (ii) (iii) Solution: (i) The given series Leibnitz s test. Now, (As p=1) is convergent by is not convergent Hence the given series is not absolutely convergent. This is an example of conditionally convergent series. (ii) The given series Also, (As p = 2 > 1) is convergent by Leibnitz s test. is convergent Hence the given series is absolutely convergent. (iii)the given series Here. Now log x is an increasing function
22 Also Hence by Leibnitz s test, the given series is convergent. Now for absolute convergence, consider It is a divergent series (as discussed earlier). Hence the given series is not absolutely convergent. This is an example of conditionally convergent series. Example 18 Test the convergence of the series: (i ) (ii) Solution:(i) The given series is Consider Now, Since is convergent (As p = 2 > 1) by Comparision test is also convergent. Hence the given series is absolutely convergent and so convergent. (ii) The given series is Consider Here Now
23 by Ratio test is convergent or the given series is absolutely convergent and hence convergent. Example 19 Find the values of x for which the series convergent. is absolutely convergent and conditionally Solution: The given series is Then Now, Thus, by Ratio test converges if, diverges if and test fails if When, we have For, the given series is test but not absolutely convergent., which is convergent by Leibnitz s For, the given series is, which is also convergent by Leibnitz s test but not absolutely convergent. Hence the given series is absolutely convergent for and conditionally convergent for.
24 Exercise 2D 1. Show that the series is convergent. 2. Show that the series is absolutely convergent. 3. Test the convergence and absolute convergence of the series Ans. Absolutely convergent 4. Show that the series is conditionally convergent. 5. Test the absolute convergence of the series Ans. Not absolutely convergent 6. Show that the series converges absolutely. 7. Find the interval of convergence of the series Ans. Taylor Series: 2.5 EXPANSION OF FUNCTIONS If a function is infinitely differentiable at the point a then can be expanded about the point as Also where is small, can be expanded as
25 Malaurin Series: It is the special case of Taylor series about the point 0. Hence the Maclaurin series of is Maclaurin series of standard functions: Example20 Expand by Maclaurin series. Solution : By Maclaurin s expansion, we have.. 1 Here Now
26 Similarly and so on. Putting these values in 1,we get Example21 Expand in powers of upto forst four terms. Solution: By Taylor s expansion, we have. 1 Here and Now Putting these values in 1,we get
27 Example22 Show that Solution: By Maclaurin s expansion, we have.. 1 Here Now Similarly and so on. Putting these values in 1,we get Example 23 Show that Solution: By Taylor s expansion, we have. 1 Here, and So 1 becomes
28 or or Example 24 Estimate the value of decimal. correct to four places of Solution: Let By Taylor s theorem, we have. 1 Taking a = 9 and h = 1 in 1, we get. =3.1623(approx.) 2.6 Approximate Error Let y be a function of x i.e. y=f(x). If resulting change in y is denoted by is a small change in x then the and is given by approximately. Example 25 Find the change in the total surface area of a right circular cone when (i) the radius is constant but there is a small change in the altitude
29 (ii) the altitude is constant but there is a small change in the radius. Solution: Let the radius of the base be r, altitude be h and the change in the altitude be the radius is constant but there is a small change in the altitude. Let S be the total surface area of the cone, then S = (i)if altitude changes then Now, (ii)if radius changes then approximately. Now, approximately. Example 26 If a, b, c are the sides of the triangle ABC and S is the semi- perimeter, show that if there is a small error in the measurement of side c then the error in the area of the triangle is given by Solution: We know that and On differentiating both the sides w.r.t. c, we get
30 = Example 27 If find the error in T corresponding to 2% error in l where is constant. Solution: Error in T is given by Now As Hence error in T is 2%. Exercise 2E 1. Expand in powers of (x-1). Ans. 2. Using Taylor s theorem find the approximate value of where Ans Show that
31 4. Show that and hence find tan 46 o Ans A soap bubble of radius 2cm shrinks to radius 1.9 cm. Finf the decrease in volume and surface area. Ans cm 3 and cm 2 6. If log 10 4 = , find the approximate value of log Ans Let A, B and C be the angles of a triangle opposite to the sides a,b and c respectively. If small errors and are made in the sides then show that where is the area of the triangle.
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