MTH101 - Calculus and Analytical Geometry- Lecture 44
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The document discusses alternating series and conditional convergence, highlighting the conditions for convergence as outlined in the alternating series test. It explains concepts such as absolute convergence, examples of converging and diverging series, and introduces power series with their criteria for convergence and radius. Key theorems are provided to guide the understanding of series behavior based on the nature of their terms.
MTH101 - Calculus and Analytical Geometry- Lecture 44
- 1. 44-Alternating Series; Conditional Convergence VU Lecture No.44 Alternating Series; Conditional Convergence lim 0k k a →+∞ = This theorem tells us when an alternating series converges. Example Use the alternating series test to show that the following series converge 1 1 1 ( 1)k k k ∞ + = −∑ Solution: The two conditions in the alternating series test are satisfied since 1 1 1 1 lim lim 0 1 k k k k k a a and a k k k + → +∞ →+∞ = > = = = + Note that this series is the Harmonic series, but alternates. It is called the Alternating Harmonic series. The Harmonic series DIVERGES. The Alternating Harmonic series CONVERGES. Now we look at the errors involved in approximating an alternating series with a partial sum. Theorem 11.7.2 If an alternating series satisfies the conditions of the alternating series test, and if the sum S of the series is approximated by the nth partial sum of Sn thereby resulting in an error of S - Sn then I S - Sn I < an+1 Moreover, the sign of erroris the same as that of the co-effient of an+1 in the series Example: The alternating series 11 1 1 1 1 .......... ( 1) ............. 2 3 4 k k + − + − + + − + Satisfies the condition of the alternating series test; hence the series has a sum S, which we know must lie between any two successive partial sums. In particular, it must lie between 7 1 1 1 1 1 1 319 1 2 3 4 5 6 7 420 S = − + − + − + = And 8 1 1 1 1 1 1 1 533 1 2 3 4 5 6 7 8 840 S = − + − + − + − = So 533 319 840 420 S< < If we take ln 2S = then 533 319 ln 2 840 420 06345 ln 2 0.7596 < < < < The value of ln2, rounded to four decimal places, is 0.6931, which is consistent with these inequalities. It follows from Theorem 11.7.2 that 7 8 319 1 ln 2 ln 2 420 8 S a− = − < = And 8 9 533 1 ln 2 ln 2 840 9 S a− = − < = © Copyright Virtual University of Pakistan 1 • So far we have seen infinite series that have positive terms only. • We have also defined the LIMIT and the SUM of such infinite series • Now we look at series that have terms which have alternating signs, known as alternating series Example 1-1+1-1+… 1+2-3+4…. 1 1 2 3 4 1 1 2 3 4 1 ( 1) ... ( 1) ... k k k k k k a a a a a a a a a a ∞ + = ∞ = − = − + − + − = − + − + − ∑ ∑ • More generally, an alternating series has one of the two following forms (1) (2) • Note that all the terms are to be taken as being positive. The following theorem is the key result on convergence of alternating series. Theorem11.7.1 (Alternating Series Test) An alternating series of either form (1) or (2) converges if the following two conditions are satisfied: (a) a1>a2>a3……..>ak>…….. (b)
- 2. 44-Alternating Series; Conditional Convergence VU Absolute and Conditional Convergence The series 2 3 4 5 6 1 1 1 1 1 1 1 ........... 2 2 2 2 2 2 − − + + − − + does not fit in any of the categories studied so far, it has mixed signs, but is not alternating. We shall now develop some convergence tests that can be applied to such series. Definition 11.7.3 A series 1 2 1 ......... ............k k k u u u u ∞ = = + + + +∑ is said to converge absolutely, if the series of absolute values 1 2 1 ......... ............k k k u u u u ∞ = = + + + +∑ converges. Example: The series 2 3 4 5 6 1 1 1 1 1 1 1 ........... 2 2 2 2 2 2 − − + + − − + converges absolutely since the series of absolute values 2 3 4 5 6 1 1 1 1 1 1 1 ........... 2 2 2 2 2 2 + + + + + + + is a convergent geometric series. On the other hand, the alternating harmonic series 1 1 1 1 1 .......... 2 3 4 5 − + − + − does not converge absolutely since the series of absolute values 1 1 1 1 1 .......... 2 3 4 5 + + + + + diverges. Absolute convergence is of importance because of the following theorem. Theorem 11.7.4 If the series 1 2 1 ......... ............k k k u u u u ∞ = = + + + +∑ Converges, then so does the series 1 2 1 ......... ............k k k u u u u ∞ = = + + + +∑ In other words, if a series converges absolutely, then it converges. Example: Since the series 2 3 4 5 6 1 1 1 1 1 1 1 ........... 2 2 2 2 2 2 − − + + − − + Converges absolutely, it follows from Theorem 11.7.4 that the given series converges. Example: Show that the series 2 1 cos k k k ∞ = ∑ converges. Solution: Since cos 1k ≤ for all k , thus 2 2 1 cos 1 k k k k ∞ = ≤∑ Thus 2 1 cos k k k ∞ = ∑ converges by the comparison test, and consequently 2 1 cos k k k ∞ = ∑ converges. If ku∑ diverges, no conclusion can be drawn about the convergence or divergence of ku∑ Example: Consider the two series 11 1 1 1 1 .......... ( 1) ............. 2 3 4 k k + − + − + + − + ----------------(A) 1 1 1 1 1 .......... ............. 2 3 4 k − − − − − − − ----------------(B) Series (A), the alternating harmonic series converges. Whereas series (B), being a constant times the harmonic series, diverges. Yet in each case the series of absolute values is 1 1 1 1 1 .......... ............. 2 3 4 k + + + + + + which diverges. A series such as (A) which is convergent but not absolutely convergent is called conditionally convergent. Theorem 11.7.5 (Ratio-test for Absolute Convergence) Let ku∑ be a series with non-zero terms and suppose that 1 lim k k k u u ρ + →+∞ = (a) If 1ρ < , the series ku∑ converges absolutely and therefore converges. © Copyright Virtual University of Pakistan 2
- 3. 44-Alternating Series; Conditional Convergence VU (b) If 1 orρ ρ> = +∞ , then the series ku∑ diverges. (c) If 1ρ = , no conclusion about convergence or absolute convergence can be drawn from this test. Example: The series 1 2 ( 1) 2! k k k ∞ = −∑ Converges absolutely since ( ) ( ) 1 1 2 ! lim lim 1 ! 2 2 lim 0 1 1 k k kk k k k u k u k k ρ + + →+∞ →+∞ →+∞ = = + = = < + Power Series in x If 0 1 2, , , ...........c c c are constants and x is a variable, then the series of the form 2 0 1 2 0 ............ ............k k k k k c x c c x c x c x ∞ = = + + + + +∑ is called a power series in x. Some examples of power series in x are 2 3 0 1 ............k k x x x x ∞ = = + + + +∑ 2 3 0 1 ............ ! 2! 3! k k x x x x k ∞ = = + + + +∑ ( ) ( ) 2 2 4 6 0 1 1 ............ 2 ! 2! 4! 6! k k k x x x x k ∞ = − = − + − +∑ Theorem 11.8.1 For any power series in x, exactly one of the following is true: a) The series converges only for x = 0 b) The series converges absolutely (and hence converges) for all real values of x. c) The series converges absolutely (and hence converges) for all x in some finite open interval (-R, R), and diverges if x < -R or x >R. At either of the points x = R or x = -R the series may converge absolutely, converge conditionally, or diverge, depending on the particular series. Radius and Interval of Convergence Theorem 11.8.1 states that the set of values for which a power series in x converges is always an interval centered at zero; we call this the interval of convergence, corresponding to this interval series has radius called radius of convergence. Example: Find the interval of convergence and radius of convergence of the following power series 1 k k x ∞ = ∑ Solution: We shall apply the ratio test for absolute convergence. We have 1 1 lim lim lim k k kk k k k u x x x u x ρ + + →+∞ →+∞ →+∞ = = = = So the ratio test for absolute convergence implies that the series converges absolutely if 1xρ = < and diverges if 1xρ = > . The test is inconclusive if 1 ( 1 1)x i e x or x= − = = − , so convergence at these points must be investigated separately. At these points the series becomes 0 1 1 1 1 1 ............ 1k k x ∞ = = + + + + =∑ ( ) 0 1 1 1 1 1 ............ 1 k k x ∞ = − = − + − + = −∑ Both of which diverge; thus, the interval of convergence for the given power series is (-1, 1) and the radius of convergence is R = 1 Power Series in x-a ( ) ( ) ( ) ( ) 2 0 1 2 0 ............ ............ k k k k k c x a c c x a c x a c x a ∞ = − = + − + − + + − + ∑ © Copyright Virtual University of Pakistan 3
- 4. 44-Alternating Series; Conditional Convergence VU This series is called power series in x-a. Some examples are: ( ) ( ) ( ) 2 0 1 1 1 1 ............ ( 1) 1 2 3 k k x x x a k ∞ = − − − = + + + = + ∑ ( ) ( ) ( ) ( ) 2 0 1 3 3 1 3 ............ ( 3) ! 2! k k x x x a k ∞ = − + + = − + + − = −∑ Theorem 11.8.1 For any power series in ( ) k kc x a−∑ , exactly one of the following is true: a) The series converges only for x a= b) The series converges absolutely (and hence converges) for all real values of x . c) The series converges absolutely (and hence converges) for all x in some finite open interval ( ),a R a R− + , and diverges if x a R< − or x a R> + . At either of the points x a R= − or x a R= + the series may converge absolutely, converge conditionally, or diverge, depending on the particular series. It follows from this theorem that now interval of convergence is centered at x a= © Copyright Virtual University of Pakistan 4