Math 0230 Calculus 2 Lectures Chapter 8 Series Numeration of sections corresponds to the text James Stewart, Essential Calculus, Early Transcendentals, Second edition. Section 8.1 Sequences A sequence is a list of numbers written in a definite order. It also can be treated as a function defined at the integer numbers only. A sequence can be finite or infinite. 1 Notation: a1, a2, a3, ... , an, ... fa1; a2; a3; :::g fang fangn=1 1 fa4; a5; a6; :::g fangn=4 1 1 1 1 1 1 Example 1. (a) an = , 1; ; ; :::; ; ::: n n=1 n 2 3 n ( p p ) n πn o1 πn 3 3 πn (b) sin an = sin , 0; ; ; 0; :::; sin ; ::: 3 n=0 3 2 2 3 (−1)nn2 1 (−1)nn2 9 25 9 (−1)nn2 (c) n an = n , 1; − ; 1; − ; ; :::; n ; ::: 2 n=2 2 8 32 16 2 (d) The Fibonacci sequence ffng cannot be represented by a formula and is defined recursively: f1 = 1, f2 = 1, fn = fn−1 + fn−2 for n ≥ 3, f1; 1; 2; 3; 5; 8; 13; 21; :::g 4 5 2 7 Example 2. Find a formula for a general term of the sequence 1; ; ; ; ; ::: . 9 27 27 243 n + 2 Solution: a = n 3n As we can see the terms of this sequence are getting smaller and smaller closser to 0. Definition A sequence fang has the limit L: lim an = L n!1 if the terms an can be made as close to L as we like by taking n sufficiently large. If the limit exists, then the sequence is called convergent, otherwise it's called divergent. If lim f(x) = L and f(n) = an when n is an integer, then lim an = L n!1 n!1 1 Definition lim an = 1 means that for any positive number M there is a positive integer N n!1 such that if n > N, then an > M. Limit Laws If an and bn are convergent sequences and c is a constant, then lim (an + bn) = lim an + lim bn n!1 n!1 n!1 lim (an − bn) = lim an − lim bn n!1 n!1 n!1 lim can = c lim an n!1 n!1 lim c = c n!1 lim (an · bn) = lim an · lim bn n!1 n!1 n!1 lim an an n!1 lim = if lim bn 6= 0 n!1 bn lim bn n!1 n!1 p p h i lim an = lim an if p > 0 and an > 0 n!1 n!1 Squeeze Theorem If an ≤ bn ≤ cn for n ≥ N and lim an = lim cn = L then lim bn = L. n!1 n!1 n!1 such that if n > N, then an > M. If lim janj = 0, then lim an = 0 n!1 n!1 n + 1 Example 3. Find lim . n!1 3n − 2 Solution: Divide numerator and denominator by the highest power of n that occurs in the denominator and then apply the Limit Laws n + 1 1 + 1=n lim 1 + lim 1=n 1 + 0 1 lim = lim = n!1 n!1 = = n!1 3n − 2 n!1 3 − 2=n lim 3 − 2 lim 1=n 3 − 0 3 n!1 n!1 (−1)n Example 4. Find lim . n!1 n + 1 n (−1) 1 Solution: lim = lim n!1 n + 1 n!1 n + 1 1 1 1 0 < < when n ≥ 1. lim 0 = 0, lim = 0. n + 1 n n!1 n!1 n 1 (−1)n Then by Squeeze theorem lim = 0 and lim = 0. n!1 n + 1 n!1 n + 1 2 Continuity and Convergence Theorem If lim an = L and the function f is continuous at n!1 L, then lim f(an) = f(L). n!1 The sequence frng is convergent if −1 < r ≤ 1 and divergent for all other values of r. lim rn = 0 if −1 < r < 1 n!1 Definition A sequence fang is called increasing if an < an+1 for all n ≥ 1. It is called decreasing if an > an+1 for all n ≥ 1. A sequence is called monotonic if it is either increasing or decreasing. Definition A sequence fang is bounded above if there is a number M such that an ≤ M for all n ≥ 1. A sequence fang is bounded below if there is a number m such that an ≥ m for all n ≥ 1. If it is bounded above and below, then fang is a bounded sequence. Monotonic Sequence Theorem Every bounded monotonic sequence is convergent. 3 Section 8.2 Series p 1 1 1 1 1 1 Example 1. 2 = 1:41421356::: = 1 + 4 · + 1 · + 4 · + 2 · + 1 · + 3 · 10 102 103 104 105 106 1 1 +5 · + 6 · + ··· 107 108 It is an infinite sum (series) of numbers. 1 In general, if we have a sequence fangn=1 then we may want to add its terms: 1 X a1 + a2 + a3 + ··· + ai + ··· = an n=1 This expression is called an infinite series (or just a series). To define a value of a series consider its partial sums: n X s1 = a1, s2 = a1 + a2, s3 = a1 + a2 + a3, ..., sn = a1 + a2 + ··· an = ai, i=1 1 where sn is called nth partial sum. The partial sums form a new sequence fsngn=1 1 1 X If the sequence fsng is convergent and lim sn = s, then the series an is called convergent n=1 n!1 n=1 1 X and an = s. The number s is called the sum of the series. n=1 1 1 X If fsngn=1 is divergent, then the series an is called divergent. n=1 1 n X X Special notation: an = lim ai n!1 n=1 i=1 1 1 X X Geometric series: arn−1 = arn, a 6= 0. n=1 n=0 1 X a It is convergent when jrj < 1. In this case its sum is arn−1 = jrj < 1. 1 − r n=1 If jrj ≥ 1, then the geometric series is divergent. Example 2. Represent the number 3:46 as a ratio of integers. 46 46 46 46 Solution: 3:46 = 3:46464646::: = 3 + + + + + ··· 100 1002 1003 1004 1 n 46 1 1 1 X 46 1 = 3 + 1 + + + + ··· = 3 + . 100 1002 1003 1004 100 100 n=0 4 46 1 It is a geometric series with a = and −1 < r = < 1. 100 100 46=100 46=100 46 297 + 46 343 Therefore, 3:46 = 3 + = 3 + = 3 + = = 1 − 1=100 99=100 99 99 99 1 X 1 Example 3. xn = if jxj < 1. 1 − x n=0 1 X 2 Example 4. Show that the series is convergent and find its limit. n(n + 2) n=1 2 1 1 Solution: = − . n(n + 2) n n + 2 The partial sum is n n X 2 X 1 1 = − n(n + 2) i i + 2 i=1 i=1 1 1 1 1 1 1 1 1 1 1 1 1 1 = 1 − + − + − + − + ··· + − + − + − 3 2 4 3 5 4 6 n − 2 n n − 1 n + 1 n n + 2 1 1 1 1 1 1 1 1 1 1 1 = 1 + + + + ··· + + + − − − · · · − − − 2 3 4 n − 2 n − 1 n 3 4 n n + 1 n + 2 1 1 1 = 1 + − − . 2 n + 1 n + 2 1 n X 2 X 2 1 1 1 3 Therefore = lim = lim 1 + − − = . n(n + 2) n!1 n(n + 2) n!1 2 n + 1 n + 2 2 n=1 i=1 1 X 1 Harmonic series: is divergent. n n=1 1 X Theorem If the series an is convergent, then lim an = 0. n!1 n=1 1 X Test for divergence If lim an DNE or lim an 6= 0, then the series an is divergent. n!1 n!1 n=1 1 X 3n2 Example 5. Show that the series is divergent. n(n + 2) n=1 3n2 Solution: lim = 3 6= 0. By the test for divergence the series is divergent. n!1 n(n + 2) 5 Section 8.3 The Integral and Comparison Tests The Integral Test Suppose an = f(n), n ≥ n0 where f is a continuous, positive function 1 X on [n0; 1) decreasing on [m; 1) for some m ≥ n0. Then the series an is convergent if and n=n0 1 Z only if the improper integral f(x) dx is convergent. In other words: n=n0 1 1 Z X If f(x) dx is convergent, then an is convergent. n=n0 n=n0 1 1 Z X If f(x) dx is divergent, then an is divergent. n=n0 n=n0 1 X 2n − 3 Example 1. Using the Integral Test determine whether the series is convergent n(n − 3) n=3 or divergent. 2x − 3 2x − 3 Solution: Consider the function f(x) = = . Define its domain as x ≥ n = 4 x(x − 3) x2 − 3x 0 on which f is continuous and positive. Is it decreasing there? Take its derivative 2(x2 − 3x) − (2x − 3)2 2x2 − 6x − 4x2 + 12x − 9 −2x2 + 6x − 9 f 0(x) = = = (x2 − 3x)2 (x2 − 3x)2 (x2 − 3x)2 The denominator is always positive inside the domain. The descriminant of the numerator is D = 9 − 18 = −9 < 0. When x = 0 the numerator is −9 < 0. Hence, it is negative for x ≥ 4. So, the function is decreasing. 1 1 t Z Z 2x − 3 Z 2x − 3 f(x) dx = dx = lim dx x2 − 3x t!1 x2 − 3x 4 4 4 u-sub: u = x2 − 3x;, du = (2x − 3) dx, u(4) = 4, u(t) = t2 − 3t.