CHAPTER 5 INFINITE SEQUENCES AND SERIES SERIES

CHAPTER INFINITE SEQUENCES AND SERIES CONTENTS 5.1 Sequences 5.2 Series 5.3 The Integral and Comparison Test 5.4 Other Convergence Test 5.5 Power Series 5.6 Representations of Functions as Power Series 5.7 Taylor and Maclaurin Series 5.8 The Binomial Series 5.9 Applications of Taylor Polynomial 5.10 Using Series to solve Differential Equations 5.11 Fourier Series 5.2 SERIES 5.2.1 The Sum of a Series 5.2.2 Geometric Series 5.2.3 The Test for Divergence 5.2.4 Properties of Convergent Series 5.2.1 The Sum of a Series  We can add the terms of a sequence {an } and get an expression of the form: a1+ a2+ a3+ …+ an + … which is called a series and denoted by  a n 1 n or a n  However what does it mean by the sum of infinitely many terms? Example We can try to add the terms of the series 1+2+3+…+n+… and get the cumulative sums 1, 3, 6, 10, …, The nth sum n(n+1)/2 becomes very large as n increases  On the other hand if we add the terms of the series 1 1    n  we get the cumulative sums , , ,  ,1  n ,  The nth sum 1-1/2n becomes closer and closer to as n increases infinitely We may say that the sum of this infinite series is in the following sense: Definition Given a series   n 1 an  a1  a2    an   Let sn denote the nth partial sum n sn    a1  a2    an i 1 If lim sn  s exists as a real number, we say that n  the series is convergent and write   n 1 an  a1  a2    an    s otherwise, it is divergent This number s is called the sum of the series So now we can write 1 1 n1 n      n     Note The fact that a series is convergent or not does not change if we suppress a finite number of terms 5.2.2 The Geometric Series Let's consider the geometric series  a  ar    ar n 1     ar n 1 n 1 a0 5.2.2 The Geometric Series Let's consider the geometric series  a  ar    ar n 1     ar n 1 a0 n 1 If r = 1, then the nth sum sn=na   as n   Hence the geometric series is divergent If r  1, then we have sn  a  ar  ar    ar n 1 and Hence rsn  ar  ar    ar sn-rsn = a - arn n 1  ar n sn-rsn = a - arn a (1  r n ) sn  1 r If –1[...]... series 40 5  103  209  27  Solution We have a =5, r=-2/3 Since r < 1 the series is convergent with the sum 5 5 10 20 40 5  3  9  27     5 3 2 1  ( 3 ) 3  Example Is the series 2 n 1 n 2  3 n 1 convergent?  Example Is the series 2 n 1 n 2  3 n 1 convergent? Solution We rewrite the nth term of the series in the form n 1 n 4 4 2 n 1 n 2 3  n 1  4  3 3 Therefore the series. .. divergent  n2  2 5 n 4 n 1 Solution We have n2 lim 2 n  5n  4 1 1  lim  0 n  4 5 5 2 n Therefore the series is divergent by the Test for Divergence Example Show that the series is divergent a 2𝑛 ∞ 𝑛=1 3𝑛+1 b ∞ 𝑛=1 sin 𝑛 c 1 1 + 3 + 1 5 (or 1 + 7 ∞ 𝑛=1 sin 𝑛 𝑥 + ⋯…+ 1 2𝑛−1 where 𝑥 ≠ 𝑘𝜋 ) +⋯ 5. 2.4 Properties of Series Theorem If  a n 1  n and  bn are convergent series n 1 and c is a constant,... 100 23 17 1147    10 990 4 95 Example Find the sum of the series  n x  , where x  1 n 0 Example Find the sum of the series  n x  , where x  1 n 0 Solution This is a geometric series with a=1 and r = x Since r < 1, the series is convergent; its sum is  1 n x   1 x n 0 Example Find the sum of the series  1  n 1 n( n  1) Solution This is not a geometric series but we can write the... geometric series  n 1 n 1 ar  a  ar    ar   n 1 is convergent if r < 1 and its sum is  a n 1 ar   1 r n 1 if r  1, the geometric series is divergent Example Find the sum of the geometric series 40 5  103  209  27  The geometric series  n 1 n 1 ar  a  ar    ar   n 1 is convergent if r < 1 and its sum is  a n 1 ar   1 r n 1 if r  1, the geometric series. .. later that the harmonic series is divergent  1 1 1 1  1     2 3 n n 1 n an The Test for Divergence If lim n  an  0 , then the series or lim n  Example Show that the series is divergent does not exist  a n 1 n  is divergent n2  2 5 n 4 n 1 an The Test for Divergence If lim n  does not exist  a an  0 , then the series or lim n  n 1 Example Show that the series is divergent... Therefore the series is a geometric series with a=4 and r = 4/3 > 1 so it is divergent Example Write 2.317  2.3171717 as a fraction Example Write 2.317  2.3171717 as a fraction Solution We may write 17 17 17 2.3171717  2.3  3  5  7   10 10 10 Apart from the first term 2.3, the rest is the sum of a geometric series with a=17/103 and r = 1/102 < 1 so it is convergent, and we have 17 1 17 1 2.317 ... then the following series are convergent and   ca n 1   n  c  an n 1    (a n  bn )   a  b  (a n  bn )   a  b n 1  n 1 n 1  n 1 n n n 1  n 1 n n Example Find the sum of the series  3 1    n   2  n 1  n( n  1)  Example Find the sum of the series  3 1    n   2  n 1  n( n  1)   1 Solution1 We have seen that the series  is convergent and its sum is... its sum is 1 n 1 n( n  1)  1 On the other hand the series  n is a geometric n 1 2 1 1 series with a  2 and r  2  1 1 2  It is convergent; its sum is  2 n 1  1  1 n 1 2 Therefore the given series is convergent; its sum is    3 1  1 1   n   3  n  2  n 1  n( n  1) n 1 n( n  1) n 1 2  3 1  1  4  Example Find the sum of the series ∞ 𝑛=1 1 2 + 3𝑛 (3𝑛 − 2)(3𝑛 + 1) ...  n  1 Therefore the series is convergent with the sum  1 1  n 1 n( n  1) Example Determine whether the series is convergent or divergent If the series is convergent, find its sum a ∞ 𝑛=1 b 1 1.4 −3 𝑛−1 23𝑛 1 + 4.7 + ⋯… + 1 3𝑛−2 3𝑛+1 + ⋯… Example Expressing the repeating decimal 0.9999999 (or 4.17326326326)… as a fraction 5. 2.3 The Test for Divergence Theorem 1 If the series then lim a  0 n ...CONTENTS 5. 1 Sequences 5. 2 Series 5. 3 The Integral and Comparison Test 5. 4 Other Convergence Test 5. 5 Power Series 5. 6 Representations of Functions as Power Series 5. 7 Taylor and Maclaurin Series 5. 8... Binomial Series 5. 9 Applications of Taylor Polynomial 5. 10 Using Series to solve Differential Equations 5. 11 Fourier Series 5. 2 SERIES 5. 2.1 The Sum of a Series 5. 2.2 Geometric Series 5. 2.3 The... Divergence 5. 2.4 Properties of Convergent Series 5. 2.1 The Sum of a Series  We can add the terms of a sequence {an } and get an expression of the form: a1+ a2+ a3+ …+ an + … which is called a series and