MATH115. Infinite Series. Paolo Lorenzo Bautista. July 17, De La Salle University. PLBautista (DLSU) MATH115 July 17, / 43

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MATH115 Infinite Series Paolo Lorenzo Bautista De La Salle University July 17, 2014 PLBautista (DLSU) MATH115 July 17, 2014 1 / 43

Infinite Series Definition If {u n } is a sequence and s n = u 1 + u 2 + u 3 + + u n, then {s n } is a sequence of partial sums called an infinite series denoted by u n = u 1 + u 2 + u 3 + + u n +. The numbers u 1, u 2,..., u n,... are called the terms of the infinite series. PLBautista (DLSU) MATH115 July 17, 2014 2 / 43

Infinite Series Definition Let u n denote a given infinite series for which {s n } is the sequence of partial sums. If lim s n exists and is equal to S, then the series is n + convergent and S is the sum of the series. If lim s n does not exist, n + then the series is divergent, and the series does not have a sum. PLBautista (DLSU) MATH115 July 17, 2014 3 / 43

Infinite Series Theorem If the infinite series u n is convergent, then lim u n = 0. n + PLBautista (DLSU) MATH115 July 17, 2014 4 / 43

Infinite Series Theorem If the infinite series Equivalently, if u n is convergent, then lim u n = 0. n + lim u n 0, then the infinite series u n is divergent. n + PLBautista (DLSU) MATH115 July 17, 2014 4 / 43

Infinite Series Definition The series is called the harmonic series. 1 n PLBautista (DLSU) MATH115 July 17, 2014 5 / 43

Infinite Series Definition The series is called the harmonic series. Theorem The harmonic sequence is divergent. 1 n PLBautista (DLSU) MATH115 July 17, 2014 5 / 43

Infinite Series Definition The series is called the geometric series. ar n 1 PLBautista (DLSU) MATH115 July 17, 2014 6 / 43

Infinite Series Definition The series is called the geometric series. Theorem ar n 1 The geometric series converges to the sum if r 1. a if r < 1 and diverges 1 r PLBautista (DLSU) MATH115 July 17, 2014 6 / 43

Infinite Series Theorem Let c be any nonzero constant. i) If the series u n is convergent and its sum if S, then the series cu n is also convergent, and its sum is cs. ii) If the series divergent. u n is divergent, then the series cu n is also PLBautista (DLSU) MATH115 July 17, 2014 7 / 43

Infinite Series Theorem If a n and T respectively, then, i) ii) b n are convergent infinite series whose sums are S and (a n + b n ) is a convergent series and its sum is S + T. (a n b n ) is a convergent series and its sum is S T. PLBautista (DLSU) MATH115 July 17, 2014 8 / 43

Infinite Series Theorem If the series the series a n is convergent and the series (a n + b n ) is divergent. b n is divergent, then PLBautista (DLSU) MATH115 July 17, 2014 9 / 43

Infinite Series Theorem If a n and b n are two infinite series, differing only in their first m terms, then either both series converge or both series diverge. PLBautista (DLSU) MATH115 July 17, 2014 10 / 43

Infinite Series of Positive Terms Theorem An infinite series of positive terms is convergent if and only if its sequence of partial sums has an upper bound. PLBautista (DLSU) MATH115 July 17, 2014 11 / 43

Infinite Series of Positive Terms Theorem ((First) Comparison Test) Let the series i) If ii) If u n be a series of positive terms. v n is a series of positive terms known to be convergent, and u n v n for all positive integers n, then u n is convergent. w n is a series of positive terms known to be divergent, and u n w n for all positive integers n, then u n is divergent. PLBautista (DLSU) MATH115 July 17, 2014 12 / 43

Infinite Series of Positive Terms Example Determine whether the following series are convergent or divergent. 1. 2. 3. 4. 4 3 n + 1 1 n2 n 1 ln(n + 1) 1 n! PLBautista (DLSU) MATH115 July 17, 2014 13 / 43

Infinite Series of Positive Terms Theorem (Limit Comparison Test) Let the u n and v n be two series of positive terms. u n i) If lim = c > 0, then the two series either both converge or n + v n both diverge. ii) If iii) If u n lim = 0, and if n + v n u n lim = +, and if n + v n v n converges, then v n diverges, then u n converges. u n diverges. PLBautista (DLSU) MATH115 July 17, 2014 14 / 43

Infinite Series of Positive Terms Example Determine whether the following series are convergent or divergent. 1. 2. 3. 4. n 2 4n 3 + 1 2n + 1 2n 2 + 3 1 n 2 + 4n sin 1 n PLBautista (DLSU) MATH115 July 17, 2014 15 / 43

Infinite Series of Positive Terms Some remarks: PLBautista (DLSU) MATH115 July 17, 2014 16 / 43

Infinite Series of Positive Terms Some remarks: Theorem If u n is a given convergent series of positive terms, its terms can be grouped in any manner, and the resulting series also will be convergent and will have the same sum as the given series. PLBautista (DLSU) MATH115 July 17, 2014 16 / 43

Infinite Series of Positive Terms Some remarks: Theorem If u n is a given convergent series of positive terms, its terms can be grouped in any manner, and the resulting series also will be convergent and will have the same sum as the given series. Theorem If u n is a given convergent series of positive terms, the order of the terms can be rearranged, and the resulting series also will be convergent and will have the same sum as the given series. PLBautista (DLSU) MATH115 July 17, 2014 16 / 43

Infinite Series of Positive Terms Definition The series = 1 1 p + 1 2 p + 1 3 p + + 1 n p + is called the p series or the hyperharmonic series. PLBautista (DLSU) MATH115 July 17, 2014 17 / 43

Infinite Series of Positive Terms Definition The series = 1 1 p + 1 2 p + 1 3 p + + 1 n p + is called the p series or the hyperharmonic series. Theorem The p series diverges if p 1, and converges if p > 1. PLBautista (DLSU) MATH115 July 17, 2014 17 / 43

Infinite Series of Positive Terms Example Determine whether the following series are convergent or divergent. 1. 2. 1 (n 2 + 2) 1/3 1 n 3 + 1 PLBautista (DLSU) MATH115 July 17, 2014 18 / 43

Infinite Series of Positive Terms Theorem (Integral Test) Let f be a function that is continuous, decreasing, and positive valued for all x 1. Then the infinite series f (n) = f (1) + f (2) + + f (n) + is convergent if the improper integral divergent if b lim b + 1 f (x)dx = +. + 1 f (x)dx exists, and is PLBautista (DLSU) MATH115 July 17, 2014 19 / 43

Infinite Series of Positive Terms Theorem (Integral Test) Let f be a function that is continuous, decreasing, and positive valued for all x 1. Then the infinite series f (n) = f (1) + f (2) + + f (n) + is convergent if the improper integral divergent if b lim b + 1 f (x)dx = +. + 1 f (x)dx exists, and is The theorem is also true if the infinite series starts at a number k > 1. PLBautista (DLSU) MATH115 July 17, 2014 19 / 43

Infinite Series of Positive Terms Example Determine whether the following series are convergent or divergent. 1. 2. ne n ln n n 2 + 2 PLBautista (DLSU) MATH115 July 17, 2014 20 / 43

Alternating Series Definition If a n > 0 for all positive integers n, then the series and the series ( 1) n+1 a n = a 1 a 2 + a 3 a 4 + + ( 1) n+1 a n + ( 1) n a n = a 1 + a 2 a 3 + a 4 + ( 1) n a n + are called alternating series. PLBautista (DLSU) MATH115 July 17, 2014 21 / 43

Alternating Series Theorem (Alternating Series Test) Let ( 1) n+1 a n (or and a n+1 < a n for all positive integers n. If series is convergent. ( 1) n a n ) be an alternating series, where a n > 0 lim a n = 0, the alternating n + PLBautista (DLSU) MATH115 July 17, 2014 22 / 43

Alternating Series Definition The infinite series is convergent. u n is absolutely convergent if the series u n PLBautista (DLSU) MATH115 July 17, 2014 23 / 43

Alternating Series Definition The infinite series is convergent. u n is absolutely convergent if the series Definition A series that is convergent, but not absolutely convergent, is conditionally convergent. u n PLBautista (DLSU) MATH115 July 17, 2014 23 / 43

Alternating Series Theorem If the series u n is convergent, then the series u n is convergent. PLBautista (DLSU) MATH115 July 17, 2014 24 / 43

Alternating Series Theorem (Ratio Test) Let the u n be a given infinite series for which every u n is nonzero. i) If lim u n+1 n + u n = L < 1, then the series is absolutely convergent. ii) If lim u n+1 n + u n = L > 1, or if lim u n+1 n + u n = +, then the series is divergent. iii) If lim u n+1 n + u n = 1, then the ratio test fails (no conclusion can be made). PLBautista (DLSU) MATH115 July 17, 2014 25 / 43

Alternating Series Theorem (Root Test) Let the i) If lim ii) If lim n + u n be a given infinite series for which every u n is nonzero. n un = L < 1, then the series is absolutely convergent. n + n un = L > 1, or if lim n + n un = +, then the series is divergent. n iii) If un = 1, then the root test fails (no conclusion can be lim n + made). PLBautista (DLSU) MATH115 July 17, 2014 26 / 43

Summary Summary of Tests for Convergence or Divergence 1. lim u n 0 implies divergence of n + 2. Alternating Series Test 3. Compare with known series (Geometric, Harmonic, p) 4. Ratio Test 5. Root Test 6. Integral Test 7. FCT, LCT u n PLBautista (DLSU) MATH115 July 17, 2014 27 / 43

Power Series Power Series Definition A power series in x a is a series of the form n=0 c n (x a) n = c 0 + c 1 (x a) + c 2 (x a) 2 + + c n (x a) n +. PLBautista (DLSU) MATH115 July 17, 2014 28 / 43

Power Series Example Find the values of x for which the following power series are convergent. 1. 2. 3. 4. n=0 n=0 ( 1) n+1 2n x n n3 n x n n! n!x n n 3 x n PLBautista (DLSU) MATH115 July 17, 2014 29 / 43

Power Series Theorem If the power series n=0 c n x n is convergent for x = x 1, (x 1 0), then it is absolutely convergent for all values of x for which x < x 1. PLBautista (DLSU) MATH115 July 17, 2014 30 / 43

Power Series Theorem If the power series n=0 c n x n is convergent for x = x 1, (x 1 0), then it is absolutely convergent for all values of x for which x < x 1. Theorem If the power series n=0 for all values of x for which x > x 2. c n x n is divergent for x = x 2, then it is divergent PLBautista (DLSU) MATH115 July 17, 2014 30 / 43

Power Series Theorem Let n=0 c n x n be a given power series. Then exactly one of the following conditions holds: i) The series converges only when x = 0. ii) The series is absolutely convergent for all values of x. iii) There exists a number R > 0 such that the series is absolutely convergent for all values of x for which x < R and is divergent for all values of x for which x > R. PLBautista (DLSU) MATH115 July 17, 2014 31 / 43

Power Series Theorem Let n=0 c n (x a) n be a given power series. Then exactly one of the following conditions holds: i) The series converges only when x = a. ii) The series is absolutely convergent for all values of x. iii) There exists a number R > 0 such that the series is absolutely convergent for all values of x for which x a < R and is divergent for all values of x for which x a > R. PLBautista (DLSU) MATH115 July 17, 2014 32 / 43

Power Series Theorem Let n=0 c n (x a) n be a given power series. Then exactly one of the following conditions holds: i) The series converges only when x = a. ii) The series is absolutely convergent for all values of x. iii) There exists a number R > 0 such that the series is absolutely convergent for all values of x for which x a < R and is divergent for all values of x for which x a > R. Definition The number R described in (iii) is called the radius of convergence of the power series. If R > 0, the set of values of x for which the power series is convergent is called the interval of convergence of the power series. PLBautista (DLSU) MATH115 July 17, 2014 32 / 43

Power Series Example Determine the interval of convergence for the following power series. 1. 2. n(x 2) n x n 2 + n 2 PLBautista (DLSU) MATH115 July 17, 2014 33 / 43

Power Series Exercise Determine the interval of convergence for the following power series. 1. 2. 3. 4. 5. n=0 n=0 n=0 n=0 n 2 x n 2 n (x + 2) n (n + 1)2 n 4 n+1 x 2n n + 1 n + 1 n 2n xn x n ln(n + 1) PLBautista (DLSU) MATH115 July 17, 2014 34 / 43

Differentiation/Integration Theorem If n=0 then c n x n is a power series having a radius of convergence of R > 0, nc n x n 1 also has R as its radius of convergence. PLBautista (DLSU) MATH115 July 17, 2014 35 / 43

Differentiation/Integration Theorem Let n=0 c n x n be a power series whose radius of convergence is R > 0. If f is the function defined by f (x) = n=0 c n x n, then f (x) exists for every x in the open interval ( R, R) and f (x) = n=0 nc n x n 1. PLBautista (DLSU) MATH115 July 17, 2014 36 / 43

Differentiation/Integration Example 1. Find a power series representation of 2. Show that e x = n=0 x n n! 1 (1 x) 2. PLBautista (DLSU) MATH115 July 17, 2014 37 / 43

Differentiation/Integration Theorem Let n=0 c n x n be a power series whose radius of convergence is R > 0. If f is the function defined by f (x) = n=0 c n x n, then f is integrable on every closed subinterval of ( R, R), and the integral of f is evaluated by integrating the given power series term by term; that is, if x is in ( R, R), then x 0 f (t)dt = n=0 c n n + 1 xn+1. Furthermore, R is the radius of convergence of the resulting series. PLBautista (DLSU) MATH115 July 17, 2014 38 / 43

Differentiation/Integration Example Obtain a power series representation of the following: 1. x 0 e t2 dt 2. tan 1 x = x 0 1 1 + t 2 dt PLBautista (DLSU) MATH115 July 17, 2014 39 / 43

Taylor Series Taylor Series Definition The Taylor series expansion of a function f (x) = given by n=0 f (n) (a) (x a) n. n! n=0 c n (x a) n at a is The Taylor series expansion of a function at a = 0 is called the Maclaurin series expansion of f given by n=0 f (n) (0) x n. n! PLBautista (DLSU) MATH115 July 17, 2014 40 / 43

Taylor Series Example 1. Find the Maclaurin series expansion of e x. 2. Find the Taylor series expansion of sin x at a = π 2. PLBautista (DLSU) MATH115 July 17, 2014 41 / 43

Taylor Series Exercise Find the Maclaurin series expansion of the following functions. 1. f (x) = 1 1 x 2. f (x) = e x2 3. f (x) = x 2 e x3 4. f (x) = sin x 5. f (x) = cos x PLBautista (DLSU) MATH115 July 17, 2014 42 / 43

Taylor Series Example 1. Find the Taylor series for e x at a = 3. 2. Obtain the Maclaurin series for sinh x = ex e x. 2 3. Obtain the Maclaurin series for cosh x. 4. Obtain the Taylor series for f (x) = ln x at a = 1. 5. Obtain the Taylor series for f (x) = 3 x at a = 8. PLBautista (DLSU) MATH115 July 17, 2014 43 / 43