P∞ n Pn n−1 P Theorem The geometric series n=0 ar = i=1 ar = a + Definition A series an is called absolutely convergent if the 2 P∞ n P ar + ar + ··· is convergent if |r| < 1 and its sum is n=0 ar = series of absolute values |an| is convergent. a P 1−r |r| < 1. If |r| ≥ 1, the geometric series is divergent. Definition A series an is called conditionally convergent if Example Find the exact value of the repeating decimal 0.156 = it is convergent but not absolutely convergent. 0.156156156 ... as a rational number. Theorem If a series converges absolutely, then is also converges. Solution Note that The Ratio Test. 156 156 1 156 1 2 an+1 P∞ 0.156 = 1000 + 1000 · 1000 + 1000 · 1000 + ··· (i) If limn→∞ = L < 1, then the series an is abso- an n=1 P∞ 156 1 n 156 = n=0 1000 1000 . This is a geometric series with a = 1000 lutely convergent (and therefore convergent). and r = 1 and has the value: 1000 an+1 an+1 156 156 (ii) If limn→∞ = L > 1 or limn→∞ = ∞, then the 1000 1000 156 an an 1 = 999 = 999 . ∞ (1− ) 1000 P 1000 series n=1 an is divergent. The Test For Divergence. If limn→∞ an does not exist or if P∞ an+1 limn→∞ an 6= 0, then the series an is divergent. (iii) If limn→∞ = 1, the Ratio Test is inconclusive. n=1 an The Integral Test. Suppose f is a continuous, positive, de- The Root Test. creasing function on [1, ∞) and let an = f(n). Then the series ∞ R ∞ pn P∞ P (i) If limn→∞ |an| = L < 1, then the series an is abso- n=1 an is convergent if and only if the improper integral 1 f(x) dx n=1 is convergent. In other words: lutely convergent (and therefore convergent). R ∞ P∞ (ii) If lim pn |a | = L > 1 or lim pn |a | = ∞, then the (i) If 1 f(x) dx is convergent, then n=1 an is convergent. n→∞ n n→∞ n ∞ ∞ ∞ P R P series n=1 an is divergent. (ii) If 1 f(x) dx is divergent, then n=1 an is divergent. ∞ pn P 1 (iii) If limn→∞ |an| = 1, the Root Test is inconclusive. The p-Series Test. A series of the form n=1 np is convergent if p > 1 and divergent if p ≤ 1. Definition A series of the form Example Test the series P∞ 1 for convergence or divergence. P∞ n 2 n=1 n2+1 n=0 cn(x − a) = c0 + c1(x − a) + c2(x − a) + ··· 1 Solution The function f(x) = (x2+1) is continuous, positive and is called a power series in (x − a) or a power series centered decreasing on [1, ∞) so we use the Integral Test: at a or a power series about a. ∞ R ∞ 1 R t 1 −1 t P n dx = lim dx = lim tan x Theorem For a given power series n=0 cn(x−a) , there are only 1 x2+1 t→∞ 1 x2+1 t→∞ 1 −1 π π π π three possibilities: = limt→∞ tan t − 4 = 2 − 4 = 4 . R ∞ 1 Thus 1 (x2+1) dx is a convergent integral and the series (i) The series converges only when x = a. P∞ 1 n=1 (n2+1) is convergent. (ii) The series converges for all x. Remainder Estimate For The Integral Test. Suppose f(k) = (iii) There is a positive number R such that the series converges if ak, where f is a continuous, positive, decreasing function for P |x − a| < R and diverges if |x − a| > R. x ≥ n and an is convergent. If Rn = s − sn, then ∞ ∞ The number R in case (iii) is called the radius of convergence R f(x) dx ≤ R ≤ R f(x) dx. n+1 n n of the power series. By convention, the radius of convergence is The Comparison Test. Suppose that P a and P b are series n n R = 0 in case (i) and R = ∞ in case (ii). The interval of with positive terms. convergence of a power series is the interval that consists of all P P (i) If bn is convergent and an ≤ bn for all n, then an is also values of x for which the series converges. convergent. Example Find the radius of convergence and the interval of con- P P n (ii) If bn is divergent and an ≥ bn for all n, then an is also P∞ x vergence for n=1 n . n n+1 divergent. x an+1 x n n Solution Let an = . Then = · n = |x| → P P n an n+1 x n+1 The Limit Comparison Test. Suppose that an and bn an |x| as n → ∞. By the Ratio Test, this series converges for |x| < 1 are series with positive terms. If limn→∞ = c where c is a finite bn and diverges for |x| > 1. So the radius of convergence is R = 1 number and c > 0, then either both series converge or both diverge. and the series converges on the interval (−1, 1). Example Test the series P∞ 1 for convergence or divergence. n=1 2n−1 We now check end points. Solution We use the Limit Comparison Test with an = P∞ 1 1 For x = 1, we have the harmonic series n=1 n which diverges 1 1 an 2n−1 n bn = n and obtain limn→∞ = limn→∞ 1 = 2 −1 2 bn n by the Integral Test or by the p-Test. For x = −1, we have the 2 ∞ (−1)n 1 2n 1 P lim = lim = 1 > 0. Since this limit exists and alternating harmonic series n=1 n . Since the numbers an = n n→∞ 2n−1 n→∞ 1− 1 2n are decreasing to zero as n → ∞, it converges by the Alternating P 1 is a convergent geometric series, the given series converges by 2n Series Test. So the interval of convergence is [−1, 1). the Limit Comparison Test. Theorem If f has a power series expansion at a, then f(x) = The Alternating Series Test. If the alternating series ∞ f (n)(a) n f 0(a) f 00(a) 2 f 000(a) ∞ P P n−1 n=0 n! (x−a) = f(a)+ 1! (x−a)+ 2! (x−a) + 3! (x− n=1(−1) bn = b1 − b2 + b3 − b4 + b5 − b6 + ··· for bn > 0 3 satisfies a) +··· . This power series is called the Taylor series of the function f(x) at a (or centered at a). If a=0, then the Taylor series of f(x) is (i) bn+1 ≤ bn for all n. P∞ f (n)(0) n called the Maclaurin series of f(x) which is equal to n=0 n! x . (ii) limn→∞ bn = 0, f 0(a) f 00(a) Definition The partial sums Tn(x) = f(a)+ 1! (x−a)+ 2! (x− then the series is convergent. (n) a)2 + ··· + f (a) (x − a)n are called the n-th-degree Taylor poly- Example The alternating harmonic series 1− 1 + 1 − 1 +··· = n! 2 3 4 nomial of f. P∞ (−1)n−1 n=1 n satisfies: Theorem If f(x) = Tn(x) + Rn(x), where Tn is the n-th-degree 1 1 Taylor polynomial of f at a and limn→∞ Rn(x) = 0 for |x − a| < (i) bn+1 < bn because n+1 < n 1 R, then f is equal to the sum of its Taylor series on the interval (ii) limn→∞ bn = limn→∞ = 0, n |x − a| < R. Rn is called the reminder. so the series is convergent by the Alternating Series Test. Taylor’s Inequality. If |f (n+1)(x)| ≤ M for |x − a| ≤ d, P n−1 Alternating Series Estimation Theorem. If s = (−1) bn then the reminder Rn(x) of the Taylor series satisfies the inequality M n+1 is the sum of an alternating series that satisfies (i) 0 ≤ bn+1 ≤ |Rn(x)| ≤ (n+1)! |x − a| for |x − a| ≤ d. bn and (ii) limn→∞ bn = 0, then |Rn| = |s − sn| ≤ bn+1. 1.