.... Analysis II: Basic knowledge of real analysis: Part IV, Series Kenichi Maruno Department of Mathematics, The University of Texas - Pan American March 1, 2011 K.Maruno (UT-Pan American) Analysis II March 1, 2011 1 / 22
The Sum of an Infinite Series Definition Let {a n } be a sequence. For each positive integer n, let s n = a 1 + a 2 + + a n = n a n. k=1 An infinite series is the ordered pair of sequences ({a n }, {s n }). The number a n is called the nth term of the infinite series, and the number s n is called the nth partial sum of the infinite series. Instead of using the cumbersone ordered pair notation for an infinite series, we will use the notation a n, or a 1 + a 2 + a 3 + to represent an infinite series. K.Maruno (UT-Pan American) Analysis II March 1, 2011 2 / 22
Convergence of an Infinite Series Definition Let a n be an infinite series. If the sequence of partial sum {s n } (s n = a 1 +a 2 + +a n ) converges to L, we say that the infinite series a n converges to L or that the infinite series a n has sum L. If the sequence {s n } diverges, we say that the infinite series a n diverges. If the infinite series a n converges, we also use the symbolism a n to denote its sum. K.Maruno (UT-Pan American) Analysis II March 1, 2011 3 / 22
Convergence of an Infinite Series If the infinite series (Proof) Suppose the infinite series Theorem a n converges, then lim n a n = 0. a n converges to L. Let s n = a 1 + + a n be the nth partial sum. Then s n+1 s n = a n+1 and so 0 = L L = lim n s n+1 lim n s n = lim n (s n+1 s n ) = lim n a n+1. Therefore lim n a n = 0. K.Maruno (UT-Pan American) Analysis II March 1, 2011 4 / 22
(Contrapositive) Convergence of an Infinite Series If lim n a n 0, the infinite series This is also true! e.g. n n + 1 n lim n n+1 a n diverges. = 1. So this series diverges. (Converse) If lim n a n = 0, the infinite series This is not true! (e.g. 1 n a n converges. 1 diverges although lim n n = 0.) K.Maruno (UT-Pan American) Analysis II March 1, 2011 5 / 22
Convergence of an Infinite Series Theorem: The Geometric Series Let a be a nonzero number. Then (i) ar n converges to if r < 1. (ii) n=0 a 1 r ar n diverges if r 1. n=0 K.Maruno (UT-Pan American) Analysis II March 1, 2011 6 / 22
Series with Nonnegative Terms Let Theorem a n be a series with nonnegative terms. Then a n converges if and only if the sequence of partial sums {s n } is bounded. (Proof) Let a n be a series with nonnegative terms. Since a n 0 for every n N, the sequence {s n } is increasing. By the monotone convergence theorem, {s n } converges if and only if {s n } is bounded. K.Maruno (UT-Pan American) Analysis II March 1, 2011 7 / 22
Series with Nonnegative Terms Example. 1 n diverges. Let s n = 1 + 1 2 + + 1 n. Then s 1 = 1 1 s 2 = 1 + 1 2 3 2 s 4 = s 2 + 1 3 + 1 4 3 2 + 1 4 + 1 4 = 2 s 8 = s 4 + 1 5 + 1 6 + 1 7 + 1 8 2 + 1 8 + 1 8 + 1 8 + 1 8 = 5 2. In general, s 2 n (n + 2)/2. Since s n is unbounded, 1 n diverges. K.Maruno (UT-Pan American) Analysis II March 1, 2011 8 / 22
Series with Nonnegative Terms Theorem: 2 n Test Let {a n } be a decreasing sequences of nonnegative numbers. Then a n converges if and only if 2 n a 2 n converges. K.Maruno (UT-Pan American) Analysis II March 1, 2011 9 / 22
Series with Nonnegative Terms Corollary The series 1 diverges if s 1 and converges if s > 1. ns (Proof) If s < 0, lim n (1/n s ) 0. So 1/ns diverges. Suppose s 0. By the 2 n test, 1/ns converges if and only if 2 n 1 (2 n ) s = 2 (1 s)n converges. This is a geometric series, so 2 (1 s)n converges if 2 (1 s) < 1 and diverges if 2 (1 s) 1. Thus 1/ns converges if 1 s < 0 and diverges if 1 s 0. K.Maruno (UT-Pan American) Analysis II March 1, 2011 10 / 22
The Alternating Series Test Theorem: Alternating Series Test Let {a n } be a decreasing sequences such that ( 1) n+1 a n converges. lim a n = 0. Then n K.Maruno (UT-Pan American) Analysis II March 1, 2011 11 / 22
Absolute Convergence Definition Let a n be an infinite series. If a n converges, we say a n converges absolutely. If a n converges and a n diverges, we say a n converges conditionally. Any convergent series with nonnegative terms converges absolutely. K.Maruno (UT-Pan American) Analysis II March 1, 2011 12 / 22
Absolute Convergence If. a n converges absolutely, then Theorem a n converges and a n a n K.Maruno (UT-Pan American) Analysis II March 1, 2011 13 / 22
Absolute Convergence Theorem: Comparison Test Let a n and b n be two series such that a n b n for every positive integer n. (i) If b n converges absolutely, then a n converges absolutely and a n b n. (ii) If a n diverges, then b n diverges. K.Maruno (UT-Pan American) Analysis II March 1, 2011 14 / 22
Absolute Convergence Example (Comparison Test): ( 1) n converges. n! (Proof) ( 1) n n! = 1 n! 1 2 n 1 1 for every positive integer n. Since converges abusolutely, by the 2n 1 ( 1) n comparison test, converges absolutely. n! K.Maruno (UT-Pan American) Analysis II March 1, 2011 15 / 22
Absolute Convergence Theorem: Ratio Test Let a n be a series such that L = lim n a n+1 a n exists (L = is allowed). (i) If L < 1, then a n converges absolutely. (ii) If L > 1, then a n diverges. The ratio test gives no information if lim n a n+1 a n = 1. K.Maruno (UT-Pan American) Analysis II March 1, 2011 16 / 22
Absolute Convergence (Examples) The series 1/n! converges since the ratio 1/(n + 1)! 1/n! = 1 n + 1 converges to 0. The series 1/n diverges, but the ratio test gives no information because of lim n (1/(n + 1))/(1/n) = lim n n/(n + 1) = 1. The series 1/n2 converges, but the ratio test gives no information because of lim n (1/(n + 1) 2 )/(1/n 2 ) = lim n n 2 /(n + 1) 2 = 1. K.Maruno (UT-Pan American) Analysis II March 1, 2011 17 / 22
Absolute Convergence Theorem: Root Test Let {a n } be a sequence and let L = lim sup n a n 1/n (L = is allowed). (i) If L < 1, then a n converges absolutely. (ii) If L > 1, then a n diverges. The root test gives no information if lim sup n a n 1/n = 1. K.Maruno (UT-Pan American) Analysis II March 1, 2011 18 / 22
Power Series Definition Let t be a fixed real number. A power series (expanded about t) is an infinite series of the form a n (x t) n n=0 where {a n } n=0 is a sequence and x is a real number. K.Maruno (UT-Pan American) Analysis II March 1, 2011 19 / 22
Power Series Let Theorem a n (x t) n be a power series. Let L = lim sup n a n 1 n. Let n=0 0 if L = 1 R = if 0 < L < L if L = 0 Then the power series a n (x t) n converges absolutely if x t < R n=0 and diverges if x t > R. The value R is called the radius of convergence of the power series a n (x t) n. n=0 K.Maruno (UT-Pan American) Analysis II March 1, 2011 20 / 22
Conditional Convergences Dirichlet s Test If the sequence of partial sums of the series a n is bounded and {b n } is a decreasing sequence with limit 0, then a nb n converges. Examples: 1 + 1 2 2 3 + 1 4 + 1 5 2 6 + Letting {a n } be the sequence 1, 1, 2, 1, 1, 2,..., and b n = 1/n, the above series is a nb n. By the Dirichlet s test, this series converges. But this series does not converge absolutely. So this converges conditionally. K.Maruno (UT-Pan American) Analysis II March 1, 2011 21 / 22
Conditional Convergences Abel s Test If a n converges and {b n } is a bounded monotone sequence, then a nb n converges. Examples: ( 1) n (1 + 1/n) n converges since the series ( 1)n /n converges and {(1 + 1/n) n } is a bounded monotone sequence. n K.Maruno (UT-Pan American) Analysis II March 1, 2011 22 / 22