Using the trigonometric identity sin x cos y 1 2 sin

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Using the trigonometric identity sin x cos y = 1 2 (sin( x + y ) + sin( x - y )) , we obtain sin 1 2 t n k =1 cos kt = n k =1 sin 1 2 t cos kt = 1 2 n k =1 sin t 2 + kt + sin t 2 - kt = 1 2 n k =1 sin k + 1 2 t - sin k - 1 2 t = 1 2 sin 1 + 1 2 t - sin 1 - 1 2 t + sin 2 + 1 2 t - sin 2 - 1 2 t + sin 3 + 1 2 t - sin 3 - 1 2 t + · · · + sin n + 1 2 t - sin n - 1 2 t = 1 2 sin n + 1 2 t - sin 1 2 t . Theorem 3.9 (Trigonometric Series Test) . Let { b n } be a sequence satisfying

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48 2. SERIES OF REAL NUMBERS (i) b 1 b 2 b 3 ≥ · · · ≥ 0 , and (ii) lim n →∞ b n = 0 . Then (a). The series k =1 b k sin kt converges for all t R , and (b). The series k =1 b k cos kt converges for all t R , except perhaps t = 2 , p Z . Proof. (a). Let a k = sin kt and let A n = n k =1 a k . If t = 2 , then a k = sin 2 kpπ = 0 and so A n = 0 for t = 2 with p Z . If t = 2 , then | A n | = n k =1 sin kt = cos 1 2 t - cos n + 1 2 t 2 sin 1 2 t cos 1 2 t + cos n + 1 2 t 2 sin 1 2 t 1 + 1 2 sin 1 2 t = 1 sin 1 2 t . Thus {| A n |} is bounded above, and the Dirichlet test applies. (b). Let a k = cos kt and let A n = n k =1 a k . The Dirichlet test applies because | A n | = n k =1 cos kt = sin n + 1 2 t - sin 1 2 t 2 sin 1 2 t sin n + 1 2 t + sin 1 2 t 2 sin 1 2 t 1 + 1 2 sin 1 2 t = 1 sin 1 2 t . Example 3.10 . Determine convergence or divergence of the series k =1 cos kt k q , t R , q > 0 . Solution. When t = 2 with p Z , the series k =1 cos kt k q = k =1 1 k q and so it converges for q > 1 and diverges for q 1.
4. ABSOLUTE AND CONDITIONAL CONVERGENCE 49 When t = 2 , then the series converges by the trigonometric series test because, for q > 0, the sequence 1 k q is positive, monotone decreasing and lim k →∞ 1 k q = 0. In conclusion, the series k =1 cos kt k q is convergent q > 1 , t R convergent 0 < q 1 , t R , t = 2 pπ, for any p Z divergent 0 < q 1 , t = 2 for some p Z 4. Absolute and Conditional Convergence Definition 4.1 . A series n =1 a n is called absolutely convergent if n =1 | a n | converges. Theorem 4.2 . Every absolutely convergent series is convergent. Proof. Suppose that n =1 a n converges absolutely, that is, n =1 | a n | converges by the definition. Let T n = n k =1 | a k | , S n = n k =1 a k . Since { T n } converges, { T n } is Cauchy. Thus, for any > 0, there is a N such that | T n - T m | < for all n, m > N . For any n, m > N , we may assume that m n , say m = n + p (as one of them should be greater than another). Then | S n - S m | = | S n - ( S n + a n +1 + a n +2 + · · · + a n + p ) | = | a n +1 + a n +2 + · · · + a n + p | ≤ | a n +1 | + | a n +2 | + · · · + | a n + p | = T m - T n = | T n - T m | < . Thus { S n } is a Cauchy sequence and so it converges. Thus the series n =1 a n converges and hence the result. Example 4.3 . Determine convergence or divergence of the series n =2 sin n + 1 2 n (ln n ) 2 . Solution. Since sin n + 1 2 n (ln n ) 2 | sin n | + 1 2 n (ln n ) 2 1 + 1 2 n (ln n ) 2

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50 2. SERIES OF REAL NUMBERS and n =2 1 + 1 2 n (ln n ) 2 = 3 2 n =2 1 n (ln n ) 2 converges by Example 2.9, the series n =2 sin n + 1 2 n (ln n ) 2 converges. Thus the series n =2 sin n + 1 2 n (ln n ) 2 converges.

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