# Thus c n ² a b n µ a 1 b n 1 µ µ a n 1 b 1 µ a n

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Thus, C n ² a 0 B n µ a 1 B n 1 µ ¸ ¸ ¸ µ a n 1 B 1 µ a n B 0 . Since 3 * j ² 0 b j ² B , for ; n ² def B B n we have that lim n ±* ; n ² 0. Substitution in the previous equation yields that C n ² a 0 ± B µ ; n ² µ a 1 ± B µ ; n 1 ² µ ¸ ¸ ¸ µ a n 1 ± B µ ; 1 ² µ a n ± B µ ; 0 ² which simpli ¿ es to C n ² A n B µ ± a 0 ; n µ a 1 ; n 1 µ ¸ ¸ ¸ µ a n 1 ; 1 µ a n ; 0 ² . Let < n ² a 0 ; n µ a 1 ; n 1 µ ¸ ¸ ¸ µ a n 1 ; 1 µ a n ; 0 Because lim n ±* A n ² A , we will be done if we can show that lim n ±* < n ² 0. In view of the absolute convergence of 3 * j ² 0 a j , we can set 3 * j ² 0 n n a j n n ² : . Suppose that µ 0 is given. From the convergence of ³ ; n ´ , there exists a positive integer M such that n M implies that · ; n · ¶ µ . For n M , it follows that · < n · ² · a 0 ; n µ a 1 ; n 1 µ ¸ ¸ ¸ µ a n M 1 ; M µ 1 µ a n M ; M µ ¸ ¸ ¸ µ a n 1 ; 1 µ a n ; 0 · From the convergence of ³ ; n ´ and 3 * j ² 0 n n a j n n , we have that · a 0 ; n µ a 1 ; n 1 µ ¸ ¸ ¸ µ a n M 1 ; M µ 1 · ¶ µ: while M being a ¿ xed number and a k ± 0 as k ± * yields that · a n M ; M µ ¸ ¸ ¸ µ a n 1 ; 1 µ a n ; 0 · ± 0 as n ± * . Hence, · < n · ² · a 0 ; n µ a 1 ; n 1 µ ¸ ¸ ¸ µ a n M 1 ; M µ 1 µ a n M ; M µ ¸ ¸ ¸ µ a n 1 ; 1 µ a n ; 0 · implies that lim sup n ±* · < n · n µ: . Since µ 0 was arbitrary, it follows that lim n ±* · < n · ² 0 as needed. The last theorem in this section asserts that if the Cauchy product of two given convergent series is known to converge and its limit must be the product of the limits of the given series.
4.5. SERIES OF COMPLEX NUMBERS 173 Theorem 4.5.36 If the series 3 * j ² 0 a j , 3 * j ² 0 b j , and 3 * j ² 0 c j are known to con- verge, 3 * j ² 0 a j ² A, 3 * j ² 0 b j ² B, and 3 * j ² 0 c j is the Cauchy product of 3 * j ² 0 a j and 3 * j ² 0 b j , then 3 * j ² 0 c j ² AB. 4.5.3 Hadamard Products and Series with Positive and Negative Terms Notice that 3 * j ² 1 1 j ± j µ 3 ² 3 can be realized as several different Hadamard prod- ucts ± letting a j ² 1 j , b j ² 1 ± j µ 3 ² 3 , c j ² 1 j ± j µ 3 ² and d j ² 1 ± j µ 3 ² 2 , gives us 3 * j ² 1 1 j ± j µ 3 ² 3 as the Hadamard product of 3 * j ² 1 a j and 3 * j ² 1 b j as well as the Hadamard product of 3 * j ² 1 c j and 3 * j ² 1 d j . Note that only 3 * j ² 1 a j diverges. The following theorem offers a useful tool for studying the nth partial sums for Hadamard products. Theorem 4.5.37 (Summation-by-Parts) Corresponding to the sequences j a j k * j ² 0 , let A n ² n ; j ² 0 a j for n + M C ³ 0 ´ , and A 1 ² 0 . Then for the sequence j b j k * j ² 0 and nonnegative integers p and q such that 0 n p n q, q ; j ² p a j b j ² q 1 ; j ² p A j b b j b j µ 1 c µ A q b q A p 1 b p Excursion 4.5.38 Fill in a proof for the claim.
174 CHAPTER 4. SEQUENCES AND SERIES–FIRST VIEW As an immediate application of this formula, we can show that the Hadamard product of a series whose nth partial sums form a bounded sequence with a series that is generated from a monotonically decreasing sequence of nonnegative terms is convergent.
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