# 2n 1 4 6 2n 2n 3 1n lim n n an 1 an lim n 2n

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Unformatted text preview: a1. P roblem Solving for Chapter 9 501 11. —CONTINUED— Now assume an &lt; a. Then a &gt; an and a 1 &gt; 1 implies 1 &gt; an 1 aa a2 a &gt; an a2 &gt; an a&gt; a an a an 1. Hence, the sequence converges to some number L. To find L, assume an L L⇒ L2 1 2 1± L 1 2 1 Hence, L 1 13. (a) n 1 4a a 4a 1 2n n 1 21 L⇒ an 1 L: . a L a 0 . 1 1 L2 1 22 1 23 1 1 24 1 1 25 ... 1 (b) 1n 2n an 1 an 2 n 2 1n 1 1 1n 1n 1 21 1 1 This sequence is 8, 2, 8, 2, . . . which diverges. S1 1 20 S1 1 1 8 9 8 S3 9 8 1 4 11 8 S4 11 8 1 32 45 32 S5 45 32 1 16 47 32 40 240 1 (c) 15. S6 130 70 S7 240 130 70 S8 440 240 130 810 440 240 1490 810 1 2n 1 1 2n 1n 2 because 2 and n 1 2 2n 1n 1 2 n 1 2, 1n 1n 1n → 1 &lt; 1 converges 2 2, 1, 2, . . . and 2 n 1 2 →1 2 → 1. 1490 S10 1 2n 810 S9 n 17. (a) n (b) Let f x fx 440 440 2740 1 diverges (harmonic) 1n sin x. By the Mean Value Theorem, fy fcx y cos c x y≤x y, where c is between x and y. Thus, 1 2n 0 ≤ sin ≤ 1 2n sin 1 2n 1 1 2n 1 2n 2n Because n 1 1 1 2n 1 2n 1 sin converges, the Comparison Theorem tells us that n 1 1 2n sin 1 2n 1 converges....
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