**Unformatted text preview: **2 converges absolutely, while ∞ X n =1 (-1) n-1 1 n converges conditionally. EXAMPLE 10.37 Does ∞ X n =2 sin n n 2 converge? In example 10.32 we saw that ∞ X n =2 | sin n | n 2 converges, so the given series converges abso-lutely. EXAMPLE 10.38 Does ∞ X n =0 (-1) n 3 n + 4 2 n 2 + 3 n + 5 converge? Taking the absolute value, ∞ X n =0 3 n + 4 2 n 2 + 3 n + 5 diverges by comparison to ∞ X n =1 3 10 n , so if the series converges it does so conditionally. It is true that lim n →∞ (3 n + 4) / (2 n 2 + 3 n + 5) = 0, so to apply the alternating series test we need to know whether the terms are decreasing. If we let f ( x ) = (3 x + 4) / (2 x 2 + 3 x + 5) then f ± ( x ) =-(6 x 2 + 16 x-3) / (2 x 2 + 3 x + 5) 2 , and it is not hard to see that this is negative for x ≥ 1, so the series is decreasing and by the alternating series test it converges....

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