1.)

Use the Limit Comparison Test to determine whether the series converges or diverges. DO 9n2 + 2n + 9 an 2n8 + 2n5 + 6n n= n=1 The comparison series is C where c = and p = 2 n=1 11=1 Then, lim an by is a divergent p-series therefore an diverges by the Limit Comparison Test. n=1 71=1Use the Limit Comparison Test to determine whether the series converges or diverges. 3n3 + 9n - 1 an n=1 (n + 2)0 The comparison series is where c = and p = 2 1=1 71=1 Then, lim an 0 n-100 by is a convergent p-series therefore M an converges by the Limit Comparison Test. 71= 1Use the Limit Comparison Test to determine whether the series converges or diverges. 4n + 11 n=1 n=1 18nll + 3n3 The comparison series is where c = 1 and p = 2 n=1 Then, lim an n-100 E by is a convergent p-series v therefore M an converges v by the Limit Comparison Test. n=1 1=1Use the Direct Comparison Test to determine whether the series converges or diverges. DO In n an The comparison series is C where c = 1 and p = which means an 2 v b, for all n > 3. DO by is a divergent p-series v therefore an diverges v by the Direct Comparison Test. n-3 71=3Use the Direct Comparison Test to determine whether the series converges or diverges. 8 + 5 sinn = 9n- + 5 n=1 1= The comparison series is Ebn = Ec( TP ) where c = and p = , which n=1 71=1 means an S vb for all n > 1. CO by is a convergent p-series v therefore an converges v by the Direct Comparison Test. n=1 n=1Use the Direct Comparison Test to determine whether the series converges or diverges. 4tan In an 32n n The comparison series is ) bn aph-1 `where a = and r = ., which 71=3 means an bn for all n > 3. by is a ? v therefore E an ? v by the Direct Comparison Test.Use the Alternating Series Test to determine whether the series converges or diverges. (For limits, enter a number, "-infnity", "infinity", or "DNE" as appropriate.) [(-1)" In(2n) 9n n=1 lim br = 0 A. {b, } is ultimately decreasing because the function f satifying f(n) = b,, is decreasing on the interval [1, infinity) Therefore the series converges by the Alternating Series test. OB. lim an = so the series diverges by the Divergence Test. 71-100Use the Alternating Series Test to determine whether the series converges or diverges. (For limits, enter a number, "-infnity", "infinity", or "DNE" as appropriate.) 100 [(-1)m 9n5 8n + 8 n=1 lim br = 0 7 +00 O A. {b, } is ultimately decreasing because the function f satifying f(n) = b,, is decreasing on the interval [2,infinity) Therefore the series converges by the Alternating Series test. OB. lim an = so the series diverges by the Divergence Test. 71-100