MATH 101 Study Guide - Quiz Guide: Ulog, Alternating Series, Absolute Convergence
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(cid:88) n=6 (a) we apply the limit comparison test to the series. (cid:88) n=1 ( 1)nn 112 n/4, is within 12 4 of its actual value. (cid:88) n=9: 4 marks does the series. Short answer question will be marked (no partial credit). (a) we apply the limit comparison test to the series or diverges and which series you would compare to. n=6. 1 n3 converges (it"s a p-series with p = 3 > 1), so by the limit comparison test the given series converges as well. Solution: by the formula for the sum of a geometric series (valid since | 3. Alternatively, the rst term in the series is 4 32. 1 = 36 and the common ratio between terms is 3. 7 , so the value of the series is ( rst term)/(1 common ratio) = 36/(1 3/7). n=0.
