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Ma1502Te2Solns-CSpring2011

Ma1502Te2Solns-CSpring2011 - MATH1502 Calculus II TEST 2 C...

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MATH1502 - Calculus II TEST 2 - C Group - March 3 - Spring 2011 Question Score Maximum 1 6 2 6 3 17 4 9 5 9 6 7 Total 54 Name _____________________________ Group (e.g. C1) _____________________ Student Number ____________________ Teaching Assistant ____________________ Answer all questions. There are 54 marks on the paper. 100% = 50 marks No °cheatsheets±or calculators are allowed. Question 1 Find the sum of the series 1 X k =3 1 (2 k + 1) (2 k ° 1) : (6 marks) Solution We use partial fractions: 1 (2 k + 1) (2 k ° 1) = 1 2 ° 1 2 k ° 1 ° 1 2 k + 1 ± : 1
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The n th partial sum is s n = n X k =3 1 (2 k + 1) (2 k ° 1) = n X k =3 1 2 ° 1 2 k ° 1 ° 1 2 k + 1 ± = 1 2 ° 1 5 ° 1 7 ± + 1 2 ° 1 7 ° 1 9 ± + 1 2 ° 1 9 ° 1 11 ± + ::: + 1 2 ° 1 2 n ° 3 ° 1 2 n ° 1 ± + 1 2 ° 1 2 n ° 1 ° 1 2 n + 1 ± = 1 2 1 5 ° 1 2 1 2 n + 1 : So lim n !1 s n = lim n !1 ² 1 10 ° 1 2 1 2 n + 1 ³ = 1 10 : Thus 1 X k =3 1 (2 k + 1) (2 k ° 1) = 1 10 : (6 marks) 2
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Question 2 Test the following series for convergence or divergence: 1 X k =2 ´ k 4 + 2 k 1 = 3 µ ´ k 1 = 3 + ln k µ k 2 ( k 2 + 1) ( k 2 ° 2) : (6 marks) Solution Recall that lim k !1 ln k k 1 = 3 = 0 We see then that for very large k , the series terms behave roughly like k 4 k 1 = 3 k 2 ( k 2 ) ( k 2 ) = 1 k 1 2 3 : Since P 1 k 1 2 3 converges ( p ° series with p = 1 2 3 > 1) , this suggests we use the limit comparison test with a k = ´ k 4 + 2 k 1 = 3 µ ´ k 1 = 3 + ln k µ k 2 ( k 2 + 1) ( k 2 ° 2) and b k = 1 k 1 2 3 : Now let us make this rigorous. lim k !1 a k =b k = lim k !1 " ´ k 4 + 2 k 1 = 3 µ ´ k 1 = 3 + ln k µ k 2 ( k 2 + 1) ( k 2 ° 2) # = ° 1 k 1 2 3 ± = lim k !1 k 4 k 1 = 3 1 + 2 k ° 3 2 3 · ´ 1 + ln k k 1 = 3 µ k 6 ´ 1 + 1 k 2 µ ´ 1 ° 2 k 2 µ k 1 2 3 1 = lim k !1 1 + 2 k ° 3 2 3 · ´ 1 + ln k k 1 = 3 µ ´ 1 + 1 k 2 µ ´ 1 ° 2 k 2 µ = 1 ; ²nite and positive. Since P 1 k 1 2 3 converges ( p ° series with p = 1 2 3 ) , by the
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