Ma1502Te2Solns-CSpring2011

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

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Unformatted text preview: 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 &cheatsheetsor 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 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 1 10 & 1 2 1 2 n + 1 = 1 10 : Thus 1 X k =3 1 (2 k + 1) (2 k & 1) = 1 10 : (6 marks) 2 Question 2 Test the following series for convergence or divergence: 1 X k =2 & k 4 + 2 k 1 = 3 & k 1 = 3 + ln k 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 & k 1 = 3 + ln k k 2 ( k 2 + 1) ( k 2 & 2) and b k = 1 k 1 2 3 : Now let us make this rigorous....
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This note was uploaded on 09/09/2011 for the course MATH 1502 taught by Professor Mcclain during the Fall '07 term at Georgia Institute of Technology.

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

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