Ma1502Te2Solns-HSpring2010

Ma1502Te2Solns-HSpring2010 - MATH1502 - Calculus II TEST 2...

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Unformatted text preview: MATH1502 - Calculus II TEST 2 - H Group - March 4 - Spring 2010 Question Score Maximum 1 7 2 7 3 18 4 9 5 10 6 7 Total 58 Name _____________________________ Group (e.g. G1 or J1) _____________________ Student Number ____________________ Teaching Assistant ____________________ Answer all questions. There are 58 marks on the paper. 100% = 50 marks No &cheatsheetsor calculators are allowed. Question 1 Let b be a number such that < b < 1 : Find the sum of the series 1 X k =1 1 ( k & b ) ( k & b + 1) : (6 marks) Deduce the value of 1 X k =1 1 & k & 1 3 & k + 2 3 : (1 mark) Solution We use partial fractions: 1 ( k & b ) ( k & b + 1) = 1 k & b & 1 k & b + 1 : 1 The n th partial sum is s n = n X k =1 1 ( k & b ) ( k & b + 1) = n X k =1 & 1 k & b & 1 k & b + 1 = & 1 1 & b & 1 2 & b + & 1 2 & b & 1 3 & b + & 1 3 & b & 1 4 & b + ::: + & 1 n & 1 & b & 1 n & b + & 1 n & b & 1 n & b + 1 = 1 1 & b & 1 n & b + 1 : So lim n !1 s n = lim n !1 1 1 & b & 1 n & b + 1 = 1 1 & b : Thus 1 X k =1 1 ( k & b ) ( k & b + 1) = 1 1 & b : (6 marks) In particular, choosing b = 1 3 ; 1 X k =1 1 k & 1 3 k + 2 3 = 1 1 & 1 3 = 3 2 : (1 mark) 2 Question 2 Test the following series for convergence or divergence: 1 X k =2 ( k 2 + 1) & k 1 = 3 + ( & 1) k k ( k + 1) ( k + 3) : (7 marks) Solution We see that for very large k , the series terms behave roughly like k 2 k 1 = 3 k ( k ) ( k ) = k & 2 = 3 : Since P k & 2 3 diverges ( p & series with p = 2 3 ) , this suggests we use the limit comparison test with a k = ( k 2 + 1) & k 1 = 3 + ( & 1) k k ( k + 1) ( k...
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Ma1502Te2Solns-HSpring2010 - MATH1502 - Calculus II TEST 2...

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