Fall 2008 Midterm &amp;Solutions

# Fall 2008 Midterm &amp;Solutions - McGill University...

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McGill University Department of Mathematics and Statistics Calculus 3 (MATH 222) Fall 2008 Instructors: Prof. K. GowriSankaran, N. Sonnerat Midterm Exam Solutions October 22nd 2008, 6pm - 8pm Instructions: This is a 2 hour, closed book/notes examination. No calculators are permitted. To get full marks, you must answer all ﬁve questions. Justify all your answers, unless the question instructs you not to. 1. (a) Compute lim n →∞ { ( - 1) n e n n ! } . Solution: The easiest way would have been to say that you saw in class that lim x n n ! = 0 for all values of x , so in particular this is true for x = - e . Alternatively, you can use the ratio test to show that the series X ( - 1) n e n n ! converges absolutely. Then the test for divergence implies that the sequence must have limit 0. As a third option, you could have shown that 0 ≤ | ( - 1) n e n n ! | = e n n ! e 2 2 e n , so lim | a n | = 0 by the squeeze theorem (see assignment solutions for more details). But you have another theorem that tells you that if | a n | tends to 0, then a n must also tend to 0. (b) Determine whether the series X k =0 2 k k ! ( k + 2)! converges or diverges. Solution: Either use l’Hopital’s Rule to show that lim 2 k k ! ( k + 2)! = lim 2 k ( k + 2)( k + 1) 6 = 0 , which implies that the series is divergent by the test for divergence. Note that d/dx (2 x ) = ln 2 · 2 x , NOT x 2 x - 1 . 1

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Alternatively, apply the ratio test to ﬁnd that lim | a n +1 | | a n | = 2 > 1 , so the series must diverge. (c) Determine whether the series X j =0 ( - 1) j +1 j j + 5 converges or diverges. Solution: Let a j = j j +5 . It is obvious that a j 0 for all j . It’s not so obvious that it’s decreasing, but if you take the derivative of f ( x ) = x x +5 , you’ll see that f 0 ( x ) < 0 for all x > 5. This implies that for j > 5, the series is decreasing. Finally, lim j j + 5 = lim 1 j + 5 / j = 0 because lim 5 / j = 0, so j + 5 / j tends to and thus 1 j +5 / j tends to 0. So all the conditions of the alternating series test are met, and therefore the series converges. (It does not converge absolutely, but we didn’t ask for that.) (d) Determine whether the series X k =1 k ln k ( k + 1) 3 converges or diverges. Solution:
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## This note was uploaded on 08/17/2011 for the course MATH 222 taught by Professor Karlpeterrussell during the Spring '08 term at McGill.

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Fall 2008 Midterm &amp;Solutions - McGill University...

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