# Exam 2 - for . Hence, for , (Method 2: Ordinary Comparison...

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MATH 1014.03MW Exam 2 Wednesday, February, 9 2000 NAME: Student Number: No. Marks Instructions : This exam contains 3 questions each of which has several parts and has a total of 100 marks. Show all of your work. (40 Marks) Evaluate each of the following limits. ( a ) ( b ) ( a ) ( b ) ( c ) . ¯ ( c ) . ( d ) . Solution: ( a ) .

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( b ) Hence, . (3) . = (20 Marks)Evaluate each improper integral or show that it diverges. ( a ) . ( b ) , p >0. ¯ ( a ) . ( b ) , p >0. Solution: ( a ) x =0 is a singularity. Let . Then . Hence, we have This implies .
( b ) Let . Then x =1 is a singularity. Let . Then We consider the following cases: ( i ) When 1- p >0, and converges. ( ii ) When 1- p <0, and diverges. ( iii ) When p =1, . Hence, . In this case, diverges. (40 Marks)Determine convergence or divergence for each of the following series. ( a ) . ¯ ( a ) . ( b ) . ( c ) . ( d )

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( c ) . ( d ) ( p >0, x >0). Solution: ( a ) (Method 1: Bounded Sum Test) Let . Note that for . Hence, for , Hence, is bounded and converges. There is another way to show is bounded. Note that

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Unformatted text preview: for . Hence, for , (Method 2: Ordinary Comparison Test) Let , Then for . Since converges, it follows that converges. Another way: for . Since converges, it follows that converges. (Method 3: Ratio Test) . It follows from Ratio Test that the series converges. (Method 4: Root Test) Noting that , we have . It follows that converges. ( b ) Since and diverges, it follows from Limit Comparison Test that diverges. ( c ) Let . Then is continuous since is continuous on and for . Also, , so f is decreasing. Moreover, Hence, and converges. It follows from Integral Test that converges. ( d ) Let . Then . Hence, . By Ratio Test, we have ( i ) If x <1, converges. ( ii ) If x >1, diverges. When x =1, the series becomes . It is known that when p >1, converges and when , diverges. • About this document . .. Next: About this document Kunquan Lan Wed Feb 9 09:35:54 EST 2000...
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## This note was uploaded on 12/16/2009 for the course MATH 1014 taught by Professor Ganong during the Fall '09 term at York University.

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Exam 2 - for . Hence, for , (Method 2: Ordinary Comparison...

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