SP21 MATH 2024 - The Integral Test.pdf - MATH 2024 The...

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MATH 2024 The Integral Test Adams Theorem: The Integral Test If f is positive, continuous, and decreasing for x 1 and a n = f ( n ), then X n =1 a n and Z 1 f ( x ) dx either both converge or both diverge. Proof: Begin by partitioning the interval [1 , n ] into n - 1 unit intervals, as illustrated in the figure below: The total areas of the inscribed rectangles and the circumscribed rectangles are as follows: n X i =2 f ( i ) = f (2) + f (3) + . . . + f ( n ) n - 1 X i =1 f ( i ) = f (1) + f (2) + . . . + f ( n - 1) The exact area under the graph of f from x = 1 to x = n lies between the inscribed and circum- scribed areas: n X i =2 f ( i ) Z n 1 f ( x ) dx n - 1 X i =1 f ( i ) Using the n th partial sum, S n = f (1) + f (2) + . . . + f ( n ), you can write this inequality as S n - f (1) Z n 1 f ( x ) dx S n - 1 Now assuming that Z 1 f ( x ) dx converges to L , it follows that for n 1, S n - f (1) L Page 1 of 6
MATH 2024 The Integral Test Adams which implies S n L + f (1) Consequently, { S n } is bounded and monotonic, which implies its convergence. So, X a n converges. For the other direction of the proof, assume that the improper integral diverges. Then Z n 1 f ( x ) dx approaches infinity as n → ∞ , and the inequality S n - 1 Z n 1 f ( x ) dx implies that { S n } diverges. So X a n diverges. Example 1. Apply the Integral Test to the series X n =1 n n 2 + 1 .