11.3.5-eg - Z ∞ 1-2 3 √ x 2 dx = lim b →∞ Z b 1-2 3...

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Example Determine if the series converges or diverges. X n =1 - 2 3 x 2 . The function f ( x ) = 2 3 x 2 is continuous, positive, and decreasing for all x 1, so we may apply the integral test to the original series. If the improper integral Z 1 - 2 3 x 2 dx converges (or diverges), the series converges (or diverges) as well.
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Unformatted text preview: Z ∞ 1-2 3 √ x 2 dx = lim b →∞ Z b 1-2 3 √ x 2 dx =-2 lim b →∞ Z b 1 x-2 / 3 dx =-2 lim b →∞ 3 x 1 / 3 ± ± ± ± ± b 1 =-6 lim b →∞ ² b 1 / 3-1 ³ =-∞ . Therefore the series diverges as well....
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