Using the Integral Test, check the convergence of the following series by verifyingthe necessary conditions of integral test.n32n + 5(i)(ii)e"/2Z (n² – 6n + 9)|n=1n=4

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Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 50E

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Using the Integral Test, check the convergence of the following series by verifying
the necessary conditions of integral test.
n3
2n + 5
(i)
(ii)
e"/2
Z (n² – 6n + 9)
|
n=1
n=4

Expert Solution

Step 1

$Integral test: If f is continuous, positive, and decreasing on [1,∞) {such that a}_{n} = f then \sum_{n = 1}^{\infty} a_{n} c o n v e r g e s \Leftrightarrow \int_{1}^{\infty} f (x) \cdot d x c o n v e r g e s$

Step 2 : part a

$f (x) = \frac{x^{3}}{e^{\frac{x}{2}}} = x^{3} e^{- \frac{x}{2}} f (x) i s c o n t i n u o u s . f (x) i s d e c r e a \sin g f o r x > 6 . f (x) i s a l w a y s > 0 \int_{1}^{\infty} f (x) \cdot d x = \int_{1}^{\infty} (x^{3} e^{- \frac{x}{2}}) d x U \sin g b y p a r t \int_{1}^{\infty} f (x) \cdot d x = {[x^{3} \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}}]}_{1}^{\infty} - \int_{1}^{\infty} (3 x^{2} \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}}) d x \int_{1}^{\infty} f (x) \cdot d x = {[- 2 x^{3} e^{- \frac{x}{2}}]}_{1}^{\infty} + 6 [{(x^{2} \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}})}_{1}^{\infty} - \int_{1}^{\infty} (2 x \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}}) d x] \int_{1}^{\infty} f (x) \cdot d x = [2 e^{- \frac{1}{2}}] - 12 {(x^{2} e^{- \frac{x}{2}})}_{1}^{\infty} + 24 \int_{1}^{\infty} (x e^{- \frac{x}{2}}) d x \int_{1}^{\infty} f (x) \cdot d x = [2 e^{- \frac{1}{2}}] - 12 (- 1 e^{- \frac{1}{2}}) + 24 {[- 2 x e^{- \frac{x}{2}}]}_{1}^{\infty} - 24 \int_{1}^{\infty} (- 2 e^{- \frac{x}{2}}) d x \int_{1}^{\infty} f (x) \cdot d x = 2 e^{- \frac{1}{2}} + 12 e^{- \frac{1}{2}} - 48 [0 - e^{- \frac{1}{2}}] + 48 {(- 2 e^{- \frac{x}{2}})}_{1}^{\infty} \int_{1}^{\infty} f (x) \cdot d x = 62 e^{- \frac{1}{2}} - 96 (0 - e^{- \frac{1}{2}}) \int_{1}^{\infty} f (x) \cdot d x = 158 e^{- \frac{1}{2}} = 95.83 S i n c e \int_{1}^{\infty} (x^{3} e^{- \frac{x}{2}}) \cdot d x i s c o n v e r g i n g s o d o e s \sum_{n = 1}^{\infty} (n^{3} e^{- \frac{n}{2}})$

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Using the Integral Test, check the convergence of the following series by verifying the necessary conditions of integral test. 2n + 5 (ii) Ln² – 6n + 9) (i) e"/2 - n=1 n=4