Comparison Test! Given the series, states whether the test applies and if it does use the test to see if the series converges or diverges. (a) If the given test can be applied to the given series, complete the following.(i) Determine a comparator series that you could use with the given test.(ii) Indicate whether or not the given series converges or diverges.(b) If the test cannot be applied to the given series, give a short reason as to why not.2n + 2"• Series: >Test: Limit Comparison Test3n + 3n'n=1(a) If the Test Applies:(i) Comparator:(ii) Choose one: Converges / Diverges(b) If the Test Does Not Apply, explain why:

It is given that ∑n = 1∞ 2n + 2n3n + 3n Let an = 2n + 2n3n + 3n , bn = 2n3n an ≥0,…

Answered: 2n + 2" • Series: Test: Limit…

Math

Calculus

2n + 2" • Series: Test: Limit Comparison Test Зп + 3п? n=1

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 49E

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Question

Comparison Test! Given the series, states whether the test applies and if it does use the test to see if the series converges or diverges.

(a) If the given test can be applied to the given series, complete the following.
(i) Determine a comparator series that you could use with the given test.
(ii) Indicate whether or not the given series converges or diverges.
(b) If the test cannot be applied to the given series, give a short reason as to why not.
2n + 2"
• Series: >
Test: Limit Comparison Test
3n + 3n'
n=1
(a) If the Test Applies:
(i) Comparator:
(ii) Choose one: Converges / Diverges
(b) If the Test Does Not Apply, explain why:

Expert Solution

Step 1

It is given that

$\sum_{n = 1}^{\infty} \frac{2 n + 2^{n}}{3 n + 3^{n}}$

Let $a_{n} = \frac{2 n + 2^{n}}{3 n + 3^{n}}, b_{n} = \frac{2^{n}}{3^{n}}$

$a_{n} \geq 0, b_{n} > 0 \forall n$

use limit caparison test

= $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}$

= $\lim_{n \to \infty} (\frac{2 n + 2^{n}}{3 n + 3^{n}} x \frac{3^{n}}{2^{n}})$

= $\lim_{n \to \infty} (\frac{\frac{2 n}{2^{n}} + \frac{2^{n}}{2^{n}}}{\frac{3 n}{3^{n}} + \frac{3^{n}}{3^{n}}}) = \frac{0 + 1}{0 + 1} = 1$ > 0

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