1. Determine the convergence or divergence of the following series an.(2n – 1)! + (-5)-+1(2n + 1)!(e) an5n+1-2n(f) an =n+1(-5)"42n+1 (n + 1)(g) an(-n)"(h) an =2n!

Answered: 1. Determine the convergence or…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 44E

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Question

1. Determine the convergence or divergence of the following series an.
(2n – 1)! + (-5)-+1
(2n + 1)!
(e) an
5n+1
-2n
(f) an =
n+1
(-5)"
42n+1 (n + 1)
(g) an
(-n)"
(h) an =
2n!

Expert Solution

Step 1

Since you have asked multiple questions, we will solve the first three questions for you. If you want any specific question to be solved then please specify the question number or post only that question.

(e) $a_{n} = \frac{(2 n - 1)! + {(- 5)}^{n + 1}}{(2 n + 1)!}$

We must check if $\sum_{n = 1}^{\infty} a_{n}$ converges or diverges.

We will first expand the given series as follows,

$a_{n} = \frac{(2 n - 1)! + {(- 5)}^{n + 1}}{(2 n + 1)!} = \frac{(2 n - 1)!}{(2 n + 1)!} + \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} a_{n} = \frac{(2 n - 1)!}{(2 n + 1) (2 n) (2 n - 1)!} + \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} a_{n} = \frac{1}{(2 n + 1) (2 n)} + \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} = \frac{1}{4 n^{2} + 2 n} + \frac{{(- 5)}^{n + 1}}{(2 n + 1)!}$

Now, applying the summation we get,

$\sum_{n = 1}^{\infty} \frac{(2 n - 1)! + {(- 5)}^{n + 1}}{(2 n + 1)!} = \sum_{n = 1}^{\infty} (\frac{1}{4 n^{2} + 2 n} + \frac{{(- 5)}^{n + 1}}{(2 n + 1)!})$

Step 2

$\sum_{n = 1}^{\infty} \frac{(2 n - 1)! + {(- 5)}^{n + 1}}{(2 n + 1)!} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2} + 2 n} + \sum_{n = 1}^{\infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!}$

$\sum_{n = 1}^{\infty} \frac{1}{4 n^{2} + 2 n} = \sum_{n = 1}^{\infty} \frac{1}{2 n^{2} (2 + \frac{1}{n})}$

Hence, $\sum_{n = 1}^{\infty} \frac{1}{4 n^{2} + 2 n}$ converges.

Consider, $\sum_{n = 1}^{\infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!}$ .

Applying Ratio test we get,

$\lim_{n \to \infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} = \lim_{n \to \infty} (\frac{{(- 5)}^{n + 2}}{(2 n + 3)!} \times \frac{(2 n + 1)!}{{(- 5)}^{n + 1}}) \lim_{n \to \infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} = \lim_{n \to \infty} (\frac{(- 5)}{(2 n + 3) (2 n + 2)}) \lim_{n \to \infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!} = \lim_{n \to \infty} (\frac{(- 5)}{2 n^{2} (2 + \frac{3}{n}) (1 + \frac{1}{n})})$

Hence, $\sum_{n = 1}^{\infty} \frac{{(- 5)}^{n + 1}}{(2 n + 1)!}$ converges.

Therefore, since sum of 2 convergent series is convergent, the series $\sum_{n = 1}^{\infty} \frac{(2 n - 1)! + {(- 5)}^{n + 1}}{(2 n + 1)!}$ converges.

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1. Determine the convergence or divergence of the following series a. (2n – 1)! + (-5)*+1 (2n + 1)! (e) an -2n n+1 (f) an n+1 (-5)" Tan+1 (n + 1) (g) an (-n)" 2n (h) an