Please show all work clearly as I often get confused. Thank you! Determine if each of the following series converge or diverge. If a series converges, then findits sum.(-1)"5n+2( a) Σ7n-2n=1(b) arctan(n)n=13(c) Dn2 +nn=1

Answered: (-1)*5n+2 (a) 7n-2 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 50E

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Question

Please show all work clearly as I often get confused. Thank you!

Determine if each of the following series converge or diverge. If a series converges, then find
its sum.
(-1)"5n+2
( a) Σ
7n-2
n=1
(b) arctan(n)
n=1
3
(c) D
n2 +n
n=1

Expert Solution

Step 1

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

a) $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2}}{7^{n - 2}}$

Given that :

The series is $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2}}{7^{n - 2}}$ .

Step 2

By using,

Geometric series :

The series $\sum_{n = 1}^{\infty} a_{1} {(r)}^{n - 1}$ is converges if and only if -1 < r < 1 and its sum is $S = \frac{a}{1 - r}$ ( a is the first term and r is the common difference).

Ratio test:

Suppose that the series $\sum_{} a_{n}$ ,

Define, L = $\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}|$

1. If L < 1 , then series is absolutely convergent.

2. If L > 1 , then series divergent.

3. If L = 1 , then inconclusive.

Step 3

To simplify the series:

$\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2}}{7^{n - 2}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2}}{7^{n - 2}} . (\frac{7^{2 - n}}{7^{2 - n}}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2} . 7^{2 - n}}{7^{n - 2 + 2 - n}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2} . 7^{2 - n}}{7^{0}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2} . 7^{2 - n}}{1} . . . . (7^{0} = 1) = \sum_{n = 1}^{\infty} {(- 1)}^{n} 5^{n + 2} . 7^{2 - n} = \sum_{n = 1}^{\infty} {(- 1)}^{n} 5^{2} . 5^{n} . 7^{2} . 7^{- n}$

Step 4

$= \sum_{n = 1}^{\infty} {(- 1)}^{n} 25 . 5^{n} . 49 . 7^{- n} = 1225 \sum_{n = 1}^{\infty} {(- 1)}^{n} 5^{n} 7^{- n} = 1225 \sum_{n = 1}^{\infty} {(- 1)}^{n} (\frac{5^{n}}{7^{n}}) = 1225 \sum_{n = 1}^{\infty} {(- 1)}^{n} {(\frac{5}{7})}^{n} = 1225 \sum_{n = 1}^{\infty} {(- \frac{5}{7})}^{n} \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} 5^{n + 2}}{7^{n - 2}} = 1225 \sum_{n = 1}^{\infty} {(- \frac{5}{7})}^{n}$

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