Answered: 2" (x-3)" 100 1.) Ση-0 Vn+3

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.5: The Binomial Theorem

Problem 16E

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Question

Determine the radius and interval of convergence for each of the following power series.

1.) Ση-0
2" (x-3)"
00
νn+3
2.) Σο.
2.) ΣΤ1(" (
–1)'
n=
.n25η
2" (x-2)"
3.) Σ-1
00
(n+2)!

Expert Solution

Step 1

Note: There are many subparts in the question and the question is so long. So I will answer the

first subpart of the question. Please send the other subparts separately.

Given:

The power series is:

$\sum_{n = 0}^{\infty} a_{n} = \sum_{n = 0}^{\infty} \frac{2^{n} {(x - 3)}^{n}}{\sqrt{n + 3}}$

We have to determine the radius and interval of convergence for the given power series.

Step 2

We know that,

When

$\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| < 1$

Then the power series is converges so we find the interval for x such that the given

power series converges.

We have,

$a_{n} = \frac{2^{n} {(x - 3)}^{n}}{\sqrt{n + 3}} and a_{n + 1} = \frac{2^{(n + 1)} {(x - 3)}^{(n + 1)}}{\sqrt{(n + 1) + 3}}$

Now

$\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| < 1 \lim_{n \to \infty} |\frac{\frac{2^{(n + 1)} {(x - 3)}^{(n + 1)}}{\sqrt{(n + 1) + 3}}}{\frac{2^{n} {(x - 3)}^{n}}{\sqrt{n + 3}}}| < 1 \lim_{n \to \infty} |2 (x - 3) \frac{\sqrt{n + 3}}{\sqrt{n + 4}}| < 1 \lim_{n \to \infty} |2 (x - 3) \frac{\sqrt{1 + \frac{3}{n}}}{\sqrt{1 + \frac{4}{n}}}| < 1 |2 (x - 3)| < 1 [\lim_{n \to \infty} \frac{1}{n} = 0] - 1 < 2 (x - 3) < 1 - \frac{1}{2} < x - 3 < \frac{1}{2} - \frac{1}{2} + 3 < x < \frac{1}{2} + 3 \frac{5}{2} < x < \frac{7}{2}$

The interval for the radius for convergence is $(\frac{5}{2}, \frac{7}{2})$

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