Chapter 8: Q 8. (page 704)

Find the interval of convergence of the power series
$\sum_{k = 1}^{\infty} (x - 3)^{k}$

Short Answer

Expert verified

The interval of convergence of the power series $\sum_{k = 1}^{\infty} (x - 3)^{k}$ is $(2, 4)$

Step by step solution

Given information

The power series is $\sum_{k = 1}^{\infty} (x - 3)^{k}$

The ratio test for absolute convergence will be used to determine the convergence interval.

Let, the first assume, therefore

\lim_{k \to \infty} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k \to \infty} |\frac{{(x - 3)}^{k + 1}}{{(x - 3)}^{k}}| = \lim_{k \to \infty} |x - 3|

The limit is $| x - 3 |$

The ratio test for absolute convergence will be used to determine the convergence interval when $| x - 3 | < 1$ that is $- 1 < x - 3 < 1$

As a result, we may write $- 1 < x - 3$ and $x - 3 < 1$

Implies that

$x > 2 a n d x < 4$

or $x \in (2, 4)$

Now, because the intervals are limited, we examine the series' behavior at the ends.

When $x = 2$

${\sum_{k = 1}^{\infty} (x - 3)^{k}|}_{x = 2} = \sum_{k = 1}^{\infty} (2 - 3)^{k}$

$= \sum_{k = 1}^{\infty} (- 1)^{k}$

The series contains the conditionally convergent alternating multiple.

When $x = 4$

${\sum_{k = 1}^{\infty} (x - 3)^{k}|}_{x = 4} = \sum_{k = 1}^{\infty} (4 - 3)^{k} = \sum_{k = 1}^{\infty} (1)^{k} = \sum_{k = 1}^{\infty} 1$

The series is a diverging constant multiple.

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Find the interval of convergence of the power series
$\sum_{k = 1}^{\infty} (x - 3)^{k}$

Short Answer

Step by step solution

Given information

The ratio test for absolute convergence will be used to determine the convergence interval.

Now, because the intervals are limited, we examine the series' behavior at the ends.

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Step by step solution

Given information

The ratio test for absolute convergence will be used to determine the convergence interval.

Now, because the intervals are limited, we examine the series' behavior at the ends.

One App. One Place for Learning.

Most popular questions from this chapter

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Find the interval of convergence of the power series
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