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Chapter 10: Problem 28

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum(-1)^{k} \frac{x^{3 k}}{27^{k}}$$

Short Answer

Expert verified
Answer: The radius of convergence (R) is 3 and the interval of convergence is (-3, 3).

Step by step solution

01

Identify the general term of the power series

First, identify the general term of the power series (a_k): $$a_k=(-1)^k \frac{x^{3k}}{27^k}$$
02

Apply the Ratio Test

Apply the Ratio Test: Calculate the limit as k approaches infinity of the absolute value of the ratio of consecutive terms, i.e.: $$\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k}\right|$$ Substitute the expression for a_k and simplify: $$\lim_{k \to \infty} \left| \frac{(-1)^{k+1} \frac{x^{3(k+1)}}{27^{k+1}}}{(-1)^k \frac{x^{3k}}{27^k}}\right|$$ Simplify the expression by cancelling out some terms: $$\lim_{k \to \infty} \left| \frac{x^3}{27} \right|$$ Observe that there is no k term left in the expression, so we can conclude: $$\left| \frac{x^3}{27} \right| < 1$$ This expression is valid for the convergence of the power series.
03

Determine the radius of convergence

We can deduce the radius of convergence (R) by solving the inequality: $$\left| \frac{x^3}{27} \right| < 1$$ Rearrange the expression: $$\left| x^3 \right| < 27$$ Take the cube root on both sides: $$\left| x \right| < 3$$ So the radius of convergence, R, is 3.
04

Test the endpoints of the interval

To determine the interval of convergence, we need to test the endpoints of the interval (-3, 3). For x = -3: $$\sum (-1)^k \frac{(-3)^{3k}}{27^k} = \sum (-1)^k 3^{3k-3k} = \sum (-1)^k$$ This series diverges due to the Alternating Series Test. For x = 3: $$\sum (-1)^k \frac{(3)^{3k}}{27^k} = \sum (-1)^k 3^{3k-3k} = \sum (-1)^k$$ This series also diverges due to the Alternating Series Test.
05

Find the interval of convergence

Since the series converges for |x| < 3 but not at the endpoints, the interval of convergence is: $$-3 < x < 3$$

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Most popular questions from this chapter

Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty} 2^{k} x^{2 k+1}$$

Assume that \(f\) has at least two continuous derivatives on an interval containing \(a\) with \(f^{\prime}(a)=0 .\) Use Taylor's Theorem to prove the following version of the Second Derivative Test: a. If \(f^{\prime \prime}(x) > 0\) on some interval containing \(a,\) then \(f\) has a local minimum at \(a\) b. If \(f^{\prime \prime}(x) < 0\) on some interval containing \(a,\) then \(f\) has a local maximum at \(a\)

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Write the Taylor series for \(f(x)=\ln (1+x)\) about 0 and find its interval of convergence. Assume the Taylor series converges to \(f\) on the interval of convergence. Evaluate \(f(1)\) to find the value of \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) (the alternating harmonic series).

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