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Chapter 10: Problem 20

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum(-1)^{k} \frac{k(x-4)^{k}}{2^{k}}$$

Short Answer

Expert verified
Question: Determine the radius of convergence and interval of convergence for the power series $$\sum(-1)^{k} \frac{k(x-4)^{k}}{2^{k}}$$. Answer: The radius of convergence is 2, and the interval of convergence is (2, 6).

Step by step solution

01

Define the power series function

The given power series is $$\sum(-1)^{k} \frac{k(x-4)^{k}}{2^{k}}$$
02

Ratio Test for Radius of Convergence

We will use the ratio test to determine the radius of convergence. The ratio test states that if $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = L$$, where \(a_k\) is the \(k^{th}\) term of the series and \(L<1\), the series converges. If \(L > 1\), the series diverges. If \(L = 1\), the test is inconclusive. So let's find the ratio \(\frac{a_{k+1}}{a_k}\): $$\frac{a_{k+1}}{a_k} = \frac{(-1)^{k+1} \frac{(k+1)(x-4)^{k+1}}{2^{k+1}}}{(-1)^{k} \frac{k(x-4)^{k}}{2^{k}}}$$ Simplify the ratio: $$\frac{a_{k+1}}{a_k} = \frac{(k+1)(x-4)}{2k}$$ Now let's find the limit of the absolute value of the ratio as \(k \to \infty\): $$L = \lim_{k \to \infty} \left|\frac{k+1}{2k}(x-4)\right| = \frac{|x-4|}{2}$$
03

Find the Radius of Convergence

Now, we need to check the conditions for the radius of convergence based on the limit L we obtained: $$\frac{|x-4|}{2} < 1 \Rightarrow |x-4| < 2$$ So, the radius of convergence is 2.
04

Test the Endpoints

Now, we need to test the endpoints to determine the interval of convergence. Let's test the lower endpoint, \(x = 4-2 = 2\). Plug in \(x=2\) back into the power series: $$\sum(-1)^{k} \frac{k(2-4)^{k}}{2^{k}} = \sum(-1)^{k} \frac{-k2^{k}}{2^{k}} = \sum(-1)^{k}(-k)$$ This series diverges since it is a harmonic series with alternating signs (Conditional convergence test). Now, let's test the upper endpoint, \(x=4+2=6\). Plug in \(x=6\) back into the power series: $$\sum(-1)^{k} \frac{k(6-4)^{k}}{2^{k}} = \sum(-1)^{k} \frac{k2^{k}}{2^{k}} = \sum(-1)^{k}k$$ This series also diverges due to the same reason (Conditional convergence test).
05

Determine the Interval of Convergence

Since the radius of convergence is 2 and both endpoints failed the convergence test, the interval of convergence for the given power series is: $$(2, 6)$$

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