91Ó°ÊÓ

The Ratio Test states that for a series \(\sum a_k\), the limit as \(k\to\infty\) of \(\left|\frac{a_{k+1}}{a_k}\right|\) can be used to determine the series' convergence. If the limit is less than 1, the series converges; if the limit is greater than 1, the series diverges; and if the limit equals 1, the test is inconclusive. Let \(a_k = (2x)^k\) for our given power series. Then, we will find the limit as \(k \to \infty\) of \(\left|\frac{a_{k+1}}{a_k}\right|\): $$\lim_{k\to\infty} \left|\frac{(2x)^{k+1}}{(2x)^k}\right| = \lim_{k\to\infty} \left|\frac{2x(2x)^k}{(2x)^k}\right| = |2x|$$ Now we want the limit to be less than 1 for convergence: $$|2x| < 1$$ Solving for x, we get the following: $$-\frac{1}{2} < x < \frac{1}{2}$$ The radius of convergence is half the length of this interval, which is \(\frac{1}{2}\).

We will now test the endpoints to determine the interval of convergence. The endpoints are \(x = -\frac{1}{2}\) and \(x = \frac{1}{2}\). 1. Check the endpoint \(x = -\frac{1}{2}\): $$\sum(2(-\frac{1}{2}))^k = \sum (-1)^k$$ This is an alternating series that does not approach zero as \(k \to \infty\). Therefore, it diverges by the Divergence Test. 2. Check the endpoint \(x = \frac{1}{2}\): $$\sum(2(\frac{1}{2}))^k = \sum 1^k = \sum 1$$ This is a series of constant terms, which diverges.

Having analyzed the endpoints, we can now state the interval of convergence for the original power series. Since neither endpoint resulted in a convergent series, we have: $$-\frac{1}{2} < x < \frac{1}{2}$$ The interval of convergence is \((-\frac{1}{2}, \frac{1}{2})\), and the radius of convergence is \(\frac{1}{2}\).