Determine the radius and interval of convergence of the following power series.
\[
\sum_{k=1}^{\infty}(-1)^{k} \frac{x^{k}}{3^{k}}
\]
The radius of convergence is $\mathrm{R}=$
Find the interval of convergence. Select the correct choice below and fill in the answer box to complete your choice.
A. The interval of convergence is $\{x: x=$ (Simplify your answer. Type an exact answer.)
B. The interval of convergence is (Simplify your answer. Type an exact answer. Type your answer in interval notation.)

Step 1 ：The given power series is \(\sum_{k=1}^\infty (-1)^k \frac{x^k}{3^k}\). We can rewrite this as \(\sum_{k=1}^\infty a_k\), where \(a_k = (-1)^k \frac{x^k}{3^k}\).

Step 2 ：To find the radius of convergence, we use the Ratio Test. The Ratio Test states that if \(\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = L\), then the series converges if \(L < 1\), diverges if \(L > 1\), and is inconclusive if \(L = 1\).

Step 3 ：We calculate \(\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|\) as follows: \[\lim_{k\to\infty} \left|\frac{(-1)^{k+1} \frac{x^{k+1}}{3^{k+1}}}{(-1)^k \frac{x^k}{3^k}}\right| = \lim_{k\to\infty} \left|\frac{-x}{3}\right| = \frac{|x|}{3}\]

Step 4 ：Setting \(\frac{|x|}{3} < 1\), we find that \(-3 < x < 3\). Therefore, the radius of convergence \(R = 3\).

Step 5 ：To find the interval of convergence, we need to check the endpoints \(x = -3\) and \(x = 3\).

Step 6 ：Substituting \(x = -3\) into the series, we get \(\sum_{k=1}^\infty (-1)^k \frac{(-3)^k}{3^k} = \sum_{k=1}^\infty (-1)^k = -1 + 1 - 1 + \ldots\), which is a divergent series.

Step 7 ：Substituting \(x = 3\) into the series, we get \(\sum_{k=1}^\infty (-1)^k \frac{3^k}{3^k} = \sum_{k=1}^\infty (-1)^k = -1 + 1 - 1 + \ldots\), which is also a divergent series.

Step 8 ：Therefore, the interval of convergence is \((-3, 3)\).

Step 9 ：\boxed{R = 3, \text{ Interval of Convergence } = (-3, 3)}