Determine the radius and interval of convergence of the following power series.
$$\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\frac{x^k}{3^k}$$
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 $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\frac{x^k}{3^k}$. We can rewrite this as $\sum _{k=1}^{\mathrm{\infty }}{a}_k$, 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 $\underset{k\to \mathrm{\infty }}{lim}\left|\frac{a_{k+1}}{a_k}\right|=L$, then the series converges if $L<1$, diverges if $L>1$, and is inconclusive if $L=1$.

Step 3 ：We calculate $\underset{k\to \mathrm{\infty }}{lim}\left|\frac{a_{k+1}}{a_k}\right|$ as follows: $$\underset{k\to \mathrm{\infty }}{lim}\left|\frac{(-1{)}^{k+1}\frac{x^{k+1}}{3^{k+1}}}{(-1{)}^k\frac{x^k}{3^k}}\right|=\underset{k\to \mathrm{\infty }}{lim}\left|\frac{-x}{3}\right|=\frac{|x|}{3}$$

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 $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\frac{(-3{)}^k}{3^k}=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k=-1+1-1+\dots$, which is a divergent series.

Step 7 ：Substituting $x=3$ into the series, we get $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\frac{3^k}{3^k}=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k=-1+1-1+\dots$, 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)}