Find the radius of convergence, $R$, of the series.
\[
\sum_{n=1}^{\infty} \frac{x^{n}}{2 n-1}
\]
\[
R=
\]
Find the interval, $I$, of convergence of the series. (Enter your answer using interval notation.)
\[
I=
\]

Step 1 ：The series given is a power series of the form \(\sum_{n=1}^\infty a_n x^n\), where \(a_n = \frac{1}{2n-1}\).

Step 2 ：The radius of convergence, \(R\), of a power series is given by \(R = \frac{1}{\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|}\).

Step 3 ：We calculate \(\frac{a_{n+1}}{a_n} = \frac{2n-1}{2n+1}\).

Step 4 ：Taking the limit as \(n\) approaches infinity, we get \(\lim_{n\to\infty} \frac{2n-1}{2n+1} = 1\).

Step 5 ：Therefore, the radius of convergence, \(R\), is \(\frac{1}{1} = 1\). So, \(R = \boxed{1}\).

Step 6 ：The interval of convergence, \(I\), is \((-R, R)\) or \((-1, 1)\).

Step 7 ：However, we need to check the endpoints \(x = -1\) and \(x = 1\) to determine if they are included in the interval of convergence.

Step 8 ：For \(x = -1\), the series becomes \(\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}\), which is an alternating series. The terms decrease in absolute value and the limit of the terms is zero, so by the Alternating Series Test, the series converges at \(x = -1\).

Step 9 ：For \(x = 1\), the series becomes \(\sum_{n=1}^\infty \frac{1}{2n-1}\), which is a harmonic series and diverges.

Step 10 ：Therefore, the interval of convergence, \(I\), is \([-1, 1)\). So, \(I = \boxed{[-1, 1)}\).