This is my first time finding the radius and interval of convergence of a series, so please bear with me. I would like to find the radius and interval of convergence of $$\sum\limits_{n=1}^{\infty}\dfrac{x^n}{2n-1}\text{.}$$
My approach: by the ratio test, the series converges if $$\left|\dfrac{\dfrac{x^{n+1}}{2n+2-1}}{\dfrac{x^n}{2n-1}}\right| = \left|x\left(\dfrac{2n-1}{2n+1}\right)\right| \to |x|<1\text{.}$$ Thus $-1 < x < 1$ and clearly, the radius of convergence is $1$.
Is my solution correct?
Your solution is correct in finding the radius of convergence.
However, since you want to find your $\textbf{interval of convergence}$ as well, you must remember to check whether or not your series converges on the endpoints as well.
In this case:
When $x=-1$, your series becomes: $$\sum_{n=1}^\infty \frac{(-1)^n}{2n-1} = \sum_{n=1}^\infty(-1)^n\underbrace{\frac{1}{2n-1}}_\text{$b_n$} $$ Notice that $b_n$ is decreasing, $b_n \ge 0 \ \forall n \in \mathbb{N}$ and $\lim_{n\to\infty}b_n=0$
Hence, by Leibniz's Test for Alternating Series, the series converges for $x=-1$.
Testing for $x=1$ (which I will leave to you) shows that the series diverges when $x=1$.
Hence our Interval of Convergence is $-1 \leq x < 1$.