Assuming the series is
[tex]\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}[/tex]
The series will converge if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|<1[/tex]
We have
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|<1[/tex]
So the series will certainly converge if [tex]-1<x<1[/tex], but we also need to check the endpoints of the interval.
If [tex]x=1[/tex], then the series is a scaled harmonic series, which we know diverges.
On the other hand, if [tex]x=-1[/tex], by the alternating series test we can show that the series converges, since
[tex]\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0[/tex]
and is strictly decreasing.
So, the interval of convergence for the series is [tex]-1\le x<1[/tex].
