# Radius of convergence: $\sum_{k=1}^\infty \frac{x^{2k-1}}{2k-1}$

It is asked to find the radius of convergence of the series

$$\sum_{k=1}^\infty \frac{x^{2k-1}}{2k-1}$$

i.e, to find the values of x such that this series converges.

Clearly, I could directly apply the root test and find the interval of convergence of the series. But, my doubt is if I can consider this a power series and evaluate

$$\lim_{k \to \infty}\left\{|\frac{1}{2k-1}|\right\}^{\frac{1}{2k-1}}$$

Is that correct? Or should we do

$\lim_{k \to \infty}\left\{\left|\frac{1}{2k-1}\right|\right\}^{\frac{1}{k}}?$

• You need $\limsup$, not $\lim$. And to consider it a power series, let the coefficients be $$a_n = \begin{cases} 0 &, n \text{ even}\\ \frac{1}{n} &, n \text{ odd}.\end{cases}$$ Since the zeros evidently can be ignored for the $\limsup$, you indeed get $$\frac{1}{R} = \limsup_{k\to\infty} \left\lvert \frac{1}{2k-1}\right\rvert^{1/(2k-1)}.$$ – Daniel Fischer Oct 21 '14 at 18:33
The root test is the correct idea here. To see which root you need to take, though, keep in mind that you could 'refill' your sum with the missing powers of $x$ by filling in a coefficient of $0$, thus we have $\sum_{k=1}^{\infty}{\frac{x^{2k-1}}{2k-1}}=\sum_{k=1}^{\infty}{x^na_n}$, where $a_k=\frac{1}{k}$ for odd $k$ and $a_k=0$ for even $k$.
Now the root test tells us to find $\limsup \sqrt[n]{|a_n|}$, where $\sqrt[n]{|a_n|}$ is $0$ for some $n$, and $\sqrt[n]{\frac{1}{n}}$ for the others.