Find the radius of convergence and the interval of convergence of the power series $$\sum\limits_{k\,=\,1}^{\infty }\dfrac{x^{2k}}{k}$$

Solution We use the Ratio Test with $a_{n}=\dfrac{x^{2n}}{n}$ and $a_{n+1}=\dfrac{x^{2(n+1)}}{n+1}$. Then $$\lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =\lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{\dfrac{x^{2(n+1)}}{n+1}}{\dfrac{x^{2n}}{n}}\right\vert =\lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{\,x^{2n+2}}{n+1}\cdot \dfrac{n}{x^{2n}}\right\vert =x^{2}\lim\limits_{n\,\rightarrow \, \infty }\dfrac{n\,}{n+1}=x^{2}$$

The series converges absolutely if $x^{2}< 1$, or equivalently, if $-1< x< 1$.

To determine the behavior at the endpoints, we investigate $x=-1$ and $x=1$. When $x=1$ or $x=-1$, $\dfrac{x^{2k}}{k}=\dfrac{( x^{2}) ^{k}}{k}= \dfrac{1^{k}}{k}=\dfrac{1}{k}$, so the series reduces to the harmonic series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}$, which diverges. Consequently, the radius of convergence is $R=1$, and the interval of convergence is $-1< x< 1$, as shown in Figure 27.