Finding the Interval of Convergence of a Power Series

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.