Finding the Interval of Convergence of a Power Series

Find the radius of convergence \(R\) and the interval of convergence of the power series \[ \sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{(x-2)^{k}}{k+1} \]

Solution \(\sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1 }\) is a power series centered at \(2\). We use the Ratio Test with \( a_{n}=(-1)^{n}\dfrac{(x-2)^{n}}{n+1}\) and \(a_{n+1}=(-1)^{n+1} \dfrac{(x-2)^{n+1}}{n+2}\). Then \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert &=& \lim\limits_{n\rightarrow \infty }\left\vert \frac{\dfrac{(-1)^{n+1}(x-2) ^{n+1}}{n+2}}{\dfrac{(-1)^{n}\,(x-2) ^{n}}{n+1}} \right\vert =\lim\limits_{n\rightarrow \infty }\left\vert \frac{(n+1)\,(x-2) }{n+2}\right\vert \nonumber \\ &=& \vert x-2\vert \,\lim\limits_{n\rightarrow \infty }\dfrac{n+1}{n+2}=\vert x-2\vert \end{eqnarray*} \]

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The series converges absolutely if \(\vert x-2\vert < 1\), or equivalently if \(1 < x < 3\). The radius of convergence is \(R=1\). We check the endpoints \(x=1\) and \(x=3\) separately.

If \(x=1\), \[ \begin{eqnarray*} \sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1}&=&\sum \limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(-1) ^{k}}{k+1} =\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{2k}}{k+1} =\sum\limits_{k\,=\,0}^{\infty }\dfrac{1}{k+1}\nonumber \\ &=&1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n+1}+\cdots \end{eqnarray*} \]

which is the divergent harmonic series.

If \(x=3\), \[ \sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1}=\sum \limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{1}{k+1}=1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4}+\cdots +\frac{(-1)^{n}}{n+1}+\cdots \]

which is the convergent alternating harmonic series.

The series \(\sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2) ^{k} }{k+1}\) converges for \(1 < x\leq 3\), as shown in Figure 28.