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.