Answer
The radius of convergence is $R=1$ and the interval of convergence is $[0,2)$ .
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\dfrac{1}{k} (x-1)^k$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{\dfrac{1}{k+1} (x-1)^{k+1}}{\dfrac{1}{k} (x-1)^k}|=|x-1| \lim\limits_{k \to \infty} \dfrac{k}{k+1}=|x-1|$
This means that the series converges for $r=|x-1| \lt 1$ . Then $-1 \lt x-1 \lt 1 \implies 0 \lt x \lt 2$
So, the radius of convergence for the power series is $R=1$
Next, we will compute the interval of convergence for the two points. Those points are $0$ and $2$.
Case 1: For $x=0$
$\sum_{k=1}^{\infty} \dfrac{(0-1)^k}{k}=\sum_{k=1}^{\infty} (-1)^k \cdot \dfrac{1}{k}$; this implies that the series converges.
Case 2: For $x=2$
$\sum_{k=1}^{\infty} \dfrac{(2-1)^k}{k}=\sum_{k=1}^{\infty} (1)^k \cdot \dfrac{1}{k}$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=1$ and the interval of convergence is $[0,2)$ .
