Answer
$0 \leq x \lt 1$
Work Step by Step
Let us consider a Ratio Test for a series $l=\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|$
1) When $l \lt 1$. then series $\Sigma a_n$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_n$ diverges.
3) When $l = 1$. then series $\Sigma a_n$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\dfrac{x^k}{k}$
Now, $l=\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{x^{k+1}}{(k+1)}}{\dfrac{x^k}{k}}|$
or, $l=\lim\limits_{k \to \infty} \dfrac{k}{k+1}= x \lim\limits_{k \to \infty} \dfrac{1}{1+1/k}$
But the series $\Sigma_{k=1}^\infty x^k$ will only converge when $l=0 \lt 1$ by the ratio test.
Since, $n \gt 0$ for all the values of $n$ and $\lim\limits_{x \to \infty} x^n=\infty$ and $\lim\limits_{x \to \infty} x^{-n}=0$
Thus. $l=x \dfrac{1}{1+0}=x \lt 1 \implies 0 \leq x \lt 1$
