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 $a_k=x^k$
Now, $l=\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty}|\dfrac{ x \cdot x^k}{x^k}|$
or, $l=\lim\limits_{k \to \infty} x= x \lim\limits_{k \to \infty} (1)=(x)(1)=x$
But the series $\Sigma_{k=1}^\infty x^k$ will only converge when $l=0 \lt 1$ by the ratio test.
Thus. $l=x \lt 1 \implies 0 \leq x \lt 1$
