Answer is: option4
\( -1 < x \le 1 \)Solution:
The general term of the series is \[ (-1)^{n+1}\frac{x^n}{n}, \qquad n\ge 1 \]
Using the Ratio Test: \[ \left|\frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n}\right| = \frac{n}{n+1}|x| \] so that \[ \lim_{n\to\infty}\frac{n}{n+1}|x|=|x| \]
Thus the series converges if
\[
|x|<1
\]
or
\[
-1 Checking endpoints:
If \(x=-1\), the series becomes
\[
\sum_{n=1}^{\infty}\frac{-1}{n}
\]
which diverges.
If \(x=1\), the series becomes
\[
\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}
\]
which is the convergent alternating harmonic series.
Therefore, the interval of convergence is
\[
-1
