Answer is: option4
\( 1 \le x < 3 \)Solution:
A radius of convergence of \(1\) implies \[ |x-2|<1 \]
Thus the series converges for
\[
1 Checking endpoints:
At \(x=1\):
\[
\sum_{n=1}^{\infty}\frac{(-1)^n}{n}
\]
which is a convergent alternating harmonic series.
At \(x=3\):
\[
\sum_{n=1}^{\infty}\frac{1}{n}
\]
which is the divergent harmonic series.
Therefore, the interval of convergence is
\[
1\le x<3
\]
